Oscillations PYQs: MHT CET, JEE Main, NEET

Q1MHT CET 2026 · 19 MAY · SHIFT 1Easy

Periodic Motion & SHM Basics · Restoring force and differential equation

The differential equation of a particle performing S.H.M. is $\frac{d^2x}{dt^2} + 64x = 0$ , where $x$ is in meter. The period of oscillation of the particle is (in second)

Show answer and solution

Answer: (b)

Compare with $\frac{d^2x}{dt^2} + \omega^2 x = 0$ : $\omega^2 = 64$ , $\omega = 8\ \text{rad/s}$ .
$T = \frac{2\pi}{\omega} = \frac{2\pi}{8} = \frac{\pi}{4}\ \text{s}$ .
So the answer is (b) $\frac{\pi}{4}$ .

Report a mistake

Q2MHT CET 2026 · 15 MAY · SHIFT 1Easy

Periodic Motion & SHM Basics · Phase and phase difference

The period of S.H.M. of a particle is $12$ s. The phase difference between the positions at $t = 3$ s and $t = 4$ s will be

Show answer and solution

Answer: (c)

Phase difference $\Delta\phi = \frac{2\pi}{T}\Delta t$ .
$\Delta\phi = \frac{2\pi}{12}\times(4 - 3) = \frac{\pi}{6}$ .
So the answer is (c) $\frac{\pi}{6}$ .

Report a mistake

Q3MHT CET 2026 · 15 APR · SHIFT 2Easy

Periodic Motion & SHM Basics · Phase and phase difference

The minimum phase difference between two simple harmonic motions is

Show answer and solution

Answer: (a)

$x_1 = \frac{1}{\sqrt{2}}\sin\omega t + \frac{1}{\sqrt{2}}\cos\omega t = \sin\left(\omega t + \frac{\pi}{4}\right)$ .
$x_2 = \sin\omega t + \cos\omega t = \sqrt{2}\sin\left(\omega t + \frac{\pi}{4}\right)$ .
Both have the same phase $\omega t + \frac{\pi}{4}$ ; only the amplitudes differ.
So the answer is (a) zero.

Report a mistake

Q4MHT CET 2026 · 11 APR · SHIFT 2Easy

Periodic Motion & SHM Basics · Time and distance in SHM

A mass attached to a spring performs S.H.M. whose displacement is $x = 3 \times 10^{-3}\cos 2\pi t$ m. The time taken to obtain maximum speed for the first time is (in s)

Show answer and solution

Answer: (c)

$\omega = 2\pi$ , so $T = 1\ \text{s}$ .
$x = A\cos 2\pi t$ starts at the extreme position ( $x = A$ ) at $t = 0$ .
Speed is maximum at the mean position, reached after $\frac{T}{4} = \frac{1}{4}\ \text{s}$ .
So the answer is (c) $\frac{1}{4}$ .

Report a mistake

Q5MHT CET 2026 · 12 MAY · SHIFT 2Medium

Periodic Motion & SHM Basics · Periodic and oscillatory motion

A ball released from height $(h)$ which makes perfectly elastic collision with ground. The frequency of periodic vibratory motion is ( $g$ = acceleration due to gravity)

Show answer and solution

Answer: (b)

Time to fall height $h$ : $h = \frac{1}{2}gt^2 \Rightarrow t = \sqrt{\frac{2h}{g}}$ .
Elastic bounce: it rises back to $h$ in the same time, so $T = 2\sqrt{\frac{2h}{g}}$ .
$f = \frac{1}{T} = \frac{1}{2}\sqrt{\frac{g}{2h}}$ .
So the answer is (b).

Report a mistake

Q6JEE Main 2026 · 5 APR · SHIFT 2Medium

Periodic Motion & SHM Basics · Periodic and oscillatory motion

Match List - I with List - II.

Choose the correct answer from the options given below:

List - I	List - II
A. $\sin^2 \omega t$	I. Periodic with time period $T = \frac{\pi}{\omega}$ but not simple harmonic motion (SHM)
B. $\sin^3(2\omega t)$	II. Periodic with time period $T = \frac{2\pi}{\omega}$ but Not SHM
C. $\sin(\omega t) + \cos(\pi\omega t)$	III. Periodic with time period $T = \frac{\pi}{\omega}$ and SHM
D. $\cos \omega t + \cos 2\omega t$	IV. Non-periodic

Show answer and solution

Answer: (a)

A: $\sin^2\omega t = \frac{1 - \cos 2\omega t}{2}$ , SHM about $\frac{1}{2}$ with $T = \frac{\pi}{\omega}$ $\to$ III.
B: $\sin^3(2\omega t)$ repeats with the period of $\sin 2\omega t$ , $T = \frac{\pi}{\omega}$ , but contains $\sin 6\omega t$ too, so not SHM $\to$ I.
C: $\sin\omega t + \cos(\pi\omega t)$ : frequency ratio $1 : \pi$ is irrational, so no common period $\to$ IV.
D: $\cos\omega t + \cos 2\omega t$ : common period $\frac{2\pi}{\omega}$ , two frequencies so not SHM $\to$ II.
So the answer is (a) A-III, B-I, C-IV, D-II.

Report a mistake

Q7MHT CET 2026 · 19 MAY · SHIFT 1Medium

Periodic Motion & SHM Basics · Displacement equation and circle of reference

Two simple harmonic motions are represented as $y_1 = 10\sin \omega t$ and $y_2 = 5\sin \omega t + 5\cos \omega t$ , both in cm. The ratio of the amplitudes of $y_2$ ( $A_2$ ) to that of $y_1$ ( $A_1$ ), is

Show answer and solution

Answer: (d)

$A_1 = 10\text{ cm}$
$y_2 = 5\sin\omega t + 5\cos\omega t$ has $A_2 = \sqrt{5^2 + 5^2} = 5\sqrt{2}\text{ cm}$
$\frac{A_2}{A_1} = \frac{5\sqrt{2}}{10} = \frac{1}{\sqrt{2}}$
So the answer is (d) $1 : \sqrt{2}$ .

Report a mistake

Q8MHT CET 2026 · 20 APR · SHIFT 2Medium

Periodic Motion & SHM Basics · Displacement equation and circle of reference

The displacement of a particle varies with time according to the relation

$x = a\sin \omega t + b\cos \omega t$

Show answer and solution

Answer: (b)

Write $a = R\cos\delta$ , $b = R\sin\delta$ : $x = R\sin(\omega t + \delta)$ .
This is SHM with $R = \sqrt{a^2 + b^2}$ .
So the answer is (b).

Report a mistake

Q9MHT CET 2026 · 17 APR · SHIFT 2Medium

Periodic Motion & SHM Basics · Displacement equation and circle of reference

The displacement of a particle performing linear S.H.M. is given by $y = A\cos[\pi(t + \phi)]$ . If at $t = 0$ , the displacement is $y = 2$ cm and velocity is $2\pi$ cm/s, the value of amplitude A in cm is

Show answer and solution

Answer: (d)

Let $\theta = \pi\phi$ . At $t = 0$ : $A\cos\theta = 2$ .
$v = -\pi A\sin\theta = 2\pi \Rightarrow A\sin\theta = -2$
$A = \sqrt{2^2 + 2^2} = 2\sqrt{2}\text{ cm}$
So the answer is (d) $2\sqrt{2}$ .

Report a mistake

Q10MHT CET 2026 · 13 MAY · SHIFT 1Medium

Periodic Motion & SHM Basics · Time and distance in SHM

A particle oscillates simple harmonically along a straight line with period $8$ s and amplitude $8\sqrt{2}$ m. It starts from mean position then the ratio of the distances travelled by it in the $2^{nd}$ second and the first second of its motion is $\left(\sin 45^\circ = \frac{1}{\sqrt{2}}\right)\ (\sin 90^\circ = 1)$

Show answer and solution

Answer: (c)

$\omega = \frac{2\pi}{8} = \frac{\pi}{4}$ , $x = A\sin\frac{\pi t}{4}$ (starts at mean).
$x(1) = \frac{A}{\sqrt{2}}$ , $x(2) = A$ ; motion is one way in both seconds.
1st second: $\frac{A}{\sqrt{2}}$ ; 2nd second: $A - \frac{A}{\sqrt{2}}$
Ratio $= \frac{A(1 - 1/\sqrt{2})}{A/\sqrt{2}} = \sqrt{2} - 1$
So the answer is (c) $\sqrt{2} - 1$ .

Report a mistake

Q11MHT CET 2026 · 12 MAY · SHIFT 1Medium

Periodic Motion & SHM Basics · Time and distance in SHM

A particle performing S.H.M., has a period of $12$ s. The particle starts from a mean position. The ratio of distances travelled by the particle in the first and second second is $\left\{\cos(\pi/6) = \frac{\sqrt{3}}{2}\right\}$

Show answer and solution

Answer: (d)

$\omega = \frac{2\pi}{12} = \frac{\pi}{6}$ , $x = A\sin\frac{\pi t}{6}$
$x(1) = \frac{A}{2}$ , $x(2) = \frac{\sqrt{3}A}{2}$
1st second: $\frac{A}{2}$ ; 2nd second: $\frac{(\sqrt{3} - 1)A}{2}$
Ratio $= \frac{1}{\sqrt{3} - 1}$
So the answer is (d).

Report a mistake

Q12MHT CET 2026 · 19 APR · SHIFT 1Medium

Periodic Motion & SHM Basics · Time and distance in SHM

A particle executes a simple harmonic motion with a periodic time $8$ second. At time $t = 0$ , it is at a mean position. The ratio of the distance traveled by a particle in the 2nd and that in the 1st second of its motion is

( $\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}},\ \sin 90^\circ = \cos 0^\circ = 1$ )

Show answer and solution

Answer: (c)

$\omega = \frac{2\pi}{8} = \frac{\pi}{4}$ , $x = A\sin\frac{\pi t}{4}$
$x(1) = \frac{A}{\sqrt{2}}$ , $x(2) = A$
1st second: $\frac{A}{\sqrt{2}}$ ; 2nd second: $A\left(1 - \frac{1}{\sqrt{2}}\right)$
Ratio (2nd : 1st) $= \sqrt{2} - 1$
So the answer is (c) $\sqrt{2} - 1$ .

Report a mistake

Q13MHT CET 2026 · 11 APR · SHIFT 1Medium

Periodic Motion & SHM Basics · Time and distance in SHM

A particle executes simple harmonic motion between $x = -A$ and $x = +A$ . If time taken by particle to go from $x = 0$ to $\frac{A}{2}$ is $2$ s, then time taken by particle in going from $x = \frac{A}{2}$ to A is

Show answer and solution

Answer: (d)

From $x = 0$ to $\frac{A}{2}$ : phase $\frac{\pi}{6}$ , time $\frac{T}{12} = 2\text{ s} \Rightarrow T = 24\text{ s}$
From $\frac{A}{2}$ to $A$ : phase $\frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$ , time $\frac{T}{6} = 4\text{ s}$
So the answer is (d) 4 s.

Report a mistake

Q14JEE Main 2026 · 6 APR · SHIFT 1Medium

Periodic Motion & SHM Basics · Time and distance in SHM

A particle is executing simple harmonic motion. Its amplitude is $A$ and time period is 5 sec. The time required by it to move from $x = A$ to $x = \frac{A}{\sqrt{2}}$ is ______ sec.

Show answer and solution

Answer: (c)

From the extreme: $x = A\cos\omega t = \frac{A}{\sqrt{2}} \Rightarrow \omega t = \frac{\pi}{4}$
$t = \frac{T}{8} = \frac{5}{8}\text{ s}$
So the answer is (c) $\frac{5}{8}$ s.

Report a mistake

Q15JEE Main 2026 · 2 APR · SHIFT 2Medium

Periodic Motion & SHM Basics · Time and distance in SHM

The equation of motion of a particle is given by $x = a\sin(50t + \pi/3)$ cm. The particle will come to rest at time $t_1$ and it will have zero acceleration at time $t_2$ . The $t_1$ and $t_2$ respectively are ________.

Show answer and solution

Answer: (a)

Rest at the extreme: $50t_1 + \frac{\pi}{3} = \frac{\pi}{2} \Rightarrow t_1 = \frac{\pi/6}{50} = \frac{\pi}{300}\text{ s}$
Zero acceleration at $x = 0$ : $50t_2 + \frac{\pi}{3} = \pi \Rightarrow t_2 = \frac{2\pi/3}{50} = \frac{\pi}{75}\text{ s}$
So the answer is (a).

Report a mistake

Q16JEE Main 2025 · 28 JAN · SHIFT 2Easy

Periodic Motion & SHM Basics · Periodic and oscillatory motion

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A): Knowing initial position $x_0$ and initial momentum $p_0$ is enough to determine the position and momentum at any time $t$ for a simple harmonic motion with a given angular frequency $\omega$ .

Reason (R): The amplitude and phase can be expressed in terms of $x_0$ and $p_0$ .

In the light of the above statements, choose the correct answer from the options given below:

Show answer and solution

Answer: (d)

SHM: $x = A\sin(\omega t + \phi)$ , $p = m\omega A\cos(\omega t + \phi)$ .
At $t = 0$ : $x_0 = A\sin\phi$ , $p_0 = m\omega A\cos\phi$ , so $A = \sqrt{x_0^2 + \frac{p_0^2}{m^2\omega^2}}$ and $\tan\phi = \frac{m\omega x_0}{p_0}$ .
Once $A$ and $\phi$ are known, $x$ and $p$ are known at every $t$ . So R is true and explains A.
So the answer is (d).

Report a mistake

Q17MHT CET 2025 · 20 APR · SHIFT 1Easy

Periodic Motion & SHM Basics · Phase and phase difference

The period of S. H.M. of a particle is 16 second. The phase difference between the positions at $t = 2\text{ s}$ and $t = 4\text{ s}$ will be

Show answer and solution

Answer: (c)

Phase difference $\Delta\phi = \frac{2\pi}{T}\Delta t$ .
$\Delta\phi = \frac{2\pi}{16}\times(4 - 2) = \frac{\pi}{4}$ .
So the answer is (c) $\frac{\pi}{4}$ .

Report a mistake

Q18MHT CET 2025 · 25 APR · SHIFT 1Easy

Periodic Motion & SHM Basics · Time and distance in SHM

The displacement of particle in S.H.M. is $x = A\cos(\omega t + \pi/6)$ . Its speed will be maximum at time $(\sin 90^\circ = 1)$

Show answer and solution

Answer: (a)

$v = -A\omega\sin(\omega t + \pi/6)$ ; speed is maximum when $\sin(\omega t + \pi/6) = 1$ .
$\omega t + \frac{\pi}{6} = \frac{\pi}{2} \Rightarrow \omega t = \frac{\pi}{3}$ .
$t = \frac{\pi}{3\omega}\ \text{s}$ .
So the answer is (a).

Report a mistake

Q19MHT CET 2025 · 20 APR · SHIFT 2Easy

Periodic Motion & SHM Basics · Time and distance in SHM

A mass attached to a spring performs S.H.M. whose displacement is $x = 3 \times 10^{-3}\cos 2\pi t$ metre. The time taken to obtain maximum speed for the first time is

Show answer and solution

Answer: (c)

$\omega = 2\pi$ , so $T = 1\ \text{s}$ .
The particle starts at the extreme position ( $x = A$ ) at $t = 0$ .
Maximum speed is at the mean position, after $\frac{T}{4} = \frac{1}{4}\ \text{s}$ .
So the answer is (c) $\frac{1}{4}$ s.

Report a mistake

Q20MHT CET 2025 · 23 APR · SHIFT 2Medium

Periodic Motion & SHM Basics · Periodic and oscillatory motion

The motion of the particle is given by the equation $x = A\sin\omega t + B\cos\omega t$ .

The motion of the particle is

Show answer and solution

Answer: (c)

Write $A = R\cos\phi$ , $B = R\sin\phi$ : $x = R\sin(\omega t + \phi)$ .
This is a single sine term, so the motion is SHM.
$R = \sqrt{A^2 + B^2} = (A^2 + B^2)^{1/2}$ .
So the answer is (c).

Report a mistake

Q21MHT CET 2025 · 25 APR · SHIFT 1Medium

Periodic Motion & SHM Basics · Phase and phase difference

Two particles ' $A$ ' and ' $B$ ' execute SHMs of periods ' $T$ ' and $\frac{3T}{2}$ . If they start from the mean position then the phase difference between them, when the particle ' $A$ ' completes two oscillations will be

Show answer and solution

Answer: (c)

Phase gained $= \frac{2\pi t}{T}$ . A completes 2 oscillations, so $t = 2T$ .
$\phi_A = 2\pi(2) = 4\pi$ , $\phi_B = \frac{2\pi(2T)}{3T/2} = \frac{8\pi}{3}$
$\Delta\phi = 4\pi - \frac{8\pi}{3} = \frac{4\pi}{3}$
So the answer is (c) $\frac{4\pi}{3}$ .

Report a mistake

Q22MHT CET 2025 · 22 APR · SHIFT 2Medium

Periodic Motion & SHM Basics · Time and distance in SHM

A particle performing linear S.H.M. has period 8 seconds. At time $t = 0$ , it is in the mean position. The ratio of the distances travelled by the particle in the $1$ st and $2$ nd second is $\left(\cos 45^\circ = 1/\sqrt{2}\right)$

Show answer and solution

Answer: (a)

$\omega = \frac{\pi}{4}$ , $x = A\sin\frac{\pi t}{4}$
1st second: $\frac{A}{\sqrt{2}}$ ; 2nd second: $A - \frac{A}{\sqrt{2}} = \frac{A(\sqrt{2} - 1)}{\sqrt{2}}$
Ratio $= 1 : (\sqrt{2} - 1)$
So the answer is (a).

Report a mistake

Q23MHT CET 2025 · 20 APR · SHIFT 1Medium

Periodic Motion & SHM Basics · Time and distance in SHM

A particle oscillates in straight line simple harmonically with period 8 second and amplitude $4\sqrt{2}\text{ m}$ . Particle starts from mean position. The ratio of the distance travelled by it in $1$ st second of its motion to that in $2$ nd second is $\left(\sin 45^\circ = 1/\sqrt{2}, \sin\frac{\pi}{2} = 1\right)$

Show answer and solution

Answer: (d)

$\omega = \frac{\pi}{4}$ , $x = 4\sqrt{2}\sin\frac{\pi t}{4}$
$x(1) = 4\text{ m}$ , $x(2) = 4\sqrt{2}\text{ m}$
1st second: $4$ ; 2nd second: $4\sqrt{2} - 4 = 4(\sqrt{2} - 1)$
So the answer is (d) $1 : (\sqrt{2} - 1)$ .

Report a mistake

Q24JEE Main 2025 · 24 JAN · SHIFT 1Medium

Periodic Motion & SHM Basics · Time and distance in SHM

A particle is executing simple harmonic motion with time period 2 s and amplitude 1 cm. If D and d are the total distance and displacement covered by the particle in 12.5 s, then $\frac{D}{d}$ is

Show answer and solution

Answer: (c)

$12.5\text{ s} = 6T + \frac{T}{4}$ (taking start at the mean position).
Distance: $6\times 4A + A = 25A = 25\text{ cm}$
In $6T$ the particle returns; in the extra $\frac{T}{4}$ it goes mean to extreme, so $d = A = 1\text{ cm}$ .
$\frac{D}{d} = 25$
So the answer is (c) 25.

Report a mistake

Q25MHT CET 2024 · 15 MAY · SHIFT 2Easy

Periodic Motion & SHM Basics · Restoring force and differential equation

The motion of a particle is described by the equation $a = -bx$ where ' $a$ ' is the acceleration, x is the displacement from the equilibrium position and b is a constant. The periodic time will be

Show answer and solution

Answer: (b)

Compare $a = -bx$ with $a = -\omega^2 x$ : $\omega = \sqrt{b}$ .
$T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{b}}$ .
So the answer is (b) $\frac{2\pi}{\sqrt{b}}$ .

Report a mistake

Q26MHT CET 2024 · 3 MAY · SHIFT 2Easy

Periodic Motion & SHM Basics · Phase and phase difference

When a particle in linear S.H.M. completes two oscillations, its phase increases by

Show answer and solution

Answer: (d)

Phase increases by $2\pi$ rad in one oscillation.
For two oscillations: $2 \times 2\pi = 4\pi$ rad.
So the answer is (d) $4\pi$ rad.

Report a mistake

Q27NEET 2024Easy

Periodic Motion & SHM Basics · Displacement equation and circle of reference

If $x = 5\sin\left(\pi t + \frac{\pi}{3}\right)\text{m}$ represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are

Show answer and solution

Answer: (b)

Compare with $x = A\sin(\omega t + \phi)$ : $A = 5\ \text{m}$ , $\omega = \pi\ \text{rad/s}$ .
$T = \frac{2\pi}{\omega} = \frac{2\pi}{\pi} = 2\ \text{s}$ .
So the answer is (b) 5 m, 2 s.

Report a mistake

Q28JEE Main 2024 · 9 APR · SHIFT 2Medium

Periodic Motion & SHM Basics · Restoring force and differential equation

A particle of mass $0.50\text{ kg}$ executes simple harmonic motion under force $F = -50(\text{N m}^{-1})x$ . The time period of oscillation is $\frac{x}{35}\text{ s}$ . The value of $x$ is _________.

(Given $\pi = \frac{22}{7}$ )

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 22

$F = -kx$ with $k = 50\text{ N/m}$ .
$T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{0.5}{50}} = \frac{2\pi}{10} = \frac{\pi}{5}$
$T = \frac{22}{7\times 5} = \frac{22}{35}\text{ s}$
So $x = 22$ .

Report a mistake

Q29MHT CET 2024 · 2 MAY · SHIFT 2Medium

Periodic Motion & SHM Basics · Displacement equation and circle of reference

The displacement of a particle performing S.H.M. is given by $Y = A\cos[\pi(t + \phi)]$ . If at $t = 0$ , the displacement is $y = 2\text{ cm}$ and velocity is $2\pi\text{ cm/s}$ , the value of amplitude $A$ in cm is

Show answer and solution

Answer: (c)

Let $\theta = \pi\phi$ . At $t = 0$ : $A\cos\theta = 2$ .
$v = -\pi A\sin\theta = 2\pi \Rightarrow A\sin\theta = -2$
$A = \sqrt{2^2 + 2^2} = 2\sqrt{2}\text{ cm}$
So the answer is (c) $2\sqrt{2}$ .

Report a mistake

Q30JEE Main 2024 · 29 JAN · SHIFT 2Medium

Periodic Motion & SHM Basics · Time and distance in SHM

A simple harmonic oscillator has an amplitude $A$ and time period $6\pi$ second. Assuming the oscillation starts from its mean position, the time required by it to travel from $x =$ A to $x = \frac{\sqrt{3}}{2}$ A will be $\frac{\pi}{x}\text{ s}$ , where $x =$ _________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 2

From the extreme: $A\cos\omega t = \frac{\sqrt{3}}{2}A \Rightarrow \omega t = \frac{\pi}{6} \Rightarrow t = \frac{T}{12}$
$t = \frac{6\pi}{12} = \frac{\pi}{2}\text{ s}$
So $x = 2$ .

Report a mistake

Q31JEE Main 2023 · 15 APR · SHIFT 1Easy

Periodic Motion & SHM Basics · Restoring force and differential equation

In a linear Simple Harmonic Motion (SHM)

(A) Restoring force is directly proportional to the displacement.

(B) The acceleration and displacement are opposite in direction.

(C) The velocity is maximum at mean position.

(D) The acceleration is minimum at extreme points.

Choose the correct answer from the options given below:

Show answer and solution

Answer: (c)

(A) True: $F = -kx$ , so $|F| \propto x$ .
(B) True: $a = -\omega^2 x$ , opposite to displacement.
(C) True: $v = \omega\sqrt{A^2 - x^2}$ is maximum at $x = 0$ .
(D) False: $|a| = \omega^2 |x|$ is maximum at the extreme points.
So the answer is (c) (A), (B) and (C) only.

Report a mistake

Q32JEE Main 2023 · 8 APR · SHIFT 2Easy

Periodic Motion & SHM Basics · Displacement equation and circle of reference

For particle P revolving round the centre O with radius of circular path $r$ and angular velocity $\omega$ , as shown in below figure, the projection of OP on the $x$ -axis at time $t$ is

Show answer and solution

Answer: (c)

At $t = 0$ , OP makes $30^\circ = \frac{\pi}{6}$ with the $x$ -axis.
P moves anticlockwise, so at time $t$ the angle is $\omega t + \frac{\pi}{6}$ .
Projection on $x$ -axis: $x(t) = r\cos\left(\omega t + \frac{\pi}{6}\right)$ .
So the answer is (c).

Report a mistake

Q33JEE Main 2023 · 10 APR · SHIFT 2Medium

Periodic Motion & SHM Basics · Periodic and oscillatory motion

For a periodic motion represented by the equation

$y = \sin \omega t + \cos \omega t$

the amplitude of the motion is

Show answer and solution

Answer: (b)

For $a\sin\omega t + b\cos\omega t$ , amplitude $= \sqrt{a^2 + b^2}$ .
$\sqrt{1^2 + 1^2} = \sqrt{2}$ , i.e. $y = \sqrt{2}\sin(\omega t + \pi/4)$ .
So the answer is (b) $\sqrt{2}$ .

Report a mistake

Q34MHT CET 2023 · 13 MAY · SHIFT 1Medium

Periodic Motion & SHM Basics · Restoring force and differential equation

A body of mass $0.04\text{ kg}$ executes simple harmonic motion (SHM) about $x = 0$ under the influence of force $F$ as shown in graph. The period of oscillation is

Show answer and solution

Answer: (b)

From the graph, $F = -kx$ with $k = \frac{8}{2} = 4\text{ N/m}$ .
$T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{0.04}{4}} = 2\pi(0.1)$
So the answer is (b) $0.2\pi\text{ s}$ .

Report a mistake

Q35MHT CET 2023 · 9 MAY · SHIFT 2Medium

Periodic Motion & SHM Basics · Phase and phase difference

Two S.H.Ms. are represented by equations $y_1 = 0.1\sin\left(100\pi t + \frac{\pi}{3}\right)$ and $y_2 = 0.1\cos(100\pi t)$ The phase difference between the speeds of the two particles is

Show answer and solution

Answer: (b)

$v_1 = \frac{dy_1}{dt} \propto \cos\left(100\pi t + \frac{\pi}{3}\right)$
$v_2 \propto -\sin(100\pi t) = \cos\left(100\pi t + \frac{\pi}{2}\right)$
$\Delta\phi = \frac{\pi}{3} - \frac{\pi}{2} = -\frac{\pi}{6}$
So the answer is (b) $-\frac{\pi}{6}$ .

Report a mistake

Q36JEE Main 2023 · 10 APR · SHIFT 1Medium

Periodic Motion & SHM Basics · Displacement equation and circle of reference

A particle executes S.H.M. of amplitude A along x-axis. At t = 0, the position of the particle is $x = \frac{A}{2}$ and it moves along positive x-axis. The displacement of particle in time t is $x = A\sin(\omega t + \delta)$ , then the value of $\delta$ will be

Show answer and solution

Answer: (d)

At $t = 0$ : $\frac{A}{2} = A\sin\delta \Rightarrow \sin\delta = \frac{1}{2}$
Moving along $+x$ means $v = A\omega\cos\delta > 0$ , so $\delta = \frac{\pi}{6}$ (not $\frac{5\pi}{6}$ ).
So the answer is (d) $\frac{\pi}{6}$ .

Report a mistake

Q37JEE Main 2023 · 25 JAN · SHIFT 2Medium

Periodic Motion & SHM Basics · Time and distance in SHM

A particle executes simple harmonic motion between $x = -A$ and $x = +A$ . If time taken by particle to go from $x = 0$ to $\frac{A}{2}$ is 2 s; then time taken by particle in going from $x = \frac{A}{2}$ to A is

Show answer and solution

Answer: (a)

$0 \to \frac{A}{2}$ takes $\frac{T}{12} = 2\text{ s} \Rightarrow T = 24\text{ s}$
$\frac{A}{2} \to A$ takes $\frac{T}{4} - \frac{T}{12} = \frac{T}{6} = 4\text{ s}$
So the answer is (a) 4 s.

Report a mistake

Q38MHT CET 2023 · 10 MAY · SHIFT 2Medium

Periodic Motion & SHM Basics · Phase in waves (mixed)

A simple harmonic progressive wave is represented by $y = A\sin(100\pi t + 3x)$ . The distance between two points on the wave at a phase difference of $\frac{\pi}{3}$ radian is

Show answer and solution

Answer: (b)

Compare with $y = A\sin(\omega t + kx)$ : $k = 3\text{ m}^{-1}$ .
$\Delta x = \frac{\Delta\phi}{k} = \frac{\pi/3}{3} = \frac{\pi}{9}\text{ m}$
So the answer is (b) $\frac{\pi}{9}\text{ m}$ .

Report a mistake

Q39NEET 2022 · PHASE 2Easy

Periodic Motion & SHM Basics · Periodic and oscillatory motion

Identify the function which represents a non-periodic motion.

Show answer and solution

Answer: (b)

A periodic function repeats itself after a fixed time $T$ .
$\sin(\omega t + \pi/4)$ , $\sin\omega t$ and $\sin\omega t + \cos\omega t$ all repeat after $T = 2\pi/\omega$ .
$e^{-\omega t}$ keeps decreasing and never repeats.
So the answer is (b) $e^{-\omega t}$ .

Report a mistake

Q40JEE Main 2022 · 27 JUN · SHIFT 1Easy

Periodic Motion & SHM Basics · Time and distance in SHM

The displacement of simple harmonic oscillator after 3 seconds starting from its mean position is equal to half of its amplitude. The time period of harmonic motion is:

Show answer and solution

Answer: (d)

From the mean position: $x = A\sin\frac{2\pi t}{T}$ .
$\frac{A}{2} = A\sin\frac{2\pi (3)}{T} \Rightarrow \frac{6\pi}{T} = \frac{\pi}{6}$ .
$T = 36\ \text{s}$ .
So the answer is (d) 36 s.

Report a mistake

Q41JEE Main 2022 · 28 JUL · SHIFT 2Medium

Periodic Motion & SHM Basics · Restoring force and differential equation

The potential energy of a particle of mass $4\text{ kg}$ in motion along the x-axis is given by $U = 4(1 - \cos 4x)$ J. The time period of the particle for small oscillation $(\sin\theta \simeq \theta)$ is $\left(\frac{\pi}{K}\right)\text{ s}$ . The value of $K$ is _________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 2

For small $x$ : $1 - \cos 4x \approx \frac{(4x)^2}{2} = 8x^2$ , so $U \approx 32x^2$ .
Compare with $U = \frac{1}{2}kx^2$ : $k = 64\text{ N/m}$ .
$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{64}{4}} = 4\text{ rad/s}$ , $T = \frac{2\pi}{4} = \frac{\pi}{2}\text{ s}$
So $K = 2$ .

Report a mistake

Q42MHT CET 2022 · 11 AUG · SHIFT 2Medium

Periodic Motion & SHM Basics · Time and distance in SHM

The time taken by a particle executing simple harmonic motion of period ' $T$ ', to move from the mean position to half the maximum displacement is

Show answer and solution

Answer: (a)

$x = A\sin\omega t = \frac{A}{2} \Rightarrow \omega t = \frac{\pi}{6}$
$t = \frac{\pi/6}{2\pi/T} = \frac{T}{12}$
So the answer is (a) $\frac{T}{12}$ .

Report a mistake

Q43JEE Main 2022 · 27 JUN · SHIFT 2Medium

Periodic Motion & SHM Basics · Time and distance in SHM

A particle executes simple harmonic motion. Its amplitude is 8 cm and time period is 6 s. The time it will take to travel from its position of maximum displacement to the point corresponding to half of its amplitude, is ___________ s.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 1

From the extreme: $A\cos\omega t = \frac{A}{2} \Rightarrow \omega t = \frac{\pi}{3} \Rightarrow t = \frac{T}{6}$
$t = \frac{6}{6} = 1\text{ s}$
So the answer is 1 s.

Report a mistake

Q44MHT CET 2022 · 11 AUG · SHIFT 2Medium

Periodic Motion & SHM Basics · Phase in waves (mixed)

In a medium, the phase difference between two particles separated by a distance ' $x$ ' is $\left(\frac{\pi}{5}\right)$ c. If the frequency of the oscillation of particles is $25\text{ Hz}$ and the velocity of propagation of the waves is $75\text{ m/s}$ , then the value of $x$ is

Show answer and solution

Answer: (d)

$\lambda = \frac{v}{f} = \frac{75}{25} = 3\text{ m}$
$x = \frac{\lambda}{2\pi}\Delta\phi = \frac{3}{2\pi}\cdot\frac{\pi}{5} = 0.3\text{ m}$
So the answer is (d) 0.3 m.

Report a mistake

Q45JEE Main 2021 · 18 MAR · SHIFT 2Easy

Periodic Motion & SHM Basics · Periodic and oscillatory motion

The function of time representing a simple harmonic motion with a period of $\frac{\pi}{\omega}$ is:

Show answer and solution

Answer: (d)

SHM needs a single sine/cosine term; period $T = \frac{2\pi}{\omega'}$ .
(a) has three frequencies and (c) has period $2\pi/\omega$ .
(d) $3\cos\left(\frac{\pi}{4} - 2\omega t\right)$ has $\omega' = 2\omega$ , so $T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$ .
So the answer is (d).

Report a mistake

Q46JEE Main 2021 · 25 FEB · SHIFT 2Easy

Periodic Motion & SHM Basics · Displacement equation and circle of reference

Y = A sin( $\omega$ t + $\phi$ 0) is the time-displacement equation of a SHM. At t = 0 the displacement of the particle is $Y = \frac{A}{2}$ and it is moving along negative x-direction. Then the initial phase angle $\phi$ 0 will be:

Show answer and solution

Answer: (a)

At $t = 0$ : $\frac{A}{2} = A\sin\phi_0 \Rightarrow \sin\phi_0 = \frac{1}{2}$ , so $\phi_0 = \frac{\pi}{6}$ or $\frac{5\pi}{6}$ .
Velocity $v = A\omega\cos\phi_0$ must be negative, so $\cos\phi_0 < 0$ .
$\cos\frac{5\pi}{6} < 0$ , so $\phi_0 = \frac{5\pi}{6}$ .
So the answer is (a) $\frac{5\pi}{6}$ .

Report a mistake

Q47JEE Main 2021 · 18 MAR · SHIFT 1Easy

Periodic Motion & SHM Basics · Time and distance in SHM

A particle performs simple harmonic motion with a period of 2 second. The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is $\frac{1}{a}$ s. The value of 'a' to the nearest integer is _________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 6

From the mean position: $x = A\sin\frac{2\pi t}{T}$ with $T = 2\ \text{s}$ .
$\frac{A}{2} = A\sin(\pi t) \Rightarrow \pi t = \frac{\pi}{6}$ .
$t = \frac{1}{6}\ \text{s} = \frac{1}{a}\ \text{s}$ .
So $a = 6$ .

Report a mistake

Q48MHT CET 2021 · 20 SEP · SHIFT 1Medium

Periodic Motion & SHM Basics · Phase and phase difference

Two particles $P$ and $Q$ performs S.H.M. of same amplitude and frequency along the same straight line. At a particular instant, maximum distance between two particles is $\sqrt{2}$ a. The initial phase difference between them is

$\left[\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}\right]$

Show answer and solution

Answer: (b)

$x_1 - x_2 = a\sin\omega t - a\sin(\omega t + \phi) = 2a\sin\frac{\phi}{2}\cos(\dots)$
Maximum separation $= 2a\sin\frac{\phi}{2} = \sqrt{2}a$
$\sin\frac{\phi}{2} = \frac{1}{\sqrt{2}} \Rightarrow \frac{\phi}{2} = \frac{\pi}{4}$
So the answer is (b) $\frac{\pi}{2}$ .

Report a mistake

Q49JEE Main 2021 · 25 FEB · SHIFT 2Medium

Periodic Motion & SHM Basics · Displacement equation and circle of reference

The point A moves with a uniform speed along the circumference of a circle of radius 0.36 m and covers 30 $^\circ$ in 0.1 s. The perpendicular projection 'P' from 'A' on the diameter MN represents the simple harmonic motion of 'P'. The restoration force per unit mass when P touches M will be:

Show answer and solution

Answer: (a)

$\omega = \frac{\pi/6}{0.1} = \frac{5\pi}{3}\text{ rad/s}$
At M, $P$ is at the extreme, $|x| = R = 0.36\text{ m}$ .
Force per unit mass $= \omega^2 R = \frac{25\pi^2}{9}\times 0.36 = \pi^2 \approx 9.87$
So the answer is (a) 9.87 (unit is really N/kg).

Report a mistake

Q50JEE Main 2021 · 27 AUG · SHIFT 2Medium

Periodic Motion & SHM Basics · Displacement equation and circle of reference

Two simple harmonic motion, are represented by the equations $y_1 = 10\sin\left(3\pi t + \frac{\pi}{3}\right) \quad y_2 = 5(\sin 3\pi t + \sqrt{3}\cos 3\pi t)$ Ratio of amplitude of y1 to y2 = x: 1. The value of x is ______________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 1

$A_1 = 10$
$y_2 = 5(\sin 3\pi t + \sqrt{3}\cos 3\pi t) = 10\sin\left(3\pi t + \frac{\pi}{3}\right)$ , so $A_2 = 5\sqrt{1 + 3} = 10$
$\frac{A_1}{A_2} = 1$
So $x = 1$ .

Report a mistake

Q51JEE Main 2021 · 26 AUG · SHIFT 2Medium

Periodic Motion & SHM Basics · Displacement equation and circle of reference

Two simple harmonic motions are represented by the equations

$x_1 = 5\sin\left(2\pi t + \frac{\pi}{4}\right)$ and $x_2 = 5\sqrt{2}(\sin 2\pi t + \cos 2\pi t)$ . The amplitude of second motion is ................ times the amplitude in first motion.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 2

$A_1 = 5$
$x_2 = 5\sqrt{2}(\sin 2\pi t + \cos 2\pi t) = 5\sqrt{2}\cdot\sqrt{2}\sin\left(2\pi t + \frac{\pi}{4}\right)$ , so $A_2 = 10$
$\frac{A_2}{A_1} = 2$
So the answer is 2.

Report a mistake

Q52JEE Main 2021 · 27 JUL · SHIFT 2Medium

Periodic Motion & SHM Basics · Displacement equation and circle of reference

A particle executes simple harmonic motion represented by displacement function as x(t) = A sin( $\omega$ t + $\phi$ )

If the position and velocity of the particle at t = 0 s are 2 cm and 2 $\omega$ cm s ${}^{-1}$ respectively, then its amplitude is $x\sqrt{2}$ cm where the value of x is _________________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 2

At $t = 0$ : $A\sin\phi = 2$
$v = A\omega\cos\phi = 2\omega \Rightarrow A\cos\phi = 2$
$A = \sqrt{2^2 + 2^2} = 2\sqrt{2}\text{ cm}$
So $x = 2$ .

Report a mistake

Q53JEE Main 2021 · 26 FEB · SHIFT 2Medium

Periodic Motion & SHM Basics · Time and distance in SHM

Time period of a simple pendulum is T. The time taken to complete $\frac{5}{8}$ oscillations starting from mean position is $\frac{\alpha}{\beta}T$ . The value of $\alpha$ is _________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 7

One oscillation covers $4A$ , so $\frac{5}{8}$ oscillation means a path of $\frac{5}{8}(4A) = \frac{5A}{2}$ .
From the mean: $0 \to A \to 0$ covers $2A$ in $\frac{T}{2}$ .
Remaining $\frac{A}{2}$ from the mean takes $\frac{T}{12}$ .
Total $t = \frac{T}{2} + \frac{T}{12} = \frac{7}{12}T$
So $\alpha = 7$ .

Report a mistake

Q54MHT CET 2021 · 22 SEP · SHIFT 2Medium

Periodic Motion & SHM Basics · Phase in waves (mixed)

' $n$ ' waves are produced on a string in 1 second. When the radius of the string is doubled, keeping tension same, the number of waves produced in 1 second for the same harmonic will be

Show answer and solution

Answer: (b)

$f = \frac{p}{2L}\sqrt{\frac{T}{\mu}}$ and $\mu = \pi r^2\rho$ , so $f \propto \frac{1}{r}$ .
Radius doubled: $f' = \frac{f}{2}$
So the answer is (b) $\frac{n}{2}$ .

Report a mistake

Q55NEET 2020 · PHASE 2Easy

Periodic Motion & SHM Basics · Periodic and oscillatory motion

Identify the function which represents a periodic motion.

Show answer and solution

Answer: (d)

$e^{-\omega t}$ , $e^{\omega t}$ and $\log_e(\omega t)$ change steadily with time and never repeat.
$\sin\omega t + \cos\omega t = \sqrt{2}\sin(\omega t + \pi/4)$ repeats after $T = 2\pi/\omega$ .
So the answer is (d) $\sin\omega t + \cos\omega t$ .

Report a mistake

Q56MHT CET 2020 · 16 OCT · SHIFT 2Medium

Periodic Motion & SHM Basics · Time and distance in SHM

A particle performs simple harmonic motion with period of 3 s. The time taken by it to cover a distance equal to half the amplitude from mean position is [ $\sin 30^\circ = 0.5$ ]

Show answer and solution

Answer: (a)

$\sin\omega t = \frac{1}{2} \Rightarrow \omega t = \frac{\pi}{6} \Rightarrow t = \frac{T}{12}$
$t = \frac{3}{12} = \frac{1}{4}\text{ s}$
So the answer is (a) $\frac{1}{4}\text{ s}$ .

Report a mistake

Q57NEET 2019Easy

Periodic Motion & SHM Basics · Displacement equation and circle of reference

The radius of circle, the period of revolution, initial position and sense of revolution are indicated is the figure.

y- projection of the radius vector of rotating particle P is:

Show answer and solution

Answer: (c)

$r = 3\ \text{m}$ , $T = 4\ \text{s}$ , so $\omega = \frac{2\pi}{4} = \frac{\pi}{2}\ \text{rad/s}$ .
At $t = 0$ , P is on the $+y$ axis, so $y(0) = +3\ \text{m}$ (maximum); it moves clockwise, so $y$ decreases.
Hence $y(t) = 3\cos\left(\frac{\pi t}{2}\right)$ m.
So the answer is (c).

Report a mistake

Q58JEE Main 2019 · 12 JAN · SHIFT 2Easy

Periodic Motion & SHM Basics · Displacement equation and circle of reference

A simple harmonic motion is represented by:

y = 5 (sin 3 $\pi$ t + $\sqrt{3}$ cos 3 $\pi$ t) cm

The amplitude and time period of the motion are:

Show answer and solution

Answer: (d)

For $a\sin\theta + b\cos\theta$ , amplitude $= \sqrt{a^2 + b^2}$ .
Amplitude $= 5\sqrt{1^2 + (\sqrt{3})^2} = 5 \times 2 = 10\ \text{cm}$ .
$\omega = 3\pi$ , so $T = \frac{2\pi}{3\pi} = \frac{2}{3}\ \text{s}$ .
So the answer is (d) 10 cm, $\frac{2}{3}$ s.

Report a mistake

Q59MHT CET 2019 · 2 MAY · SHIFT 1Easy

Periodic Motion & SHM Basics · Time and distance in SHM

A particle executes the simple harmonic motion with an amplitude ' $A$ '. The distance travelled by it in one periodic time is

Show answer and solution

Answer: (d)

In one period the particle goes mean $\to$ $+A$ $\to$ mean $\to$ $-A$ $\to$ mean.
Distance $= A + A + A + A = 4A$ .
So the answer is (d) $4A$ .

Report a mistake

Q60NEET 2019Easy

Periodic Motion & SHM Basics · Time and distance in SHM

The distance covered by a particle undergoing SHM in one time period is (amplitude = $A$ ),

Show answer and solution

Answer: (a)

In one period the particle goes mean $\to$ $+A$ $\to$ mean $\to$ $-A$ $\to$ mean.
Distance $= A + A + A + A = 4A$ .
So the answer is (a) $4A$ .

Report a mistake

Q61NEET 2019Easy

Periodic Motion & SHM Basics · Time and distance in SHM

Average velocity of a particle executing SHM in one complete vibration is:

Show answer and solution

Answer: (d)

Average velocity $= \frac{\text{displacement}}{\text{time}}$ .
After one complete vibration the particle is back at its starting point, so displacement $= 0$ .
So the answer is (d) zero.

Report a mistake

Q62NEET 2019Medium

Periodic Motion & SHM Basics · Displacement equation and circle of reference

The displacement of a particle executing simple harmonic motion is given by

y = A0 + A sin $\omega$ t + B cos $\omega$ t.

Then the amplitude of its oscillation is given by:

Show answer and solution

Answer: (a)

$A_0$ only shifts the mean position; it does not change the amplitude.
$A\sin\omega t + B\cos\omega t = \sqrt{A^2 + B^2}\,\sin(\omega t + \delta)$
So the answer is (a) $\sqrt{A^2 + B^2}$ .

Report a mistake

Q63JEE Main 2018 · 16 APR · SHIFT 1Difficult

Periodic Motion & SHM Basics · Time and distance in SHM

A particle executes simple harmonic motion and is located at x = a, b and c at times t0, 2t0 and 3t0 respectively. The frequency of the oscillation is:

Show answer and solution

Answer: (a)

Let $x = A\sin(\omega t + \delta)$ , so $a = A\sin(\omega t_0 + \delta)$ , $b = A\sin(2\omega t_0 + \delta)$ , $c = A\sin(3\omega t_0 + \delta)$ .
$a + c = 2A\sin(2\omega t_0 + \delta)\cos(\omega t_0) = 2b\cos\omega t_0$
$\cos\omega t_0 = \frac{a+c}{2b} \Rightarrow \omega = \frac{1}{t_0}\cos^{-1}\left(\frac{a+c}{2b}\right)$
$f = \frac{\omega}{2\pi} = \frac{1}{2\pi t_0}\cos^{-1}\left(\frac{a+c}{2b}\right)$
So the answer is (a).

Report a mistake

Q64JEE Main 2016 · 9 APR · SHIFT 1Medium

Periodic Motion & SHM Basics · Phase and phase difference

Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to A and T, respectively. At time t = 0 one particle has displacement A while the other one has displacement $\frac{-A}{2}$ and they are moving towards each other. If they cross each other at time t, then t is:

Show answer and solution

Answer: (a)

Take $x = A\sin(\omega t + \phi)$ . Particle 1 at $x = A$ : $\phi_1 = \frac{\pi}{2}$ .
Particle 2 at $-\frac{A}{2}$ moving towards $+x$ : $\phi_2 = -\frac{\pi}{6}$ .
They cross when $\sin(\omega t + \frac{\pi}{2}) = \sin(\omega t - \frac{\pi}{6})$ , i.e. $\omega t + \frac{\pi}{2} = \pi - \omega t + \frac{\pi}{6}$ .
$2\omega t = \frac{2\pi}{3} \Rightarrow \omega t = \frac{\pi}{3} \Rightarrow t = \frac{T}{6}$
So the answer is (a) $\frac{T}{6}$ .

Report a mistake

Q65JEE Main 2014 · 12 APR · SHIFT 1Easy

Periodic Motion & SHM Basics · Restoring force and differential equation

Which of the following expressions corresponds to simple harmonic motion along a straight line, where x is the displacement and a, b, c are positive constants?

Show answer and solution

Answer: (d)

For SHM the force (or acceleration) must be $F = -kx$ only.
Options (a), (b), (c) have constant or $x^2$ terms.
$-bx$ is the only linear restoring expression.
So the answer is (d) $-bx$ .

Report a mistake

Q66JEE Main 2014 · 19 APR · SHIFT 1Medium

Periodic Motion & SHM Basics · Time and distance in SHM

A body is in simple harmonic motion with time period half second (T = 0.5 s) and amplitude one cm (A = 1 cm). Find the average velocity in the interval in which it moves from equilibrium position to half of its amplitude.

Show answer and solution

Answer: (c)

Mean to $\frac{A}{2}$ takes $t = \frac{T}{12} = \frac{0.5}{12} = \frac{1}{24}\text{ s}$
Displacement $= \frac{A}{2} = 0.5\text{ cm}$
$v_{avg} = \frac{0.5}{1/24} = 12\text{ cm/s}$
So the answer is (c) 12 cm/s.

Report a mistake

Q67JEE Main 2014Difficult

Periodic Motion & SHM Basics · Time and distance in SHM

A particle moves with simple harmonic motion in a straight line. In first $\tau s,$ after starting from rest it travels a distance $a,$ and in next $\tau s$ it travels $2a,$ in same direction, then:

Show answer and solution

Answer: (d)

Starting from rest means starting at an extreme: distance covered in time $t$ is $A(1 - \cos\omega t)$ .
After $\tau$ : $A(1 - \cos\omega\tau) = a$ ; after $2\tau$ : $A(1 - \cos 2\omega\tau) = 3a$
With $c = \cos\omega\tau$ : $\frac{1 - \cos2\omega\tau}{1 - c} = \frac{2(1-c)(1+c)}{1-c} = 2(1+c) = 3 \Rightarrow c = \frac{1}{2}$
$\omega\tau = \frac{\pi}{3} \Rightarrow T = \frac{2\pi}{\omega} = 6\tau$ (and $A = 2a$ )
So the answer is (d) $T = 6\tau$ .

Report a mistake

Q68NEET 2011 · AIPMT · PRELIMSEasy

Periodic Motion & SHM Basics · Periodic and oscillatory motion

Out of the following functions representing motion of a particle which represents SHM (1) y = sin $\omega$ t $-$ cos $\omega$ t

(2) y = sin3 $\omega$ t

(3) y = 5cos $\left(\frac{3\pi}{4} - 3\omega t\right)$

(4) y = 1 + $\omega$ t + $\omega$ 2t2

Show answer and solution

Answer: (c)

(1) $\sin\omega t - \cos\omega t = \sqrt{2}\sin(\omega t - \pi/4)$ : a single sine term, so SHM.
(2) $\sin^3\omega t = \frac{3\sin\omega t - \sin 3\omega t}{4}$ : two different frequencies, periodic but not SHM.
(3) $5\cos\left(\frac{3\pi}{4} - 3\omega t\right)$ : a single cosine term, so SHM.
(4) $1 + \omega t + \omega^2 t^2$ grows without limit, not even periodic.
So the answer is (c) only (1) and (3).

Report a mistake

Q69NEET 2011 · AIPMT · MAINSMedium

Periodic Motion & SHM Basics · Phase and phase difference

Two particles are oscillating along two close parallel straight lines side by side, with the same frequency and amplitudes. They pass each other, moving in opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie on a straight line perpendicular to the paths of the two particles. The phase difference is

Show answer and solution

Answer: (c)

At $x = \frac{A}{2}$ : $\sin\phi = \frac{1}{2}$ , so $\phi = \frac{\pi}{6}$ or $\frac{5\pi}{6}$ .
Moving in opposite directions means one has $\cos\phi > 0$ ( $\frac{\pi}{6}$ ) and the other $\cos\phi < 0$ ( $\frac{5\pi}{6}$ ).
$\Delta\phi = \frac{5\pi}{6} - \frac{\pi}{6} = \frac{2\pi}{3}$
So the answer is (c) $\frac{2\pi}{3}$ .

Report a mistake

Q70JEE Main 2011 · AIEEEDifficult

Periodic Motion & SHM Basics · Phase and phase difference

Two particles are executing simple harmonic motion of the same amplitude $A$ and frequency $\omega$ along the $x$ -axis. Their mean position is separated by distance $X_0\ (X_0 > A)$ . If the maximum separation between them is $(X_0 + A),$ the phase difference between their motion is:

Show answer and solution

Answer: (a)

Take $x_1 = A\sin\omega t$ and $x_2 = X_0 + A\sin(\omega t + \phi)$ .
Separation $= X_0 + A[\sin(\omega t+\phi) - \sin\omega t] = X_0 + 2A\sin\frac{\phi}{2}\cos\left(\omega t + \frac{\phi}{2}\right)$
Maximum separation $= X_0 + 2A\sin\frac{\phi}{2} = X_0 + A \Rightarrow \sin\frac{\phi}{2} = \frac{1}{2}$
$\frac{\phi}{2} = \frac{\pi}{6} \Rightarrow \phi = \frac{\pi}{3}$
So the answer is (a) $\frac{\pi}{3}$ .

Report a mistake

Q71NEET 2010 · AIPMT · PRELIMSEasy

Periodic Motion & SHM Basics · Periodic and oscillatory motion

The displacement of a particle along the x-axis is given by x = asin2 $\omega$ t. The motion of the particle corresponds to

Show answer and solution

Answer: (c)

$x = a\sin^2\omega t = \frac{a}{2} - \frac{a}{2}\cos 2\omega t$ .
This is SHM of amplitude $a/2$ about the shifted mean position $x = a/2$ , with angular frequency $2\omega$ .
Frequency $f = \frac{2\omega}{2\pi} = \frac{\omega}{\pi}$ .
So the working gives (a) SHM of frequency $\omega/\pi$ ; the key's (c) treats the shifted-mean motion as non-SHM.

Report a mistake

Q72NEET 2009 · AIPMTEasy

Periodic Motion & SHM Basics · Restoring force and differential equation

Which one of the following equations of motion represents simple harmonic motion? where k, k0, k1 and a are all positive.

Show answer and solution

Answer: (c)

SHM condition: acceleration $= -\omega^2 x$ (proportional to displacement and opposite to it).
(b) has the wrong sign and (d) has an $x^2$ term.
(a) is SHM only about a shifted mean $x=-a$ ; (c) is the standard form.
(c) acceleration $= -\omega^2 x$ matches the definition exactly.
So the answer is (c).

Report a mistake

Q73JEE Main 2007 · AIEEEEasy

Periodic Motion & SHM Basics · Restoring force and differential equation

A point mass oscillates along the $x$ -axis according to the law $x = x_0 \cos(\omega t - \pi/4).$ If the acceleration of the particle is written as $a = A\cos(\omega t + \delta),$ then

Show answer and solution

Answer: (a)

$a = \frac{d^2x}{dt^2} = -\omega^2 x_0\cos(\omega t - \pi/4)$ .
Use $-\cos\theta = \cos(\theta + \pi)$ : $a = x_0\omega^2\cos(\omega t - \pi/4 + \pi) = x_0\omega^2\cos(\omega t + 3\pi/4)$ .
So $A = x_0\omega^2$ , $\delta = 3\pi/4$ .
So the answer is (a).

Report a mistake

Q74NEET 2007 · AIPMTEasy

Periodic Motion & SHM Basics · Time and distance in SHM

A particle executes simple harmonic oscillation with an amplitude $a$ . The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is

Show answer and solution

Answer: (b)

From the mean position: $x = a\sin\frac{2\pi t}{T}$ .
$\frac{a}{2} = a\sin\frac{2\pi t}{T} \Rightarrow \frac{2\pi t}{T} = \frac{\pi}{6}$ .
$t = \frac{T}{12}$ .
So the answer is (b) $T/12$ .

Report a mistake

Q75JEE Main 2007 · AIEEEEasy

Periodic Motion & SHM Basics · Time and distance in SHM

The displacement of an object attached to a spring and executing simple harmonic motion is given by $x = 2 \times 10^{-2}\cos \pi t$ metre. The time at which the maximum speed first occurs is

Show answer and solution

Answer: (b)

$\omega = \pi$ , so $T = \frac{2\pi}{\pi} = 2\ \text{s}$ .
$x = A\cos\pi t$ starts at the extreme position; maximum speed is at the mean position.
First reached after $\frac{T}{4} = 0.5\ \text{s}$ .
So the answer is (b) 0.5 s.

Report a mistake

Q76NEET 2005 · AIPMTEasy

Periodic Motion & SHM Basics · Periodic and oscillatory motion

The circular motion of a particle with constant speed is

Show answer and solution

Answer: (a)

Uniform circular motion repeats after every revolution, so it is periodic.
It is a 2-D motion, not a to-and-fro motion along a line with $F \propto -x$ ; only its projection on a diameter is SHM.
So the answer is (a) periodic but not simple harmonic.

Report a mistake

Q77JEE Main 2005 · AIEEEEasy

Periodic Motion & SHM Basics · Displacement equation and circle of reference

If a simple harmonic motion is represented by $\frac{d^2x}{dt^2} + \alpha x = 0.$ its time period is

Show answer and solution

Answer: (a)

Compare with $\frac{d^2x}{dt^2} + \omega^2 x = 0$ : $\omega = \sqrt{\alpha}$ .
$T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{\alpha}}$ .
So the answer is (a).

Report a mistake

Q78JEE Main 2005 · AIEEEMedium

Periodic Motion & SHM Basics · Periodic and oscillatory motion

The function $\sin^2(\omega t)$ represents

Show answer and solution

Answer: (a)

$\sin^2\omega t = \frac{1}{2} - \frac{1}{2}\cos 2\omega t$
It repeats when $2\omega t$ changes by $2\pi$ , so period $T = \frac{\pi}{\omega}$ .
Displacement from the origin does not satisfy $\frac{d^2y}{dt^2} = -\omega'^2 y$ (it has a constant term), so it is periodic but not SHM.
So the answer is (a).

Report a mistake

Q79JEE Main 2005 · AIEEEMedium

Periodic Motion & SHM Basics · Phase and phase difference

Two simple harmonic motions are represented by the equations $y_1 = 0.1\sin\left(100\pi t + \frac{\pi}{3}\right)$ and $y_2 = 0.1\cos \pi t.$ The phase difference of the velocity of particle $1$ with respect to the velocity of particle $2$ is

Show answer and solution

Answer: (b)

Take $y_2 = 0.1\cos 100\pi t$ (same $\omega$ as $y_1$ ).
$v_1 \propto \cos\left(100\pi t + \frac{\pi}{3}\right)$ , $v_2 \propto -\sin 100\pi t = \cos\left(100\pi t + \frac{\pi}{2}\right)$
$\Delta\phi = \frac{\pi}{3} - \frac{\pi}{2} = -\frac{\pi}{6}$
So the answer is (b) $-\frac{\pi}{6}$ .

Report a mistake

Q80JEE Main 2003 · AIEEEMedium

Periodic Motion & SHM Basics · Displacement equation and circle of reference

The displacement of particle varies according to the relation

$x = 4(\cos \pi t + \sin \pi t).$ The amplitude of the particle is

Show answer and solution

Answer: (c)

$x = 4\cos\pi t + 4\sin\pi t$
Amplitude $= \sqrt{4^2 + 4^2} = 4\sqrt{2}$
So the answer is (c) $4\sqrt{2}$ .

Report a mistake

Q81NEET 1994 · AIPMTEasy

Periodic Motion & SHM Basics · Restoring force and differential equation

A particle executes S.H.M. along $x$ -axis. The force acting on it is given by

Show answer and solution

Answer: (d)

In SHM the restoring force is $F = -kx$ : proportional to $x$ and directed towards the mean position.
Only $-Akx$ has this form.
So the answer is (d) $-Akx$ .

Report a mistake

Q82NEET 1992 · AIPMTEasy

Periodic Motion & SHM Basics · Time and distance in SHM

A simple harmonic oscillator has an amplitude $A$ and time period $T$ . The time required by it to travel from $x = A$ to $x = A/2$ is

Show answer and solution

Answer: (a)

From the extreme position: $x = A\cos\frac{2\pi t}{T}$ .
$\frac{A}{2} = A\cos\frac{2\pi t}{T} \Rightarrow \frac{2\pi t}{T} = \frac{\pi}{3}$ .
$t = \frac{T}{6}$ .
So the answer is (a) $T/6$ .

Report a mistake

Q83NEET 1988 · AIPMTEasy

Periodic Motion & SHM Basics · Restoring force and differential equation

A particle executes S.H.M. along $x$ -axis. The force acting on it is given by

Show answer and solution

Answer: (d)

In SHM the restoring force is $F = -kx$ : proportional to $x$ and directed towards the mean position.
Only $-Akx$ has this form.
So the answer is (d) $-Akx$ .

Report a mistake

Q84MHT CET 2026 · 20 MAY · SHIFT 1Easy

Velocity & Acceleration in SHM · Velocity at a displacement

A simple pendulum performs simple harmonic motion about $x = 0$ with an amplitude 'A' and angular frequency $\omega$ . The speed of pendulum at $x = \frac{A}{8}$ will be.

Show answer and solution

Answer: (c)

$v = \omega\sqrt{A^2 - x^2}$
$v = \omega\sqrt{A^2 - \frac{A^2}{64}} = \omega A\sqrt{\frac{63}{64}}$
So the answer is (c) $\frac{\sqrt{63}A\omega}{8}$ .

Report a mistake

Q85MHT CET 2026 · 20 APR · SHIFT 2Easy

Velocity & Acceleration in SHM · Velocity at a displacement

A particle is performing S.H.M. about $x = 0$ , with an amplitude $a$ and time period $T$ . The speed of the particle at $x = \frac{a}{3}$ will be

Show answer and solution

Answer: (d)

$v = \omega\sqrt{a^2 - x^2}$ with $\omega = \frac{2\pi}{T}$
$\sqrt{a^2 - \frac{a^2}{9}} = \frac{2\sqrt{2}a}{3}$
$v = \frac{2\pi}{T}\cdot\frac{2\sqrt{2}a}{3} = \frac{4\sqrt{2}\pi a}{3T}$
So the answer is (d) $\frac{4\sqrt{2}\pi a}{3T}$ .

Report a mistake

Q86MHT CET 2026 · 18 APR · SHIFT 1Easy

Velocity & Acceleration in SHM · Velocity at a displacement

A particle is performing simple harmonic motion about $x = 0$ with an amplitude ' $a$ ' and periodic time T. The speed of the particle at $x = \frac{a}{3}$ will be

Show answer and solution

Answer: (c)

$v = \omega\sqrt{a^2 - x^2}$ with $\omega = \frac{2\pi}{T}$
$\sqrt{a^2 - \frac{a^2}{9}} = \frac{2\sqrt{2}a}{3}$
$v = \frac{2\pi}{T}\cdot\frac{2\sqrt{2}a}{3} = \frac{4\sqrt{2}\pi a}{3T}$
So the answer is (c) $\frac{4\sqrt{2}\pi a}{3T}$ .

Report a mistake

Q87MHT CET 2026 · 17 APR · SHIFT 1Easy

Velocity & Acceleration in SHM · Velocity at a displacement

A particle executing simple harmonic motion starts from mean position with amplitude ' $A$ ' and periodic time ' $T$ '. At what displacement is its speed one-fourth of the maximum speed?

Show answer and solution

Answer: (d)

$v = \omega\sqrt{A^2 - x^2}$ , $v_{max} = \omega A$
$\omega\sqrt{A^2 - x^2} = \frac{\omega A}{4} \Rightarrow A^2 - x^2 = \frac{A^2}{16}$
$x^2 = \frac{15A^2}{16} \Rightarrow x = \frac{\sqrt{15}A}{4}$
So the answer is (d) $\frac{A\sqrt{15}}{4}$ .

Report a mistake

Q88JEE Main 2026 · 4 APR · SHIFT 1Easy

Velocity & Acceleration in SHM · Velocity at a displacement

The velocity of a particle executing simple harmonic motion along $x$ -axis is described as $v^2 = 50 - x^2$ , where $x$ represents displacement. If the time period of motion is $\frac{x}{7}\text{ s}$ , the value of $x$ is ______.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 44

For SHM $v^2 = \omega^2(A^2 - x^2) = \omega^2A^2 - \omega^2x^2$
Compare with $v^2 = 50 - x^2$ : $\omega^2 = 1 \Rightarrow \omega = 1$ rad/s
$T = \frac{2\pi}{\omega} = 2\pi = 2\times\frac{22}{7} = \frac{44}{7}$ s
$\frac{x}{7} = \frac{44}{7} \Rightarrow x = 44$

Report a mistake

Q89MHT CET 2026 · 14 MAY · SHIFT 2Easy

Velocity & Acceleration in SHM · Maximum velocity and acceleration

The maximum velocity and maximum acceleration of a particle performing a linear S.H.M. are $\alpha$ and $\beta$ respectively. Then, path length of the particle is

Show answer and solution

Answer: (c)

$v_{max} = A\omega = \alpha$ , $a_{max} = A\omega^2 = \beta$
$\frac{\alpha^2}{\beta} = \frac{A^2\omega^2}{A\omega^2} = A$
Path length $= 2A = \frac{2\alpha^2}{\beta}$
So the answer is (c) $\frac{2\alpha^2}{\beta}$ .

Report a mistake

Q90MHT CET 2026 · 12 MAY · SHIFT 1Easy

Velocity & Acceleration in SHM · Maximum velocity and acceleration

Two particles performing simple harmonic motions of maximum accelerations, $1$ m/s $2$ and $100$ m/s $2$ , have the same amplitudes. The ratio of their angular frequencies is

Show answer and solution

Answer: (b)

$a_{max} = A\omega^2$ ; same $A$ , so $\omega \propto \sqrt{a_{max}}$
$\frac{\omega_1}{\omega_2} = \sqrt{\frac{1}{100}} = \frac{1}{10}$
So the answer is (b) $1 : 10$ .

Report a mistake

Q91MHT CET 2026 · 15 APR · SHIFT 2Easy

Velocity & Acceleration in SHM · Maximum velocity and acceleration

A body of mass $0.4$ kg performs simple harmonic motion. It experiences a restoring force of $0.4$ N when its displacement from the mean position is 4 cm. The force constant and magnitude of acceleration respectively are

Show answer and solution

Answer: (b)

$k = \frac{F}{x} = \frac{0.4}{0.04} = 10$ N/m
$a = \frac{F}{m} = \frac{0.4}{0.4} = 1\ \text{m/s}^2$
So the answer is (b) $10$ N/m, $1\ \text{m/s}^2$ .

Report a mistake

Q92MHT CET 2026 · 14 MAY · SHIFT 2Medium

Velocity & Acceleration in SHM · Velocity at a displacement

A particle executes S.H.M. with time period $16$ s. At time $t = 2$ s, the particle crosses the mean position while at $t = 4$ s, velocity is $4$ m/s. The amplitude of motion (in m) is $\left(\sin 45^\circ = \frac{1}{\sqrt{2}}\right)$

Show answer and solution

Answer: (d)

$\omega = \frac{2\pi}{T} = \frac{2\pi}{16} = \frac{\pi}{8}$ rad/s.
Particle is at the mean position at $t = 2$ s, so $x = A\sin\omega(t - 2)$ and $v = A\omega\cos\omega(t-2)$ .
At $t = 4$ s: $\omega(t-2) = \frac{\pi}{4}$ , so $4 = A\cdot\frac{\pi}{8}\cdot\frac{1}{\sqrt{2}}$ .
$A = \frac{32\sqrt{2}}{\pi}$ m.
So the answer is (d) $\frac{32\sqrt{2}}{\pi}$ .

Report a mistake

Q93MHT CET 2026 · 15 APR · SHIFT 1Medium

Velocity & Acceleration in SHM · Velocity at a displacement

A particle executes linear S.H.M. with amplitude $4$ cm. The magnitude of velocity and acceleration is equal when it is at $3$ cm from mean position. Time period is

Show answer and solution

Answer: (b)

$|v| = \omega\sqrt{A^2 - x^2}$ , $|a| = \omega^2 x$ .
$\omega\sqrt{16 - 9} = \omega^2(3) \Rightarrow \omega = \frac{\sqrt{7}}{3}$ rad/s.
$T = \frac{2\pi}{\omega} = \frac{6\pi}{\sqrt{7}}$ s.
So the answer is (b) $\frac{6\pi}{\sqrt{7}}$ s.

Report a mistake

Q94MHT CET 2026 · 13 APR · SHIFT 2Medium

Velocity & Acceleration in SHM · Velocity at a displacement

The velocity of a particle executing SHM varies with displacement ( $x$ ) as $4v^2 = 50 - x^2$ . The time period of oscillations is $x/7$ s.

The value of $x$ is (Take $\pi = 22/7$ )

Show answer and solution

Answer: (c)

Compare $v^2 = \frac{1}{4}(50 - x^2)$ with $v^2 = \omega^2(A^2 - x^2)$ : $\omega^2 = \frac{1}{4}$ , $\omega = \frac{1}{2}$ rad/s.
$T = \frac{2\pi}{\omega} = 4\pi = 4\times\frac{22}{7} = \frac{88}{7}$ s.
$\frac{x}{7} = \frac{88}{7} \Rightarrow x = 88$ .
So the answer is (c) 88.

Report a mistake

Q95MHT CET 2026 · 19 APR · SHIFT 2Medium

Velocity & Acceleration in SHM · Velocities at two positions

A particle executing linear S.H.M. has velocities $V_1$ and $V_2$ at distance $x_1$ and $x_2$ respectively, from the mean position, its angular velocity is

Show answer and solution

Answer: (a)

$V_1^2 = \omega^2(A^2 - x_1^2)$ and $V_2^2 = \omega^2(A^2 - x_2^2)$ .
Subtract: $V_2^2 - V_1^2 = \omega^2(x_1^2 - x_2^2)$ .
$\omega = \sqrt{\frac{V_2^2 - V_1^2}{x_1^2 - x_2^2}}$ .
So the answer is (a).

Report a mistake

Q96NEET 2026 · RE-NEETMedium

Velocity & Acceleration in SHM · Graphs and phase relations

Consider a spring-mass simple harmonic oscillator in one dimension. The mass of the particle is $m\text{ kg}$ and the spring constant is $k\text{ N m}^{-1}$ . At a given instant, the extension of the spring is $x$ -meter and the speed of the particle is $v\text{ m s}^{-1}$ . On the $x - v$ plane, if the graph of $v$ as a function of $x$ is a circle, then the correct option is:

Show answer and solution

Answer: (c)

$v^2 = \frac{k}{m}(A^2 - x^2) \Rightarrow v^2 + \frac{k}{m}x^2 = \frac{k}{m}A^2$ .
This is a circle only if the coefficients of $x^2$ and $v^2$ are equal: $\frac{k}{m} = 1$ .
So the answer is (c) $k = m$ .

Report a mistake

Q97MHT CET 2025 · 5 MAY · SHIFT 2Easy

Velocity & Acceleration in SHM · Velocity at a displacement

A simple pendulum starts oscillating simple harmonically from its mean position ( $x = 0$ ) with amplitude ' $a$ ' and periodic time ' $T$ '. The magnitude of velocity of pendulum at $x = \frac{a}{2}$ is

Show answer and solution

Answer: (d)

$v = \omega\sqrt{a^2 - x^2}$ with $\omega = \frac{2\pi}{T}$
$\sqrt{a^2 - \frac{a^2}{4}} = \frac{\sqrt{3}a}{2}$
$v = \frac{2\pi}{T}\cdot\frac{\sqrt{3}a}{2} = \frac{\sqrt{3}\pi a}{T}$
So the answer is (d) $\frac{\sqrt{3}\pi a}{T}$ .

Report a mistake

Q98MHT CET 2025 · 22 APR · SHIFT 2Easy

Velocity & Acceleration in SHM · Velocity at a displacement

A particle executes S.H.M. starting from the mean position. Its amplitude is ' a ' and its periodic time is ' $T$ '. At a certain instant, its speed ' $u$ ' is half that of maximum speed $V$ max. The displacement of the particle at that instant is

Show answer and solution

Answer: (d)

$v = \omega\sqrt{a^2 - x^2}$ , $V_{max} = \omega a$
$\omega\sqrt{a^2 - x^2} = \frac{\omega a}{2} \Rightarrow a^2 - x^2 = \frac{a^2}{4}$
$x = \frac{\sqrt{3}a}{2}$
So the answer is (d) $\frac{\sqrt{3}a}{2}$ .

Report a mistake

Q99MHT CET 2025 · 21 APR · SHIFT 1Easy

Velocity & Acceleration in SHM · Maximum velocity and acceleration

Two simple harmonic motions of angular frequency $300\text{ rad/s}$ and $3000\text{ rad/s}$ have same amplitude. The ratio of their maximum accelerations is

Show answer and solution

Answer: (b)

$a_{max} = A\omega^2$ ; same $A$ , so $a_{max} \propto \omega^2$
$\frac{a_1}{a_2} = \left(\frac{300}{3000}\right)^2 = \frac{1}{100}$
So the answer is (b) $1 : 10^2$ .

Report a mistake

Q100MHT CET 2025 · 26 APR · SHIFT 2Easy

Velocity & Acceleration in SHM · Graphs and phase relations

A particle is performing S.H.M. starting from extreme position. Graphical representation shows that between displacement and acceleration, there is a phase difference of

Show answer and solution

Answer: (a)

In SHM $a = -\omega^2 x$ : acceleration is always opposite to displacement.
$-\cos\theta = \cos(\theta + \pi)$ , so the acceleration graph is the displacement graph shifted by $\pi$ .
So the answer is (a) $\pi$ rad.

Report a mistake

Q101MHT CET 2025 · 19 APR · SHIFT 2Easy

Velocity & Acceleration in SHM · Graphs and phase relations

For a particle performing S.H.M. the displacement - time graph is shown.

For that particle the force - time graph is correctly shown in graph

Show answer and solution

Answer: (c)

The graph starts at $y = 0$ and rises, so $y = A\sin\omega t$ .
Restoring force $F = -ky = -kA\sin\omega t$ .
So $F$ starts at zero and first goes negative, then positive: an inverted sine curve.
So the answer is (c).

Report a mistake

Q102MHT CET 2025 · 20 APR · SHIFT 2Difficult

Velocity & Acceleration in SHM · Velocities at two positions

A particle executes a linear S.H.M. In two of its positions the velocities are $V_1, V_2$ and accelerations are $a_1$ and $a_2$ respectively $(0 < a_1 < a_2)$ . The distance between the positions is

Show answer and solution

Answer: (c)

$V^2 = \omega^2(A^2 - x^2)$ and $a = \omega^2 x$ (magnitudes).
$V_1^2 - V_2^2 = \omega^2(x_2^2 - x_1^2) = \omega^2(x_2 - x_1)(x_2 + x_1)$
$a_1 + a_2 = \omega^2(x_1 + x_2)$
Dividing: $x_2 - x_1 = \dfrac{V_1^2 - V_2^2}{a_1 + a_2}$
So the answer is (c).

Report a mistake

Q103MHT CET 2024 · 15 MAY · SHIFT 2Easy

Velocity & Acceleration in SHM · Velocity at a displacement

For a particle executing S.H.M. having amplitude A, the speed of the particle is $\left(\frac{1}{3}\right)$ rd of its maximum speed when the displacement from the mean position is

Show answer and solution

Answer: (c)

$v = \omega\sqrt{A^2 - x^2}$ , $v_{max} = \omega A$
$\sqrt{A^2 - x^2} = \frac{A}{3} \Rightarrow x^2 = A^2 - \frac{A^2}{9} = \frac{8A^2}{9}$
$x = \frac{2\sqrt{2}}{3}A$
So the answer is (c) $\frac{2\sqrt{2}}{3}A$ .

Report a mistake

Q104MHT CET 2024 · 11 MAY · SHIFT 2Easy

Velocity & Acceleration in SHM · Velocity at a displacement

A particle is performing S.H.M. about its mean position with an amplitude ' $a$ ' and periodic time ' $T$ '. The speed of the particle when its displacement from mean position is $\frac{a}{3}$ will be

Show answer and solution

Answer: (b)

$v = \omega\sqrt{a^2 - x^2}$ with $\omega = \frac{2\pi}{T}$
$\sqrt{a^2 - \frac{a^2}{9}} = \frac{2\sqrt{2}a}{3}$
$v = \frac{2\pi}{T}\cdot\frac{2\sqrt{2}a}{3} = \frac{4\sqrt{2}\pi a}{3T}$
So the answer is (b) $\frac{4\sqrt{2}\pi a}{3T}$ .

Report a mistake

Q105MHT CET 2024 · 9 MAY · SHIFT 1Easy

Velocity & Acceleration in SHM · Velocity at a displacement

A particle is performing S.H.M. with an amplitude 4 cm. At the mean position the velocity of the particle is $12\text{ cm/s}$ . When the speed of the particle becomes $6\text{ cm/s}$ , the distance of the particle from mean position is

Show answer and solution

Answer: (c)

$v_{max} = A\omega \Rightarrow \omega = \frac{12}{4} = 3\ \text{rad/s}$
$v = \omega\sqrt{A^2 - x^2}$ : $6 = 3\sqrt{16 - x^2} \Rightarrow 16 - x^2 = 4$
$x = \sqrt{12} = 2\sqrt{3}\ \text{cm}$
So the answer is (c) $2\sqrt{3}$ cm.

Report a mistake

Q106MHT CET 2024 · 15 MAY · SHIFT 1Easy

Velocity & Acceleration in SHM · Maximum velocity and acceleration

The maximum velocity of a particle, executing S.H.M. with an amplitude 7 mm is $4.4\text{ m s}^{-1}$ The period of oscillation is $\left[\pi = \frac{22}{7}\right]$

Show answer and solution

Answer: (d)

$v_{max} = A\omega = \frac{2\pi A}{T} \Rightarrow T = \frac{2\pi A}{v_{max}}$
$T = \frac{2\times\frac{22}{7}\times 7\times10^{-3}}{4.4} = \frac{0.044}{4.4}$
$T = 0.01$ s
So the answer is (d) 0.01 s.

Report a mistake

Q107MHT CET 2024 · 10 MAY · SHIFT 1Easy

Velocity & Acceleration in SHM · Maximum velocity and acceleration

A particle performing S.H.M. with maximum velocity ' $V$ '. If the amplitude double and periodic time is made, $\left(\frac{1}{3}\right)$ rd then the maximum velocity is

Show answer and solution

Answer: (a)

$v_{max} = A\omega = \frac{2\pi A}{T}$ , so $v_{max} \propto \frac{A}{T}$
$A' = 2A$ , $T' = \frac{T}{3}$ : $v' = \frac{2}{1/3}V = 6V$
So the answer is (a) $6V$ .

Report a mistake

Q108MHT CET 2024 · 4 MAY · SHIFT 2Easy

Velocity & Acceleration in SHM · Maximum velocity and acceleration

The maximum velocity and maximum acceleration of a particle performing a linear S.H.M. is ' $\alpha$ ' and ' $\beta$ ' respectively. Then the path length of the particle is

Show answer and solution

Answer: (c)

$v_{max} = A\omega = \alpha$ , $a_{max} = A\omega^2 = \beta$
$\frac{\alpha^2}{\beta} = \frac{A^2\omega^2}{A\omega^2} = A$
Path length $= 2A = \frac{2\alpha^2}{\beta}$
So the answer is (c) $\frac{2\alpha^2}{\beta}$ .

Report a mistake

Q109JEE Main 2024 · 6 APR · SHIFT 1Easy

Velocity & Acceleration in SHM · Maximum velocity and acceleration

A particle is doing simple harmonic motion of amplitude $0.06\text{ m}$ and time period $3.14\text{ s}$ . The maximum velocity of the particle is _________ $\text{cm/s}$ .

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 12

$\omega = \frac{2\pi}{T} = \frac{2\times 3.14}{3.14} = 2$ rad/s
$v_{max} = A\omega = 0.06\times 2 = 0.12$ m/s
$v_{max} = 12$ cm/s

Report a mistake

Q110MHT CET 2024 · 11 MAY · SHIFT 1Medium

Velocity & Acceleration in SHM · Velocity at a displacement

A particle performing S.H.M. starts from equilibrium position and its time period is 12 second. After 2 seconds its velocity is $\pi\text{ m/s}$ . Amplitude of the oscillation is

$\left[\sin 30^\circ = \cos 60^\circ = 0.5,\ \sin 60^\circ = \cos 30^\circ = \sqrt{3}/2\right]$

Show answer and solution

Answer: (b)

$\omega = \frac{2\pi}{12} = \frac{\pi}{6}$ rad/s; starting from the mean position, $v = A\omega\cos\omega t$ .
At $t = 2$ s: $\omega t = \frac{\pi}{3}$ , so $\pi = A\cdot\frac{\pi}{6}\cdot\cos 60^\circ = \frac{A\pi}{12}$ .
$A = 12$ m.
So the answer is (b) 12 m.

Report a mistake

Q111MHT CET 2024 · 4 MAY · SHIFT 1Medium

Velocity & Acceleration in SHM · Velocity at a displacement

The velocity of particle executing S.H.M. varies with displacement $(x)$ as $4V^2 = 50 - x^2$ . The time period of oscillation is $\frac{x}{7}$ second. The value of ' $x$ ' is (Take $\pi = \frac{22}{7}$ )

Show answer and solution

Answer: (d)

$v^2 = \frac{1}{4}(50 - x^2)$ compared with $v^2 = \omega^2(A^2 - x^2)$ gives $\omega = \frac{1}{2}$ rad/s.
$T = \frac{2\pi}{\omega} = 4\pi = \frac{88}{7}$ s.
$\frac{x}{7} = \frac{88}{7} \Rightarrow x = 88$ .
So the answer is (d) 88.

Report a mistake

Q112JEE Main 2024 · 4 APR · SHIFT 2Medium

Velocity & Acceleration in SHM · Velocity at a displacement

The displacement of a particle executing SHM is given by $x = 10\sin\left(\omega t + \frac{\pi}{3}\right)\text{ m}$ . The time period of motion is $3.14\text{ s}$ . The velocity of the particle at $t = 0$ is _______ $\text{m/s}$ .

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 10

$\omega = \frac{2\pi}{T} = \frac{2\times 3.14}{3.14} = 2$ rad/s.
$v = \frac{dx}{dt} = 10\omega\cos(\omega t + \frac{\pi}{3})$ .
At $t = 0$ : $v = 10\times 2\times\cos 60^\circ = 10$ m/s.
Answer: 10 m/s.

Report a mistake

Q113JEE Main 2024 · 31 JAN · SHIFT 1Medium

Velocity & Acceleration in SHM · Velocity at a displacement

A particle performs simple harmonic motion with amplitude $A$ . Its speed is increased to three times at an instant when its displacement is $\frac{2A}{3}$ . The new amplitude of motion is $\frac{nA}{3}$ . The value of $n$ is ___________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 7

At $x = \frac{2A}{3}$ : $v = \omega\sqrt{A^2 - \frac{4A^2}{9}} = \frac{\sqrt{5}}{3}\omega A$ .
New speed $v' = 3v = \sqrt{5}\,\omega A$ with the same $\omega$ .
$A'^2 = \frac{4A^2}{9} + 5A^2 = \frac{49A^2}{9} \Rightarrow A' = \frac{7A}{3}$ .
So $n = 7$ .

Report a mistake

Q114JEE Main 2024 · 27 JAN · SHIFT 1Medium

Velocity & Acceleration in SHM · Velocity at a displacement

A particle executes simple harmonic motion with an amplitude of $4\text{ cm}$ . At the mean position, velocity of the particle is $10\text{ cm/s}$ . The distance of the particle from the mean position when its speed becomes $5\text{ cm/s}$ is $\sqrt{\alpha}\text{ cm}$ , where $\alpha =$ ________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 12

$v_{max} = A\omega \Rightarrow 10 = 4\omega \Rightarrow \omega = 2.5$ rad/s.
$v = \omega\sqrt{A^2 - x^2} \Rightarrow 5 = 2.5\sqrt{16 - x^2}$ .
$16 - x^2 = 4 \Rightarrow x^2 = 12$ , so $x = \sqrt{12}$ cm.
So $\alpha = 12$ .

Report a mistake

Q115MHT CET 2024 · 16 MAY · SHIFT 1Medium

Velocity & Acceleration in SHM · Velocities at two positions

A particle executing S.H.M. has velocities ' $V_1$ ' and ' $V_2$ ' at distances ' $x_1$ ' and ' $x_2$ ' respectively, from the mean position. Its frequency is

Show answer and solution

Answer: (c)

$V_1^2 = \omega^2(A^2 - x_1^2)$ , $V_2^2 = \omega^2(A^2 - x_2^2)$ .
Subtract: $\omega = \sqrt{\frac{V_2^2 - V_1^2}{x_1^2 - x_2^2}}$ .
$f = \frac{\omega}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{V_2^2 - V_1^2}{x_1^2 - x_2^2}}$ .
So the answer is (c).

Report a mistake

Q116MHT CET 2024 · 10 MAY · SHIFT 2Medium

Velocity & Acceleration in SHM · Velocities at two positions

A particle performs linear S.H.M. When the displacement of the particle from mean position is 3 cm and 4 cm, corresponding velocities are $8\text{ cm/s}$ and $6\text{ cm/s}$ respectively. Its periodic time is

Show answer and solution

Answer: (b)

$64 = \omega^2(A^2 - 9)$ and $36 = \omega^2(A^2 - 16)$ .
Subtract: $28 = 7\omega^2 \Rightarrow \omega = 2$ rad/s.
$T = \frac{2\pi}{\omega} = \pi$ s.
So the answer is (b) $\pi$ s.

Report a mistake

Q117MHT CET 2024 · 9 MAY · SHIFT 2Medium

Velocity & Acceleration in SHM · Velocities at two positions

A particle is executing a linear simple harmonic motion. Let ' $V_1$ ' and ' $V_2$ ' are its speed at distance ' $x_1$ ' and ' $x_2$ ' from the equilibrium position. The amplitude of oscillation is

Show answer and solution

Answer: (c)

$\frac{V_1^2}{\omega^2} = A^2 - x_1^2$ and $\frac{V_2^2}{\omega^2} = A^2 - x_2^2$ .
Divide: $V_1^2(A^2 - x_2^2) = V_2^2(A^2 - x_1^2)$ .
$A^2(V_1^2 - V_2^2) = V_1^2x_2^2 - V_2^2x_1^2$ .
$A = \sqrt{\frac{V_1^2x_2^2 - V_2^2x_1^2}{V_1^2 - V_2^2}}$ , so the answer is (c).

Report a mistake

Q118JEE Main 2024 · 9 APR · SHIFT 1Medium

Velocity & Acceleration in SHM · Velocities at two positions

The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes of $4\text{ m}, 2\text{ m s}^{-1}$ and $16\text{ m s}^{-2}$ at a certain instant. The amplitude of the motion is $\sqrt{x}\text{ m}$ where $x$ is _________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 17

$|a| = \omega^2 x \Rightarrow 16 = 4\omega^2 \Rightarrow \omega = 2$ rad/s.
$A^2 = x^2 + \frac{v^2}{\omega^2} = 16 + \frac{4}{4} = 17$ .
$A = \sqrt{17}$ m, so $x = 17$ .

Report a mistake

Q119MHT CET 2024 · 15 MAY · SHIFT 1Medium

Velocity & Acceleration in SHM · Vertical SHM and loss of contact

A horizontal platform with a small object placed on it executes a linear S.H.M. in the vertical direction. The amplitude of oscillation is 40 cm. What should be the least period of these oscillations, so that the object is not detached from the platform? [Take $g = 10\text{ m/s}^2$ ]

Show answer and solution

Answer: (c)

The object stays on the platform if the largest downward acceleration $\omega^2 A \le g$ .
Limit: $\omega^2 = \dfrac{g}{A} = \dfrac{10}{0.4} = 25 \Rightarrow \omega = 5\text{ rad/s}$
$T_{min} = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{5} = 0.4\pi\text{ s}$
So the answer is (c) $0.4\pi\text{ s}$ .

Report a mistake

Q120MHT CET 2024 · 16 MAY · SHIFT 2Difficult

Velocity & Acceleration in SHM · Velocities at two positions

A particle performs linear S.H.M. At a particular instant, velocity of the particle is ' $u$ ' and acceleration is ' $\alpha$ ' while at another instant, velocity is ' $v$ ' and acceleration is ' $\beta$ ' $(0 < \alpha < \beta)$ . The distance between the two positions is

Show answer and solution

Answer: (a)

$u^2 = \omega^2(A^2 - x_1^2)$ , $v^2 = \omega^2(A^2 - x_2^2)$ , $\alpha = \omega^2 x_1$ , $\beta = \omega^2 x_2$ .
$u^2 - v^2 = \omega^2(x_2 - x_1)(x_2 + x_1)$ and $\alpha + \beta = \omega^2(x_1 + x_2)$
$x_2 - x_1 = \dfrac{u^2 - v^2}{\alpha + \beta}$
So the answer is (a).

Report a mistake

Q121MHT CET 2024 · 11 MAY · SHIFT 1Difficult

Velocity & Acceleration in SHM · Velocities at two positions

A particle performs linear S.H.M. At a particular instant, velocity of the particle is ' $u$ ' and acceleration is ' $a_1$ ' while at another instant velocity is ' $v$ ' and acceleration is ' $a_2$ ' $(0 < a_1 < a_2)$ . The distance between the two positions is

Show answer and solution

Answer: (d)

$u^2 = \omega^2(A^2 - x_1^2)$ , $v^2 = \omega^2(A^2 - x_2^2)$ , $a_1 = \omega^2 x_1$ , $a_2 = \omega^2 x_2$ .
$u^2 - v^2 = \omega^2(x_2 - x_1)(x_2 + x_1)$ and $a_1 + a_2 = \omega^2(x_1 + x_2)$
$x_2 - x_1 = \dfrac{u^2 - v^2}{a_1 + a_2}$
So the answer is (d).

Report a mistake

Q122MHT CET 2023 · 14 MAY · SHIFT 1Easy

Velocity & Acceleration in SHM · Velocity at a displacement

The displacement of a particle executing S.H.M. is $x = a\sin(\omega t - \phi)$ . Velocity of the particle at time $t = \frac{\phi}{\omega}$ is $(\cos 0^\circ = 1)$

Show answer and solution

Answer: (b)

$v = \frac{dx}{dt} = a\omega\cos(\omega t - \phi)$
At $t = \frac{\phi}{\omega}$ : $\omega t - \phi = 0$ , so $v = a\omega\cos 0^\circ = a\omega$
So the answer is (b) $a\omega$ .

Report a mistake

Q123MHT CET 2023 · 12 MAY · SHIFT 1Easy

Velocity & Acceleration in SHM · Velocity at a displacement

A simple pendulum performs simple harmonic motion about $x = 0$ with an amplitude ' $a$ ' and time period ' $T$ '. The speed of the pendulum at $x = \frac{a}{2}$ is

Show answer and solution

Answer: (c)

$v = \omega\sqrt{a^2 - x^2}$ with $\omega = \frac{2\pi}{T}$
$\sqrt{a^2 - \frac{a^2}{4}} = \frac{\sqrt{3}a}{2}$
$v = \frac{2\pi}{T}\cdot\frac{\sqrt{3}a}{2} = \frac{\pi a\sqrt{3}}{T}$
So the answer is (c) $\frac{\pi a\sqrt{3}}{T}$ .

Report a mistake

Q124MHT CET 2023 · 9 MAY · SHIFT 1Easy

Velocity & Acceleration in SHM · Maximum velocity and acceleration

The maximum velocity of a particle performing S.H.M. is ' $V$ '. If the periodic time is made $\left(\frac{1}{3}\right)$ rd and the amplitude is doubled, then the new maximum velocity of the particle will be

Show answer and solution

Answer: (d)

$v_{max} = \frac{2\pi A}{T}$ , so $v_{max} \propto \frac{A}{T}$
$A' = 2A$ , $T' = \frac{T}{3}$ : $v' = 2\times 3\,V = 6V$
So the answer is (d) $6V$ .

Report a mistake

Q125JEE Main 2023 · 30 JAN · SHIFT 2Medium

Velocity & Acceleration in SHM · Velocity at a displacement

The velocity of a particle executing SHM varies with displacement $(x)$ as $4v^2 = 50 - x^2$ . The time period of oscillations is $\frac{x}{7}\text{ s}$ . The value of $x$ is ___________. $($ Take $\pi = \frac{22}{7})$

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 88

$v^2 = \frac{1}{4}(50 - x^2)$ compared with $v^2 = \omega^2(A^2 - x^2)$ gives $\omega = \frac{1}{2}$ rad/s.
$T = \frac{2\pi}{\omega} = 4\pi = \frac{88}{7}$ s.
$\frac{x}{7} = \frac{88}{7} \Rightarrow x = 88$ .
Answer: 88.

Report a mistake

Q126JEE Main 2023 · 10 APR · SHIFT 2Medium

Velocity & Acceleration in SHM · Maximum velocity and acceleration

A rectangular block of mass $5\text{ kg}$ attached to a horizontal spiral spring executes simple harmonic motion of amplitude $1\text{ m}$ and time period $3.14\text{ s}$ . The maximum force exerted by spring on block is _________ N

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 20

$\omega = \frac{2\pi}{T} = \frac{2\times 3.14}{3.14} = 2$ rad/s.
Maximum force at the extreme: $F_{max} = m\omega^2 A$ .
$F_{max} = 5\times 4\times 1 = 20$ N.
Answer: 20 N.

Report a mistake

Q127JEE Main 2023 · 29 JAN · SHIFT 2Medium

Velocity & Acceleration in SHM · Maximum velocity and acceleration

A particle of mass 250 g executes a simple harmonic motion under a periodic force $F = (-25x)\text{ N}$ . The particle attains a maximum speed of 4 m/s during its oscillation. The amplitude of the motion is ___________ cm.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 40

$F = -25x \Rightarrow k = 25$ N/m; $m = 0.25$ kg.
$\omega = \sqrt{\frac{k}{m}} = \sqrt{100} = 10$ rad/s.
$A = \frac{v_{max}}{\omega} = \frac{4}{10} = 0.4$ m.
Answer: 40 cm.

Report a mistake

Q128NEET 2023Medium

Velocity & Acceleration in SHM · Graphs and phase relations

The $x\text{-}t$ graph of a particle performing simple harmonic motion is shown in the figure. The acceleration of the particle at $t = 2\text{ s}$ is:

Show answer and solution

Answer: (c)

From the graph: $A = 1$ m, $T = 8$ s, so $\omega = \frac{2\pi}{8} = \frac{\pi}{4}$ rad/s.
At $t = 2$ s the particle is at the extreme, $x = +1$ m.
$a = -\omega^2 x = -\frac{\pi^2}{16}\times 1$ .
So the answer is (c) $-\frac{\pi^2}{16}$ m s $^{-2}$ .

Report a mistake

Q129MHT CET 2023 · 14 MAY · SHIFT 2Medium

Velocity & Acceleration in SHM · Vertical SHM and loss of contact

A block of mass ' $M$ ' rests on a piston executing S.H.M. of period one second. The amplitude of oscillations, so that the mass is separated from the piston, is (acceleration due to gravity, $g = 10\text{ m s}^{-2},\ \pi^2 = 10$ )

Show answer and solution

Answer: (a)

The block separates at the top when $\omega^2 A = g$ .
$\omega = \dfrac{2\pi}{T} = 2\pi\text{ rad/s} \Rightarrow \omega^2 = 4\pi^2 = 40$
$A = \dfrac{g}{\omega^2} = \dfrac{10}{40} = 0.25\text{ m}$
So the answer is (a) 0.25 m.

Report a mistake

Q130JEE Main 2022 · 27 JUN · SHIFT 2Easy

Velocity & Acceleration in SHM · Velocity at a displacement

The equation of a particle executing simple harmonic motion is given by $x = \sin\pi\left(t + \frac{1}{3}\right)\text{ m}$ . At t = 1s, the speed of particle will be

(Given: $\pi$ = 3.14)

Show answer and solution

Answer: (b)

$v = \frac{dx}{dt} = \pi\cos\pi\left(t + \frac{1}{3}\right)$ m/s
At $t = 1$ s: $\cos\frac{4\pi}{3} = -\frac{1}{2}$ , so $|v| = \frac{\pi}{2} = 1.57$ m/s
So the answer is (b) $157\ \text{cm s}^{-1}$ .

Report a mistake

Q131JEE Main 2022 · 30 JUL · SHIFT 2Easy

Velocity & Acceleration in SHM · Maximum velocity and acceleration

A particle executing SHM has its velocity $20\text{ cm s}^{-1}$ at mean position and acceleration of $25\text{ cm s}^{-2}$ at one of its extreme position. The amplitude of the particle will be ____ cm.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 16

$v_{max} = A\omega = 20$ cm/s and $a_{max} = A\omega^2 = 25\ \text{cm/s}^2$
$\omega = \frac{25}{20} = 1.25$ rad/s
$A = \frac{20}{1.25} = 16$ cm

Report a mistake

Q132JEE Main 2022 · 26 JUL · SHIFT 1Easy

Velocity & Acceleration in SHM · Graphs and phase relations

When a particle executes Simple Harmonic Motion, the nature of graph of velocity as a function of displacement will be:

Show answer and solution

Answer: (b)

$v = \omega\sqrt{A^2 - x^2} \Rightarrow v^2 = \omega^2(A^2 - x^2)$ .
$\frac{x^2}{A^2} + \frac{v^2}{\omega^2A^2} = 1$ , an ellipse with semi-axes $A$ and $\omega A$ .
So the answer is (b) Elliptical.

Report a mistake

Q133JEE Main 2022 · 29 JUN · SHIFT 1Medium

Velocity & Acceleration in SHM · Velocity at a displacement

A body is performing simple harmonic with an amplitude of 10 cm. The velocity of the body was tripled by air jet when it is at 5 cm from its mean position. The new amplitude of vibration is $\sqrt{x}$ cm. The value of x is _____________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 700

At $x = 5$ cm: $v = \omega\sqrt{10^2 - 5^2} = \omega\sqrt{75}$ .
Tripled speed $v' = 3\omega\sqrt{75}$ , same $\omega$ .
$A'^2 = x^2 + \frac{v'^2}{\omega^2} = 25 + 9\times 75 = 700$ .
$A' = \sqrt{700}$ cm, so $x = 700$ .

Report a mistake

Q134MHT CET 2021 · 23 SEP · SHIFT 1Easy

Velocity & Acceleration in SHM · Velocity at a displacement

A particle executing S.H.M starts from the mean position. Its amplitude is 'A' and time period ' $T$ '. At what displacement its speed is one-fourth of the maximum speed?

Show answer and solution

Answer: (d)

$v = \omega\sqrt{A^2 - x^2}$ , $v_{max} = \omega A$
$\sqrt{A^2 - x^2} = \frac{A}{4} \Rightarrow x^2 = \frac{15A^2}{16}$
$x = \frac{A\sqrt{15}}{4}$
So the answer is (d) $\frac{A\sqrt{15}}{4}$ .

Report a mistake

Q135JEE Main 2021 · 26 FEB · SHIFT 2Easy

Velocity & Acceleration in SHM · Velocity at a displacement

A particle executes S.H.M. with amplitude 'a', and time period 'T'. The displacement of the particle when its speed is half of maximum speed is $\frac{\sqrt{x}a}{2}$ . The value of x is __________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 3

$v = \omega\sqrt{a^2 - y^2}$ , $v_{max} = \omega a$
$\sqrt{a^2 - y^2} = \frac{a}{2} \Rightarrow y = \frac{\sqrt{3}a}{2}$
Comparing with $\frac{\sqrt{x}a}{2}$ : $x = 3$

Report a mistake

Q136MHT CET 2021 · 24 SEP · SHIFT 2Easy

Velocity & Acceleration in SHM · Maximum velocity and acceleration

The displacement of a particle performing S.H.M. is given by $x = 5\sin(3t + 3)$ , where $x$ is in $\text{cm}$ and $t$ is in second. The maximum acceleration of the particle will be

Show answer and solution

Answer: (c)

From $x = 5\sin(3t + 3)$ : $A = 5$ cm, $\omega = 3$ rad/s
$a_{max} = A\omega^2 = 5\times 9 = 45\ \text{cm s}^{-2}$
So the answer is (c) $45\ \text{cm s}^{-2}$ .

Report a mistake

Q137MHT CET 2021 · 21 SEP · SHIFT 2Easy

Velocity & Acceleration in SHM · Maximum velocity and acceleration

A particle is performing S.H.M. with maximum velocity ' $v$ '. If the amplitude is tripled and periodic time is doubled then maximum velocity will be

Show answer and solution

Answer: (a)

$v_{max} = \frac{2\pi A}{T}$ , so $v_{max} \propto \frac{A}{T}$
$A' = 3A$ , $T' = 2T$ : $v' = \frac{3}{2}v = 1.5v$
So the answer is (a) $1.5v$ .

Report a mistake

Q138MHT CET 2021 · 20 SEP · SHIFT 1Easy

Velocity & Acceleration in SHM · Maximum velocity and acceleration

A particle of mass 5 g is executing S.H.M. with an amplitude 0.3 m and time period $\frac{\pi}{5}$ s. The maximum value of the force acting on the particle is

Show answer and solution

Answer: (a)

$\omega = \frac{2\pi}{T} = \frac{2\pi}{\pi/5} = 10$ rad/s
$F_{max} = mA\omega^2 = 0.005\times 0.3\times 100$
$F_{max} = 0.15$ N
So the answer is (a) 0.15 N.

Report a mistake

Q139JEE Main 2021 · 26 FEB · SHIFT 2Easy

Velocity & Acceleration in SHM · Graphs and phase relations

A particle executes S.H.M., the graph of velocity as a function of displacement is:

Show answer and solution

Answer: (c)

$v^2 = \omega^2(A^2 - x^2) \Rightarrow \frac{x^2}{A^2} + \frac{v^2}{\omega^2A^2} = 1$ .
This is the equation of an ellipse (a circle only in the special case $\omega = 1$ ).
So the answer is (c) an ellipse.

Report a mistake

Q140JEE Main 2021 · 24 FEB · SHIFT 2Easy

Velocity & Acceleration in SHM · Graphs and phase relations

When a particle executes SHM, the nature of graphical representation of velocity as a function of displacement is:

Show answer and solution

Answer: (d)

$v^2 = \omega^2(A^2 - x^2) \Rightarrow \frac{x^2}{A^2} + \frac{v^2}{\omega^2A^2} = 1$ .
This is the equation of an ellipse.
So the answer is (d) elliptical.

Report a mistake

Q141MHT CET 2021 · 20 SEP · SHIFT 2Medium

Velocity & Acceleration in SHM · Velocity at a displacement

A particle executes S.H.M. of period $\frac{2\pi}{\sqrt{3}}$ second along a straight line $4\text{ cm}$ long. The displacement of the particle at which the velocity is numerically equal to the acceleration is

Show answer and solution

Answer: (b)

Path length 4 cm, so $A = 2$ cm; $\omega = \frac{2\pi}{T} = \sqrt{3}$ rad/s.
$\omega\sqrt{A^2 - x^2} = \omega^2 x \Rightarrow \sqrt{4 - x^2} = \sqrt{3}\,x$ .
$4 - x^2 = 3x^2 \Rightarrow x = 1$ cm.
So the answer is (b) 1 cm.

Report a mistake

Q142MHT CET 2021 · 24 SEP · SHIFT 1Medium

Velocity & Acceleration in SHM · Maximum velocity and acceleration

A particle performing linear S.H.M. of amplitude $0.1\text{ m}$ has displacement $0.02\text{ m}$ and acceleration $0.5\text{ m/s}^2$ . The maximum velocity of the particle in $\text{m/s}$ is

Show answer and solution

Answer: (b)

$a = \omega^2 x \Rightarrow \omega^2 = \frac{0.5}{0.02} = 25$ , $\omega = 5$ rad/s.
$v_{max} = A\omega = 0.1\times 5 = 0.5$ m/s.
So the answer is (b) 0.50.

Report a mistake

Q143JEE Main 2021 · 20 JUL · SHIFT 2Medium

Velocity & Acceleration in SHM · Velocities at two positions

A particle is making simple harmonic motion along the X-axis. If at a distances x1 and x2 from the mean position the velocities of the particle are v1 and v2 respectively. The time period of its oscillation is given as:

Show answer and solution

Answer: (d)

$v_1^2 = \omega^2(A^2 - x_1^2)$ , $v_2^2 = \omega^2(A^2 - x_2^2)$ .
Subtract: $v_1^2 - v_2^2 = \omega^2(x_2^2 - x_1^2)$ .
$T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{x_2^2 - x_1^2}{v_1^2 - v_2^2}}$ .
So the answer is (d).

Report a mistake

Q144NEET 2020 · PHASE 1Easy

Velocity & Acceleration in SHM · Graphs and phase relations

The phase difference between displacement and acceleration of a particle in a simple harmonic motion is:

Show answer and solution

Answer: (d)

In SHM $a = -\omega^2 x$ .
The minus sign means $a$ is always opposite to $x$ , i.e. phase difference $\pi$ .
So the answer is (d) $\pi$ rad.

Report a mistake

Q145JEE Main 2020 · 2 SEP · SHIFT 2Medium

Velocity & Acceleration in SHM · Graphs and phase relations

The displacement time graph of a particle executing S.H.M is given in figure:

(sketch is schematic and not to scale)

Which of the following statements is/are true for this motion?

(A) The force is zero at t = $\frac{3T}{4}$

(B) The acceleration is maximum at t = T

(C) The speed is maximum at t = $\frac{T}{4}$

(D) The P.E. is equal to K.E. of the oscillation at t = $\frac{T}{2}$

Show answer and solution

Answer: (b)

Graph starts at the positive extreme: $x = A\cos\omega t$ .
(A) $t = \frac{3T}{4}$ : $x = 0$ , so force $= -kx = 0$ . True.
(B) $t = T$ : $x = A$ (extreme), so acceleration is maximum. True.
(C) $t = \frac{T}{4}$ : $x = 0$ (mean position), so speed is maximum. True.
(D) $t = \frac{T}{2}$ : $x = -A$ , K.E. $= 0$ and P.E. is maximum, not equal. False.
So the answer is (b) (A), (B) and (C).

Report a mistake

Q146MHT CET 2019 · 2 MAY · SHIFT 2Medium

Velocity & Acceleration in SHM · Velocity at a displacement

A particle is performing a linear simple harmonic motion of amplitude ' $A$ '. When it is midway between its mean and extreme position, the magnitudes of its velocity and acceleration are equal. What is the periodic time of the motion?

Show answer and solution

Answer: (a)

At $x = \frac{A}{2}$ : $v = \omega\sqrt{A^2 - \frac{A^2}{4}} = \frac{\sqrt{3}}{2}\omega A$ and $a = \omega^2\frac{A}{2}$ .
$\frac{\sqrt{3}}{2}\omega A = \frac{\omega^2 A}{2} \Rightarrow \omega = \sqrt{3}$ rad/s.
$T = \frac{2\pi}{\sqrt{3}}$ s.
So the answer is (a) $\frac{2\pi}{\sqrt{3}}$ s.

Report a mistake

Q147JEE Main 2019 · 10 JAN · SHIFT 2Medium

Velocity & Acceleration in SHM · Velocity at a displacement

A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is -

Show answer and solution

Answer: (d)

$\omega\sqrt{A^2 - x^2} = \omega^2 x$ with $A = 5$ , $x = 4$ (same units cancel).
$\sqrt{25 - 16} = 4\omega \Rightarrow \omega = \frac{3}{4}$ rad/s.
$T = \frac{2\pi}{\omega} = \frac{8\pi}{3}$ s.
So the answer is (d) $\frac{8\pi}{3}$ .

Report a mistake

Q148MHT CET 2019 · 2 MAY · SHIFT 2Medium

Velocity & Acceleration in SHM · Graphs and phase relations

If ' $x$ ', $v$ ' and ' $a$ ' denote the displacement, velocity and acceleration of a particle respectively executing SHM of periodic time $T$ then which one of the following does not change with time?

Show answer and solution

Answer: (a)

In SHM $a = -\omega^2 x = -\frac{4\pi^2}{T^2}x$ .
$\frac{aT}{x} = -\frac{4\pi^2}{T}$ , a constant.
The other expressions contain $a$ , $v$ or $t$ in combinations that vary with time.
So the answer is (a) $\frac{aT}{x}$ .

Report a mistake

Q149NEET 2018Easy

Velocity & Acceleration in SHM · Maximum velocity and acceleration

A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 m s–2 at a distance of 5 m from the mean position. The time period of oscillation is

Show answer and solution

Answer: (b)

$|a| = \omega^2 x \Rightarrow \omega^2 = \frac{20}{5} = 4$ , $\omega = 2$ rad/s
$T = \frac{2\pi}{\omega} = \frac{2\pi}{2} = \pi$ s
So the answer is (b) $\pi$ s.

Report a mistake

Q150NEET 2017Medium

Velocity & Acceleration in SHM · Velocity at a displacement

A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in second is

Show answer and solution

Answer: (b)

$\omega\sqrt{A^2 - x^2} = \omega^2 x$ with $A = 3$ , $x = 2$ .
$\sqrt{9 - 4} = 2\omega \Rightarrow \omega = \frac{\sqrt{5}}{2}$ rad/s.
$T = \frac{2\pi}{\omega} = \frac{4\pi}{\sqrt{5}}$ s.
So the answer is (b) $\frac{4\pi}{\sqrt{5}}$ .

Report a mistake

Q151JEE Main 2017 · 8 APR · SHIFT 1Medium

Velocity & Acceleration in SHM · Maximum velocity and acceleration

The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is 10 s−1. At, t = 0 the displacement is 5 m. What is the maximum acceleration? The initial phase is $\frac{\pi}{4}$ .

Show answer and solution

Answer: (b)

$\frac{a_{max}}{v_{max}} = \frac{\omega^2 A}{\omega A} = \omega = 10$ rad/s.
$x = A\sin(\omega t + \frac{\pi}{4})$ ; at $t = 0$ : $5 = \frac{A}{\sqrt{2}} \Rightarrow A = 5\sqrt{2}$ m.
$a_{max} = \omega^2 A = 100\times 5\sqrt{2} = 500\sqrt{2}$ m/s $^2$ .
So the answer is (b) $500\sqrt{2}$ m/s $^2$ .

Report a mistake

Q152JEE Main 2016Medium

Velocity & Acceleration in SHM · Velocity at a displacement

A particle performs simple harmonic motion with amplitude $A.$ Its speed is trebled at the instant that it is at a distance $\frac{2A}{3}$ from equilibrium position. The new amplitude of the motion is:

Show answer and solution

Answer: (b)

At $x = \frac{2A}{3}$ : $v = \omega\sqrt{A^2 - \frac{4A^2}{9}} = \frac{\sqrt{5}}{3}\omega A$ .
New speed $v' = 3v = \sqrt{5}\,\omega A$ ; $\omega$ is unchanged (same mass and spring).
$A'^2 = x^2 + \frac{v'^2}{\omega^2} = \frac{4A^2}{9} + 5A^2 = \frac{49A^2}{9}$ .
$A' = \frac{7A}{3}$ , so the answer is (b).

Report a mistake

Q153JEE Main 2016 · 10 APR · SHIFT 1Medium

Velocity & Acceleration in SHM · Vertical SHM and loss of contact

In an engine the piston undergoes vertical simple harmonic motion with amplitude 7 cm. A washer rests on top of the piston and moves with it. The motor speed is slowly increased. The frequency of the piston at which the washer no longer stays in contact with the piston, is close to:

Show answer and solution

Answer: (d)

The washer loses contact when $\omega^2 A = g$ .
$\omega = \sqrt{\dfrac{g}{A}} = \sqrt{\dfrac{10}{0.07}} \approx 11.95\text{ rad/s}$
$f = \dfrac{\omega}{2\pi} = \dfrac{11.95}{6.28} \approx 1.9\text{ Hz}$
So the answer is (d) 1.9 Hz.

Report a mistake

Q154NEET 2015 · AIPMTEasy

Velocity & Acceleration in SHM · Maximum velocity and acceleration

A particle is executing a simple harmonic motion. Its maximum acceleration is $\alpha$ and maximum velocity is $\beta$ . Then, its time period of vibration will be

Show answer and solution

Answer: (b)

$\alpha = A\omega^2$ , $\beta = A\omega$
$\omega = \frac{\alpha}{\beta}$
$T = \frac{2\pi}{\omega} = \frac{2\pi\beta}{\alpha}$
So the answer is (b) $\frac{2\pi\beta}{\alpha}$ .

Report a mistake

Q155NEET 2015 · AIPMT · CANCELLED PAPERMedium

Velocity & Acceleration in SHM · Velocities at two positions

A particle is executing SHM along a straight line. Its velocities at distances x1 and x2 from the mean position are V1 and V2 respectively. Its time period is

Show answer and solution

Answer: (d)

$V_1^2 = \omega^2(A^2 - x_1^2)$ , $V_2^2 = \omega^2(A^2 - x_2^2)$ .
Subtract: $V_1^2 - V_2^2 = \omega^2(x_2^2 - x_1^2)$ .
$T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{x_2^2 - x_1^2}{V_1^2 - V_2^2}}$ .
So the answer is (d).

Report a mistake

Q156NEET 2014 · AIPMTEasy

Velocity & Acceleration in SHM · Graphs and phase relations

The oscillation of a body on a smooth horizontal surface is represented by the equation, X = A cos $(\omega t)$

where X = displacement at time t

$\omega$ = frequency of oscillation

Which one of the following graphs shows correctly the variation of a with t?

Here a = acceleration at time t

T = time period

Show answer and solution

Answer: (c)

$X = A\cos\omega t \Rightarrow a = \frac{d^2X}{dt^2} = -\omega^2 A\cos\omega t$ .
At $t = 0$ : $a = -\omega^2 A$ (negative maximum); at $t = T/2$ : $a = +\omega^2 A$ ; at $t = T$ : $a = -\omega^2 A$ again.
This is an inverted cosine curve starting from the negative maximum.
So the answer is (c).

Report a mistake

Q157NEET 2013Easy

Velocity & Acceleration in SHM · Graphs and phase relations

A particle of mass m oscillates along x-axis according to equation x = asin $\omega$ t. The nature of the graph between momentum and displacement of the particle is

Show answer and solution

Answer: (c)

$x = a\sin\omega t$ , $p = mv = ma\omega\cos\omega t$ .
$\frac{x^2}{a^2} + \frac{p^2}{m^2a^2\omega^2} = \sin^2\omega t + \cos^2\omega t = 1$ .
This is the equation of an ellipse.
So the answer is (c) Ellipse.

Report a mistake

Q158JEE Main 2013 · 22 APR · SHIFT 1Medium

Velocity & Acceleration in SHM · Vertical SHM and loss of contact

A mass $m = 1.0$ kg is put on a flat pan attached to a vertical spring fixed on the ground. The mass of the spring and the pan is negligible. When pressed slightly and released, the mass executes simple harmonic motion. The spring constant is 500 N/m. What is the amplitude A of the motion, so that the mass m tends to get detached from the pan? (Take $g = 10$ m/s $2$ ). The spring is stiff enough so that it does not get distorted during the motion.

Show answer and solution

Answer: (a)

The mass tends to leave the pan when the downward acceleration at the top reaches $g$ : $\omega^2 A \ge g$ .
$\omega^2 = \dfrac{k}{m} = \dfrac{500}{1} = 500\text{ s}^{-2}$
$A \ge \dfrac{g}{\omega^2} = \dfrac{10}{500} = 0.02\text{ m} = 2\text{ cm}$
So the answer is (a) $A \ge 2.0$ cm.

Report a mistake

Q159NEET 2009 · AIPMTEasy

Velocity & Acceleration in SHM · Velocity at a displacement

A simple pendulum performs simple harmonic motion about x = 0 with an amplitude a and time period T. The speed of the pendulum at x = a/2 will be

Show answer and solution

Answer: (c)

$v = \omega\sqrt{a^2 - x^2}$ with $\omega = \frac{2\pi}{T}$
$\sqrt{a^2 - \frac{a^2}{4}} = \frac{\sqrt{3}a}{2}$
$v = \frac{2\pi}{T}\cdot\frac{\sqrt{3}a}{2} = \frac{\pi a\sqrt{3}}{T}$
So the answer is (c) $\frac{\pi a\sqrt{3}}{T}$ .

Report a mistake

Q160JEE Main 2009 · AIEEEMedium

Velocity & Acceleration in SHM · Graphs and phase relations

If $x, v$ and $a$ denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period $T,$ then, which of the following does not change with time?

Show answer and solution

Answer: (a)

In SHM $a = -\omega^2 x$ with $\omega = \dfrac{2\pi}{T}$ .
$\dfrac{aT}{x} = -\omega^2 T = -\dfrac{4\pi^2}{T}$ , a constant.
$aT + 2\pi v$ and $aT/v$ change with time because $a$ and $v$ change differently.
So the answer is (a) $aT/x$ .

Report a mistake

Q161NEET 2008 · AIPMTEasy

Velocity & Acceleration in SHM · Maximum velocity and acceleration

$-$ 1

Two simple harmonic motions of angular frequency 100 and 1000 rad s have the same displacement amplitude. The ratio of their maximum acceleration is

Show answer and solution

Answer: (d)

$a_{max} = A\omega^2$ ; same $A$ , so $a_{max} \propto \omega^2$
$\frac{a_1}{a_2} = \left(\frac{100}{1000}\right)^2 = \frac{1}{100}$
So the answer is (d) $1 : 10^2$ .

Report a mistake

Q162NEET 2007 · AIPMTEasy

Velocity & Acceleration in SHM · Graphs and phase relations

The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is

Show answer and solution

Answer: (d)

$x = A\sin\omega t$ , $v = A\omega\cos\omega t = A\omega\sin(\omega t + \frac{\pi}{2})$ .
$a = -A\omega^2\sin\omega t = A\omega^2\sin(\omega t + \pi)$ .
Phase difference between $v$ and $a$ $= \pi - \frac{\pi}{2} = \frac{\pi}{2} = 0.5\pi$ .
So the answer is (d) $0.5\pi$ .

Report a mistake

Q163NEET 2007 · AIPMTMedium

Velocity & Acceleration in SHM · Vertical SHM and loss of contact

A mass of 2.0 kg is put on a flat pan attached to a vertical spring fixed on the ground as shown in the figure. The mass of the spring and the pan is negligible.

When pressed slightly and released the mass executes a simple harmonic motion. The spring constant is 200 N/m. What should be the minimum amplitude of the motion so that the mass gets detached from the pan (take g = 10 m/s2).

Show answer and solution

Answer: (a)

The mass leaves the pan at the top when $\omega^2 A = g$ .
$\omega^2 = \dfrac{k}{m} = \dfrac{200}{2} = 100\text{ s}^{-2}$
$A = \dfrac{g}{\omega^2} = \dfrac{10}{100} = 0.1\text{ m} = 10\text{ cm}$
So the answer is (a) 10.0 cm.

Report a mistake

Q164JEE Main 2006 · AIEEEEasy

Velocity & Acceleration in SHM · Maximum velocity and acceleration

The maximum velocity of a particle, executing simple harmonic motion with an amplitude $7\text{ mm},$ is $4.4\text{ m/s}.$ The period of oscillation is

Show answer and solution

Answer: (a)

$v_{max} = \frac{2\pi A}{T} \Rightarrow T = \frac{2\pi A}{v_{max}}$
$T = \frac{2\times 3.14\times 7\times10^{-3}}{4.4} \approx \frac{0.044}{4.4}$
$T = 0.01$ s
So the answer is (a) 0.01 s.

Report a mistake

Q165JEE Main 2006 · AIEEEMedium

Velocity & Acceleration in SHM · Vertical SHM and loss of contact

A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency $\omega.$ The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time

Show answer and solution

Answer: (b)

At the highest point the platform's acceleration is $\omega^2 A$ downward.
The normal force $N = m(g - \omega^2 A)$ becomes zero first when $\omega^2 A = g$ .
So contact is first lost for amplitude $A = \dfrac{g}{\omega^2}$ .
So the answer is (b) $\dfrac{g}{\omega^2}$ .

Report a mistake

Q166NEET 2005 · AIPMTEasy

Velocity & Acceleration in SHM · Maximum velocity and acceleration

A particle executing simple harmonic motion of amplitude $5\text{ cm}$ has maximum speed of $31.4\text{ cm/s}$ . The frequency of its oscillation is

Show answer and solution

Answer: (d)

$v_{max} = A\omega \Rightarrow \omega = \frac{31.4}{5} = 6.28$ rad/s
$f = \frac{\omega}{2\pi} = \frac{6.28}{6.28} = 1$ Hz
So the answer is (d) 1 Hz.

Report a mistake

Q167NEET 2004 · AIPMTEasy

Velocity & Acceleration in SHM · Graphs and phase relations

Which one of the following statements is true for the speed v and the acceleration a of a particle executing simple harmonic motion?

Show answer and solution

Answer: (d)

$v = \omega\sqrt{A^2 - x^2}$ and $a = -\omega^2 x$ .
At the mean position $x = 0$ : $v$ is maximum and $a = 0$ .
At the extreme $x = \pm A$ : $v = 0$ and $|a|$ is maximum.
So the answer is (d): when $v$ is maximum, $a$ is zero.

Report a mistake

Q168NEET 1996 · AIPMTEasy

Velocity & Acceleration in SHM · Velocity at a displacement

A particle starts with S.H.M. from the mean position as shown in the figure. Its amplitude is $A$ and its time period is $T$ . At one time, its speed is half that of the maximum speed. What is this displacement?

Show answer and solution

Answer: (d)

$v = \omega\sqrt{A^2 - x^2}$ , $v_{max} = \omega A$
$\omega\sqrt{A^2 - x^2} = \frac{\omega A}{2} \Rightarrow A^2 - x^2 = \frac{A^2}{4}$
$x = \frac{\sqrt{3}A}{2}$
So the answer is (d) $\frac{\sqrt{3}A}{2}$ .

Report a mistake

Q169NEET 1992 · AIPMTEasy

Velocity & Acceleration in SHM · Maximum velocity and acceleration

If a simple harmonic oscillator has got a displacement of $0.02\text{ m}$ and acceleration equal to $2.0\text{ m/s}^2$ at any time, the angular frequency of the oscillator is equal to

Show answer and solution

Answer: (a)

$|a| = \omega^2 x$
$\omega^2 = \frac{2.0}{0.02} = 100 \Rightarrow \omega = 10$ rad/s
So the answer is (a) 10 rad/s.

Report a mistake

Q170NEET 1991 · AIPMTMedium

Velocity & Acceleration in SHM · Velocities at two positions

A body is executing simple harmonic motion. When the displacements from the mean position is $4\text{ cm}$ and $5\text{ cm}$ , the corresponding velocities of the body is $10\text{ cm/sec}$ and $8\text{ cm/sec}$ . Then the time period of the body is

Show answer and solution

Answer: (c)

$100 = \omega^2(A^2 - 16)$ and $64 = \omega^2(A^2 - 25)$ .
Subtract: $36 = 9\omega^2 \Rightarrow \omega = 2$ rad/s.
$T = \frac{2\pi}{2} = \pi$ s.
So the answer is (c) $\pi$ sec.

Report a mistake

Q171MHT CET 2026 · 19 MAY · SHIFT 2Easy

Energy in SHM · Total energy

The total energy of a particle executing SHM is $E$ . If the amplitude is doubled and the frequency is halved, the new total energy will be

Show answer and solution

Answer: (b)

$E = \dfrac{1}{2}m\omega^2A^2 \propto f^2A^2$
$E' \propto \left(\dfrac{f}{2}\right)^2(2A)^2 = f^2A^2$
So $E' = E$ ; the answer is (b) $E$ .

Report a mistake

Q172MHT CET 2026 · 11 APR · SHIFT 1Easy

Energy in SHM · Total energy

There is a body having mass $m$ and performing SHM with amplitude ' $a$ '. There is a restoring force $F = -Kx$ , where $x$ is the displacement. The total energy of the body depends upon which of the following?

Show answer and solution

Answer: (d)

Total energy $E = \dfrac{1}{2}Ka^2$ .
It depends only on the force constant $K$ and amplitude $a$ , not on $x$ or $v$ .
So the answer is (d) $K, a$ .

Report a mistake

Q173MHT CET 2026 · 20 APR · SHIFT 2Easy

Energy in SHM · KE and PE at a displacement

For a particle executing S.H.M., the potential energy is $n$ times the kinetic energy when its displacement from mean position is $\left(\frac{2\sqrt{2}}{3}\right)A$ , where $A$ is the amplitude of S.H.M. The value of $n$ is

Show answer and solution

Answer: (d)

$\dfrac{PE}{KE} = \dfrac{x^2}{A^2 - x^2}$
$x^2 = \dfrac{8}{9}A^2 \Rightarrow A^2 - x^2 = \dfrac{1}{9}A^2$
$n = \dfrac{8/9}{1/9} = 8$
So the answer is (d) 8.

Report a mistake

Q174MHT CET 2026 · 17 APR · SHIFT 2Easy

Energy in SHM · KE and PE at a displacement

For a particle performing linear S.H.M. of amplitude 'r', the potential energy is ' $\lambda$ ' times its total energy. The displacement of particle is

Show answer and solution

Answer: (c)

$\dfrac{PE}{E} = \dfrac{\frac{1}{2}kx^2}{\frac{1}{2}kr^2} = \dfrac{x^2}{r^2} = \lambda$
$x = r\sqrt{\lambda}$
So the answer is (c).

Report a mistake

Q175MHT CET 2026 · 17 APR · SHIFT 1Easy

Energy in SHM · KE and PE at a displacement

For a particle executing S.H.M., its potential energy is 8 times its kinetic energy at a certain displacement ' $x$ ' from the mean position. If ' $A$ ' is the amplitude of S.H.M., the value of ' $x$ ' is

Show answer and solution

Answer: (c)

$PE = 8\,KE \Rightarrow x^2 = 8(A^2 - x^2)$
$9x^2 = 8A^2 \Rightarrow x = \dfrac{2\sqrt{2}}{3}A$
So the answer is (c).

Report a mistake

Q176MHT CET 2026 · 16 APR · SHIFT 2Easy

Energy in SHM · KE and PE at a displacement

The maximum potential energy of a block executing S.H.M. is E and amplitude of oscillation is 'A'. The kinetic energy of the block at amplitude A/2 is

Show answer and solution

Answer: (a)

Maximum PE = total energy $E = \dfrac{1}{2}kA^2$ .
At $x = A/2$ : $PE = \dfrac{1}{2}k\dfrac{A^2}{4} = \dfrac{E}{4}$
$KE = E - \dfrac{E}{4} = \dfrac{3E}{4}$
So the answer is (a) $\dfrac{3E}{4}$ .

Report a mistake

Q177MHT CET 2026 · 16 APR · SHIFT 1Easy

Energy in SHM · KE and PE at a displacement

For a particle P performing S.H.M., when displacement is 'x', potential energy and restoring force acting on it is denoted by 'E' and 'F' respectively. The relation between x, E and F is

Show answer and solution

Answer: (b)

$E = \dfrac{1}{2}kx^2$ and $F = -kx$ .
$\dfrac{2E}{F} = \dfrac{kx^2}{-kx} = -x$
$\dfrac{2E}{F} + x = 0$
So the answer is (b).

Report a mistake

Q178MHT CET 2026 · 13 APR · SHIFT 2Easy

Energy in SHM · KE and PE at a displacement

The amplitude of a particle executing SHM is $3$ cm. The displacement at which its kinetic energy will be $25\%$ more than the potential energy is (in cm)

Show answer and solution

Answer: (b)

$KE = 1.25\,PE \Rightarrow A^2 - x^2 = 1.25x^2$
$2.25x^2 = A^2 = 9 \Rightarrow x^2 = 4$
$x = 2\text{ cm}$
So the answer is (b) 2.

Report a mistake

Q179NEET 2026Easy

Energy in SHM · Energy graphs and energy frequency

For a simple pendulum, having time period $T$ , the variation of kinetic energy (K.E.) with time $(t)$ is represented by:

Show answer and solution

Answer: (c)

Pendulum released from extreme: $\theta = \theta_0\cos\omega t$ , so $KE = E\sin^2\omega t = \frac{E}{2}(1 - \cos 2\omega t)$ .
KE is never negative, starts at zero and is zero again at $t = T/2, T$ (the extremes).
KE repeats every $T/2$ , with maxima at $T/4$ and $3T/4$ .
So the answer is (c).

Report a mistake

Q180MHT CET 2026 · 19 MAY · SHIFT 2Medium

Energy in SHM · Total energy

A linear harmonic oscillator of force constant $2 \times 10^{6}$ N/m has amplitude $0.01$ m. Its

Show answer and solution

Answer: (a)

Maximum PE $= \frac{1}{2}kA^2$ (at the extreme).
$= \frac{1}{2}(2 \times 10^6)(0.01)^2 = 100$ J
Maximum KE is also 100 J (at the mean position).
So the answer is (a) maximum potential energy is 100 J.

Report a mistake

Q181MHT CET 2026 · 11 APR · SHIFT 2Medium

Energy in SHM · Total energy

A mass $0.4$ kg performs S.H.M. with a frequency $\frac{16}{\pi}$ Hz. At a certain displacement it has kinetic energy $2$ J and potential energy $1.2$ J. The amplitude of oscillation is

Show answer and solution

Answer: (a)

$\omega = 2\pi f = 2\pi \times \frac{16}{\pi} = 32$ rad/s
$E = KE + PE = 2 + 1.2 = 3.2$ J
$E = \frac{1}{2}m\omega^2A^2 \Rightarrow 3.2 = \frac{1}{2}(0.4)(1024)A^2 = 204.8A^2$
$A^2 = 0.015625 \Rightarrow A = 0.125$ m
So the answer is (a) 0.125 m.

Report a mistake

Q182MHT CET 2026 · 14 MAY · SHIFT 1Medium

Energy in SHM · KE and PE at a displacement

Maximum kinetic energy of a particle of mass $1$ kg in S.H.M. is $8$ J and the time period is $4$ s. Maximum potential energy during the motion is $10$ J. Then

Show answer and solution

Answer: (c)

$\omega = \frac{2\pi}{T} = \frac{\pi}{2}$ rad/s; PE has a constant offset: $PE_{min} = 10 - 8 = 2$ J, so (b) is false.
Minimum KE is 0 at the extremes, so (d) is false.
$\frac{1}{2}m\omega^2A^2 = 8 \Rightarrow A^2 = \frac{16}{\pi^2/4} = \frac{64}{\pi^2}$ , $A = \frac{8}{\pi} \approx 2.55$ m, so (a) is false.
$a_{max} = \omega^2A = \frac{\pi^2}{4} \times \frac{8}{\pi} = 2\pi$ m/s $^2$
So the answer is (c).

Report a mistake

Q183MHT CET 2026 · 18 APR · SHIFT 2Medium

Energy in SHM · KE and PE at a displacement

A particle of mass ' $m$ ' is executing S.H.M. about the origin on x-axis with frequency $\sqrt{\frac{Ka}{\pi m}}$ , where K is a constant and a is the amplitude of S.H.M. If ' $x$ ' is the displacement of a particle at time ' $t$ ', the potential energy of a particle will be

Show answer and solution

Answer: (c)

$\omega = 2\pi f \Rightarrow \omega^2 = 4\pi^2 \cdot \frac{Ka}{\pi m} = \frac{4\pi Ka}{m}$
$U = \frac{1}{2}m\omega^2x^2 = \frac{1}{2}m \cdot \frac{4\pi Ka}{m}x^2$
$U = 2\pi Kax^2$
So the answer is (c) $2\pi Kax^2$ .

Report a mistake

Q184MHT CET 2026 · 13 APR · SHIFT 2Medium

Energy in SHM · KE and PE at a displacement

A block is fastened to a horizontal spring. The block is pulled to a distance $x = 10$ cm from its equilibrium position (at $x = 0$ ) on a frictionless surface from rest. The kinetic energy of the block at $x = 5$ cm is $0.25$ J. The spring constant of the spring is nearly (in Nm $-1$ )

Show answer and solution

Answer: (c)

Released from rest at 10 cm, so $A = 0.1$ m.
$KE = \frac{1}{2}k(A^2 - x^2) = \frac{1}{2}k(0.01 - 0.0025) = 0.00375k$
$0.00375k = 0.25 \Rightarrow k = 66.7 \approx 67$ N/m
So the answer is (c) 67.

Report a mistake

Q185MHT CET 2026 · 14 MAY · SHIFT 2Medium

Energy in SHM · KE and PE at a given time

A particle starts oscillating simple harmonically from its equilibrium position with time period T. The ratio of kinetic energy to potential energy of the particle at time $t = T/12$ is

Show answer and solution

Answer: (c)

Starting from equilibrium: $x = A\sin\omega t$ .
At $t = T/12$ : $\omega t = \frac{2\pi}{T}\cdot\frac{T}{12} = \frac{\pi}{6}$ , so $x = \frac{A}{2}$ .
$\frac{KE}{PE} = \frac{A^2 - x^2}{x^2} = \frac{3A^2/4}{A^2/4} = 3$
So the answer is (c) 3 : 1.

Report a mistake

Q186MHT CET 2026 · 15 APR · SHIFT 1Medium

Energy in SHM · KE and PE at a given time

A particle starts from mean position and performs S.H.M. with period $6$ second. At what time its kinetic energy is $50\%$ of total energy? ( $\cos 45^\circ = 1/\sqrt{2}$ )

Show answer and solution

Answer: (a)

$K = \frac{1}{2}m\omega^2(A^2 - x^2)$ , so $K = 50\%$ of $E$ when $x^2 = \frac{A^2}{2}$ , i.e. $x = \frac{A}{\sqrt{2}}$ .
From mean position $x = A\sin\omega t \Rightarrow \sin\omega t = \frac{1}{\sqrt{2}} \Rightarrow \omega t = \frac{\pi}{4}$ .
$t = \frac{T}{8} = \frac{6}{8} = 0.75$ s
So the answer is (a) $0.75$ s.

Report a mistake

Q187JEE Main 2026 · 28 JAN · SHIFT 1Medium

Energy in SHM · KE and PE at a given time

The displacement of a particle, executing simple harmonic motion with time period $T$ , is expressed as $x(t) = A\sin\omega t$ , where $A$ is the amplitude. The maximum value of potential energy of this oscillator is found at $t = T/2\beta$ . The value of $\beta$ is ______.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 2

$U = \frac{1}{2}kA^2\sin^2\omega t$ is maximum when $\sin\omega t = 1$ , i.e. $\omega t = \frac{\pi}{2}$ .
First time: $t = \frac{\pi/2}{2\pi/T} = \frac{T}{4}$
$\frac{T}{2\beta} = \frac{T}{4} \Rightarrow \beta = 2$
So $\beta = 2$ .

Report a mistake

Q188JEE Main 2026 · 21 JAN · SHIFT 2Medium

Energy in SHM · Energy graphs and energy frequency

The kinetic energy of a simple harmonic oscillator is oscillating with angular frequency of 176 rad/s. The frequency of this simple harmonic oscillator is ______ Hz. [ take $\pi = \frac{22}{7}$ ]

Show answer and solution

Answer: (b)

$K = \frac{1}{2}m\omega^2A^2\cos^2\omega t = \frac{1}{4}m\omega^2A^2(1 + \cos 2\omega t)$ , so KE oscillates at $2\omega$ .
$2\omega = 176 \Rightarrow \omega = 88$ rad/s
$f = \frac{\omega}{2\pi} = \frac{88 \times 7}{2 \times 22} = 14$ Hz
So the answer is (b) 14 Hz.

Report a mistake

Q189MHT CET 2025 · 23 APR · SHIFT 2Easy

Energy in SHM · KE and PE at a displacement

A particle is executing S.H.M. of amplitude ' $A$ '. When the potential energy of the particle is half of its maximum value during the oscillation, its displacement from the equilibrium position is

Show answer and solution

Answer: (d)

$\dfrac{1}{2}kx^2 = \dfrac{1}{2}\cdot\dfrac{1}{2}kA^2$
$x^2 = \dfrac{A^2}{2} \Rightarrow x = \pm\dfrac{A}{\sqrt{2}}$
So the answer is (d).

Report a mistake

Q190MHT CET 2025 · 23 APR · SHIFT 1Easy

Energy in SHM · KE and PE at a displacement

A particle is executing linear S.H.M. starting from mean position. The ratio of the kinetic energy to the potential energy of the particle at a point of half the amplitude is

Show answer and solution

Answer: (b)

$\dfrac{KE}{PE} = \dfrac{A^2 - x^2}{x^2}$
At $x = A/2$ : $\dfrac{A^2 - A^2/4}{A^2/4} = \dfrac{3/4}{1/4} = 3$
So the answer is (b) $3 : 1$ .

Report a mistake

Q191MHT CET 2025 · 5 MAY · SHIFT 2Medium

Energy in SHM · Total energy

A point particle of mass 200 gram is executing S.H.M. of amplitude 0.2 m. When the particle passes through the mean position, its kinetic energy is $16 \times 10^{-3}\text{ J}$ . The equation of motion of this particle is (Initial phase of oscillation $= 0^\circ$ )

Show answer and solution

Answer: (d)

KE at mean position = total energy: $\frac{1}{2}m\omega^2A^2 = 16 \times 10^{-3}$ J.
$\frac{1}{2}(0.2)\omega^2(0.2)^2 = 0.004\,\omega^2 = 0.016 \Rightarrow \omega = 2$ rad/s
$Y = A\sin(\omega t) = 0.2\sin(2t)$
So the answer is (d) $Y = 0.2\sin(2t)$ .

Report a mistake

Q192MHT CET 2025 · 23 APR · SHIFT 2Medium

Energy in SHM · Total energy

An object of mass 0.2 kg executes simple harmonic oscillations along the x -axis with frequency of $\left(\frac{25}{\pi}\right)\text{Hz}$ . At the position $x = 0.04\text{ m}$ , the object has kinetic energy 1 J and potential energy 0.6 J. The amplitude of oscillation is

Show answer and solution

Answer: (c)

$\omega = 2\pi \times \frac{25}{\pi} = 50$ rad/s
$E = 1 + 0.6 = 1.6$ J
$E = \frac{1}{2}m\omega^2A^2 = \frac{1}{2}(0.2)(2500)A^2 = 250A^2$
$A^2 = \frac{1.6}{250} = 0.0064 \Rightarrow A = 0.08$ m
So the answer is (c) 0.08 m.

Report a mistake

Q193MHT CET 2025 · 22 APR · SHIFT 1Medium

Energy in SHM · KE and PE at a displacement

For a particle performing S.H.M.; the total energy is ' $n$ ' times the kinetic energy, when the displacement of a particle from mean position is $\frac{\sqrt{3}}{2}A$ , where A is the amplitude of S.H.M. The value of ' $n$ ' is

Show answer and solution

Answer: (c)

$\frac{KE}{E} = \frac{A^2 - x^2}{A^2}$
At $x = \frac{\sqrt{3}}{2}A$ : $\frac{KE}{E} = 1 - \frac{3}{4} = \frac{1}{4}$
$E = 4\,KE \Rightarrow n = 4$
So the answer is (c) 4.

Report a mistake

Q194MHT CET 2025 · 25 APR · SHIFT 2Medium

Energy in SHM · KE and PE at a given time

A particle starts oscillating simple harmonically from its mean position with time period ' $T$ '. At time $t = \frac{T}{6}$ , the ratio of the potential energy to kinetic energy of the particle is

$\left[\sin 30^\circ = \cos 60^\circ = 0.5, \cos 30^\circ = \sin 60^\circ = \sqrt{3}/2\right]$

Show answer and solution

Answer: (d)

$x = A\sin\omega t$ ; at $t = \frac{T}{6}$ , $\omega t = \frac{2\pi}{T}\cdot\frac{T}{6} = \frac{\pi}{3} = 60^\circ$ .
$U \propto \sin^2\omega t$ and $K \propto \cos^2\omega t$ , so $\frac{U}{K} = \tan^2 60^\circ = 3$ .
So the answer is (d) $3 : 1$ .

Report a mistake

Q195JEE Main 2025 · 24 JAN · SHIFT 2Medium

Energy in SHM · Energy graphs and energy frequency

A particle oscillates along the $x$ -axis according to the law, $x(t) = x_0\sin^2\left(\frac{t}{2}\right)$ where $x_0 = 1\text{ m}$ . The kinetic energy $(K)$ of the particle as a function of $x$ is correctly represented by the graph

Show answer and solution

Answer: (d)

$x = x_0\sin^2\frac{t}{2} = \frac{1}{2} - \frac{1}{2}\cos t$ (with $x_0 = 1$ m): SHM about $x = \frac{1}{2}$ with amplitude $\frac{1}{2}$ and $\omega = 1$ .
$K = \frac{1}{2}m\omega^2\left[A^2 - \left(x - \frac{1}{2}\right)^2\right]$ , an inverted parabola in $x$ .
$K = 0$ at the extremes $x = 0$ and $x = 1$ ; $K$ is maximum at the mean position $x = \frac{1}{2}$ .
So the answer is (d).

Report a mistake

Q196JEE Main 2024 · 4 APR · SHIFT 2Easy

Energy in SHM · Total energy

In simple harmonic motion, the total mechanical energy of given system is $E$ . If mass of oscillating particle $P$ is doubled then the new energy of the system for same amplitude is:

Show answer and solution

Answer: (d)

$E = \dfrac{1}{2}kA^2$ for the spring-mass system.
Doubling $m$ changes $\omega$ but not $k$ or $A$ .
So the energy stays $E$ ; the answer is (d).

Report a mistake

Q197NEET 2024Easy

Energy in SHM · KE and PE at a displacement

A particle executing simple harmonic motion with amplitude A has the same potential and kinetic energies at the displacement

Show answer and solution

Answer: (c)

$KE = PE \Rightarrow \dfrac{1}{2}k(A^2 - x^2) = \dfrac{1}{2}kx^2$
$x^2 = \dfrac{A^2}{2} \Rightarrow x = \dfrac{A}{\sqrt{2}}$
So the answer is (c).

Report a mistake

Q198JEE Main 2024 · 29 JAN · SHIFT 1Easy

Energy in SHM · KE and PE at a displacement

When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is $\frac{x}{8}$ , where $x =$ _________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 9

$\frac{E}{KE} = \frac{A^2}{A^2 - x^2}$
At $x = A/3$ : $\frac{E}{KE} = \frac{A^2}{A^2 - A^2/9} = \frac{9}{8}$
$\frac{x}{8} = \frac{9}{8} \Rightarrow x = 9$
So the answer is 9.

Report a mistake

Q199MHT CET 2024 · 16 MAY · SHIFT 2Medium

Energy in SHM · Total energy

The kinetic energy of a particle, executing simple harmonic motion is 16 J when it is in mean position. If amplitude of motion is 25 cm and the mass of the particle is 5.12 kg, the period of oscillation is

Show answer and solution

Answer: (a)

KE at mean position = total energy: $\frac{1}{2}m\omega^2A^2 = 16$ J.
$\frac{1}{2}(5.12)\omega^2(0.25)^2 = 0.16\,\omega^2 = 16 \Rightarrow \omega = 10$ rad/s
$T = \frac{2\pi}{\omega} = \frac{2\pi}{10} = \frac{\pi}{5}$ s
So the answer is (a) $\frac{\pi}{5}$ s.

Report a mistake

Q200JEE Main 2024 · 8 APR · SHIFT 2Medium

Energy in SHM · Total energy

An object of mass $0.2\text{ kg}$ executes simple harmonic motion along $x$ axis with frequency of $\left(\frac{25}{\pi}\right)\text{Hz}$ . At the position $x = 0.04\text{ m}$ the object has kinetic energy $0.5\text{ J}$ and potential energy $0.4\text{ J}$ . The amplitude of oscillation is ________ $\text{cm}$ .

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 6

$\omega = 2\pi \times \frac{25}{\pi} = 50$ rad/s
$E = 0.5 + 0.4 = 0.9$ J
$0.9 = \frac{1}{2}(0.2)(2500)A^2 = 250A^2 \Rightarrow A^2 = 0.0036$
$A = 0.06$ m $= 6$ cm
So the answer is 6 cm.

Report a mistake

Q201MHT CET 2024 · 9 MAY · SHIFT 1Medium

Energy in SHM · KE and PE at a displacement

In S.H.M. the displacement of a particle at an instant is $Y = A\cos 30^\circ$ , where $A = 40\text{ cm}$ and kinetic energy is 200 J. If force constant is $1 \times 10^{x}\text{ N/m}$ , then x will be $\left(\cos 30^\circ = \sqrt{3}/2\right)$

Show answer and solution

Answer: (a)

$KE = \frac{1}{2}k(A^2 - Y^2)$ with $Y^2 = \frac{3}{4}A^2$ , so $KE = \frac{kA^2}{8}$ .
$200 = \frac{k(0.4)^2}{8} = 0.02k \Rightarrow k = 10^4$ N/m
So $x = 4$ ; the answer is (a) 4.

Report a mistake

Q202MHT CET 2024 · 2 MAY · SHIFT 2Medium

Energy in SHM · KE and PE at a displacement

For a body performing simple harmonic motion, its potential energy is $E_x$ at displacement x and $E_y$ at displacement y from mean position. The potential energy $E_0$ at displacement $(x+y)$ is

Show answer and solution

Answer: (d)

$E = \frac{1}{2}kx^2 \Rightarrow$ displacement $\propto \sqrt{E}$ .
$\sqrt{E_0} = \sqrt{E_x} + \sqrt{E_y}$
$E_0 = E_x + E_y + 2\sqrt{E_xE_y}$
So the answer is (d).

Report a mistake

Q203MHT CET 2024 · 15 MAY · SHIFT 1Medium

Energy in SHM · KE and PE at a given time

Starting from mean position, a body oscillates simple harmonically with a period ' $T$ '. After what time will its kinetic energy be $75\%$ of the total energy? $(\sin 30^\circ = 0.5)$

Show answer and solution

Answer: (b)

$K = 75\%$ of $E$ means $U = 25\%$ of $E$ : $\frac{x^2}{A^2} = \frac{1}{4} \Rightarrow x = \frac{A}{2}$ .
$x = A\sin\omega t \Rightarrow \sin\omega t = \frac{1}{2} \Rightarrow \omega t = \frac{\pi}{6}$ .
$t = \frac{\pi/6}{2\pi/T} = \frac{T}{12}$
So the answer is (b) $\frac{T}{12}$ .

Report a mistake

Q204MHT CET 2024 · 10 MAY · SHIFT 2Medium

Energy in SHM · KE and PE at a given time

A particle starts oscillating simple harmonically from its equilibrium position with time period ' T '. What is the ratio of potential energy to kinetic energy of the particle at time $t = \frac{T}{12}$ ? $\left(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\right)$

Show answer and solution

Answer: (c)

At $t = \frac{T}{12}$ , $\omega t = \frac{2\pi}{T}\cdot\frac{T}{12} = \frac{\pi}{6}$ .
$\frac{U}{K} = \frac{\sin^2\omega t}{\cos^2\omega t} = \tan^2 30^\circ = \frac{1}{3}$
So the answer is (c) $1 : 3$ .

Report a mistake

Q205MHT CET 2023 · 11 MAY · SHIFT 1Easy

Energy in SHM · KE and PE at a displacement

For a particle executing S.H.M., its potential energy is 8 times its kinetic energy at certain displacement ' $x$ ' from the mean position. If ' $A$ ' is the amplitude of S.H.M the value of ' $x$ ' is

Show answer and solution

Answer: (c)

$PE = 8\,KE \Rightarrow x^2 = 8(A^2 - x^2)$
$9x^2 = 8A^2 \Rightarrow x = \dfrac{2\sqrt{2}A}{3}$
So the answer is (c).

Report a mistake

Q206MHT CET 2023 · 10 MAY · SHIFT 1Easy

Energy in SHM · KE and PE at a displacement

The amplitude of a particle executing S.H.M. is $3\text{ cm}$ . The displacement at which its kinetic energy will be $25\%$ more than the potential energy is

Show answer and solution

Answer: (b)

$KE = 1.25\,PE \Rightarrow A^2 - x^2 = 1.25x^2$
$2.25x^2 = 9 \Rightarrow x = 2\text{ cm}$
So the answer is (b) 2 cm.

Report a mistake

Q207MHT CET 2023 · 9 MAY · SHIFT 2Easy

Energy in SHM · KE and PE at a displacement

A particle is vibrating in S.H.M. with an amplitude of $4\text{ cm}$ . At what displacement from the equilibrium position is its energy half potential and half kinetic?

Show answer and solution

Answer: (d)

$KE = PE \Rightarrow A^2 - x^2 = x^2$
$x = \dfrac{A}{\sqrt{2}} = \dfrac{4}{\sqrt{2}} = 2\sqrt{2}\text{ cm}$
So the answer is (d) $2\sqrt{2}$ cm.

Report a mistake

Q208JEE Main 2023 · 13 APR · SHIFT 2Easy

Energy in SHM · KE and PE at a displacement

A particle executes SHM of amplitude A. The distance from the mean position when its kinetic energy becomes equal to its potential energy is:

Show answer and solution

Answer: (a)

$KE = PE \Rightarrow PE = \frac{E}{2}$ .
$\frac{1}{2}kx^2 = \frac{1}{2}\cdot\frac{1}{2}kA^2 \Rightarrow x^2 = \frac{A^2}{2}$
$x = \frac{A}{\sqrt{2}}$
So the answer is (a) $\frac{1}{\sqrt{2}}A$ .

Report a mistake

Q209JEE Main 2023 · 12 APR · SHIFT 1Easy

Energy in SHM · KE and PE at a displacement

A particle is executing simple harmonic motion (SHM). The ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude will be

Show answer and solution

Answer: (d)

$PE = \frac{1}{2}kx^2$ , $KE = \frac{1}{2}k(A^2 - x^2)$ , so $\frac{PE}{KE} = \frac{x^2}{A^2 - x^2}$ .
At $x = A/2$ : $\frac{PE}{KE} = \frac{A^2/4}{3A^2/4} = \frac{1}{3}$
So the answer is (d) 1 : 3.

Report a mistake

Q210JEE Main 2023 · 31 JAN · SHIFT 1Easy

Energy in SHM · KE and PE at a displacement

The maximum potential energy of a block executing simple harmonic motion is $25\text{ J}$ . A is amplitude of oscillation. At $A/2$ , the kinetic energy of the block is

Show answer and solution

Answer: (c)

Maximum PE = total energy $E = 25$ J.
At $x = A/2$ : $PE = \frac{E}{4} = 6.25$ J.
$KE = 25 - 6.25 = 18.75$ J
So the answer is (c) 18.75 J.

Report a mistake

Q211JEE Main 2023 · 1 FEB · SHIFT 1Easy

Energy in SHM · KE and PE at a displacement

The amplitude of a particle executing SHM is $3\text{ cm}$ . The displacement at which its kinetic energy will be $25\%$ more than the potential energy is: __________ $\text{cm}$

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 2

$KE = 1.25\,PE$ and $KE + PE = E \Rightarrow 2.25\,PE = E$ .
$\frac{x^2}{A^2} = \frac{PE}{E} = \frac{1}{2.25} \Rightarrow \frac{x}{A} = \frac{2}{3}$
$x = \frac{2}{3} \times 3 = 2$ cm
So the answer is 2 cm.

Report a mistake

Q212JEE Main 2023 · 13 APR · SHIFT 1Easy

Energy in SHM · Energy graphs and energy frequency

Which graph represents the difference between total energy and potential energy of a particle executing SHM vs its distance from mean position?

Show answer and solution

Answer: (a)

$E - PE = KE = \frac{1}{2}k(A^2 - x^2)$
This is a downward parabola: maximum at $x = 0$ , zero at $x = \pm A$ , rounded (not pointed) at the top.
So the answer is (a).

Report a mistake

Q213JEE Main 2023 · 11 APR · SHIFT 1Easy

Energy in SHM · Energy graphs and energy frequency

The variation of kinetic energy (KE) of a particle executing simple harmonic motion with the displacement $(x)$ starting from mean position to extreme position (A) is given by

Show answer and solution

Answer: (c)

$KE = \frac{1}{2}k(A^2 - x^2)$
From $x = 0$ to $x = A$ it falls from its maximum to zero along a downward parabola (flat at $x = 0$ , steep near $A$ ).
So the answer is (c).

Report a mistake

Q214MHT CET 2023 · 10 MAY · SHIFT 2Medium

Energy in SHM · KE and PE at a displacement

A body is executing a linear S.H.M. Its potential energies at the displacement ' $x$ ' and ' $y$ ' are ' $E_1$ ' and ' $E_2$ ' respectively. Its potential energy at displacement $(x+y)$ will be

Show answer and solution

Answer: (b)

$x = \sqrt{\frac{2E_1}{k}}$ , $y = \sqrt{\frac{2E_2}{k}}$
$E = \frac{1}{2}k(x + y)^2 = \left(\sqrt{E_1} + \sqrt{E_2}\right)^2$
So the answer is (b).

Report a mistake

Q215JEE Main 2023 · 13 APR · SHIFT 1Medium

Energy in SHM · KE and PE at a displacement

At a given point of time the value of displacement of a simple harmonic oscillator is given as $y = A\cos(30^\circ)$ . If amplitude is $40\text{ cm}$ and kinetic energy at that time is $200\text{ J}$ , the value of force constant is $1.0 \times 10^{x}\text{ N m}^{-1}$ . The value of $x$ is ____________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 4

$y^2 = A^2\cos^2 30^\circ = \frac{3}{4}A^2$
$KE = \frac{1}{2}k(A^2 - y^2) = \frac{kA^2}{8}$
$200 = \frac{k(0.4)^2}{8} = 0.02k \Rightarrow k = 1.0 \times 10^4$ N m $^{-1}$
So $x = 4$ .

Report a mistake

Q216MHT CET 2023 · 12 MAY · SHIFT 2Medium

Energy in SHM · KE and PE at a given time

A particle starts from mean position and performs S.H.M. with period 4 second. At what time its kinetic energy is $50\%$ of total energy?

$\left(\cos 45^\circ = \frac{1}{\sqrt{2}}\right)$

Show answer and solution

Answer: (d)

$K = 50\%$ of $E$ at $x = \frac{A}{\sqrt{2}}$ , so $\sin\omega t = \frac{1}{\sqrt{2}} \Rightarrow \omega t = \frac{\pi}{4}$ .
$t = \frac{T}{8} = \frac{4}{8} = 0.5$ s
So the answer is (d) $0.5$ s.

Report a mistake

Q217JEE Main 2023 · 30 JAN · SHIFT 1Difficult

Energy in SHM · KE and PE at a given time

The general displacement of a simple harmonic oscillator is $x = A\sin\omega t$ . Let T be its time period. The slope of its potential energy (U) - time (t) curve will be maximum when $t = \frac{T}{\beta}$ . The value of $\beta$ is ______________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 8

$U = \frac{1}{2}kA^2\sin^2\omega t$ , so $\frac{dU}{dt} = \frac{1}{2}kA^2\omega\sin 2\omega t$ .
The slope is maximum when $\sin 2\omega t = 1$ : $2\omega t = \frac{\pi}{2} \Rightarrow t = \frac{\pi}{4\omega} = \frac{T}{8}$ .
So $\beta = 8$ .

Report a mistake

Q218MHT CET 2022 · 11 AUG · SHIFT 2Medium

Energy in SHM · KE and PE at a given time

A particle starts oscillating simple harmonically from its mean position with time period ' $T$ '. At time $t = \frac{T}{12}$ , the ratio of the potential energy to kinetic energy of the particle is

$\left(\sin 30^\circ = \cos 60^\circ = 0.5, \cos 30^\circ = \sin 60^\circ = \sqrt{3}/2\right)$

Show answer and solution

Answer: (d)

At $t = \frac{T}{12}$ , $\omega t = \frac{\pi}{6} = 30^\circ$ .
$\frac{U}{K} = \tan^2 30^\circ = \frac{(0.5)^2}{(\sqrt{3}/2)^2} = \frac{1}{3}$
So the answer is (d) $1 : 3$ .

Report a mistake

Q219NEET 2022 · PHASE 2Medium

Energy in SHM · Energy graphs and energy frequency

Match List-I with List-II

Choose the correct answer from the options given below

Show answer and solution

Answer: (c)

(a) Amplitude decreasing with time: damped oscillation, pendulum with air friction $\to$ (iv).
(b) Straight line through origin with negative slope: $F = -kx$ , restoring force of a spring $\to$ (iii).
(c) Constant amplitude: undamped, pendulum with negligible air friction $\to$ (ii).
(d) KE and PE curves whose sum stays constant (dotted line): total mechanical energy conserved $\to$ (i).
So the answer is (c) (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i).

Report a mistake

Q220MHT CET 2021 · 22 SEP · SHIFT 2Easy

Energy in SHM · Total energy

A body of mass ' $m$ ' performs linear S.H.M. given by equation $x = P\sin\omega t + Q\sin\left(\omega t + \frac{\pi}{2}\right)$ . The total energy of the particle at any instant is

Show answer and solution

Answer: (c)

$x = P\sin\omega t + Q\cos\omega t$ : two perpendicular-phase SHMs of the same $\omega$ .
Resultant amplitude $A = \sqrt{P^2 + Q^2}$
$E = \dfrac{1}{2}m\omega^2A^2 = \dfrac{1}{2}m\omega^2(P^2 + Q^2)$
So the answer is (c).

Report a mistake

Q221MHT CET 2021 · 22 SEP · SHIFT 1Easy

Energy in SHM · Total energy

If the amplitude of linear S.H.M. is decreased then

Show answer and solution

Answer: (d)

Period $T = 2\pi\sqrt{m/k}$ does not depend on amplitude.
$E = \dfrac{1}{2}kA^2$ , so a smaller $A$ gives smaller energy.
So the answer is (d): period unchanged, total energy decreases.

Report a mistake

Q222JEE Main 2021 · 31 AUG · SHIFT 2Easy

Energy in SHM · Total energy

For a body executing S.H.M.:

(1) Potential energy is always equal to its K.E.

(2) Average potential and kinetic energy over any given time interval are always equal. (3) Sum of the kinetic and potential energy at any point of time is constant.

(4) Average K.E. in one time period is equal to average potential energy in one time period. Choose the most appropriate option from the options given below:

Show answer and solution

Answer: (a)

(1) False: PE = KE only at $x = A/\sqrt{2}$ .
(2) False: over an arbitrary interval the averages differ.
(3) True: KE + PE $= \dfrac{1}{2}kA^2$ is constant.
(4) True: over one period, average KE = average PE $= \dfrac{1}{4}kA^2$ .
So the answer is (a) (3) and (4).

Report a mistake

Q223MHT CET 2021 · 23 SEP · SHIFT 1Easy

Energy in SHM · KE and PE at a displacement

A particle performing S.H.M. when displacement is ' $x$ ', the potential energy and restoring force acting on it are denoted by ' $E$ ' and ' $F$ ' respectively. The relation between $x, E$ and $F$ is

Show answer and solution

Answer: (c)

$E = \dfrac{1}{2}kx^2$ and $F = -kx$ .
$\dfrac{2E}{F} = \dfrac{kx^2}{-kx} = -x$
$\dfrac{2E}{F} + x = 0$
So the answer is (c).

Report a mistake

Q224MHT CET 2021 · 23 SEP · SHIFT 1Easy

Energy in SHM · KE and PE at a displacement

A body is performing S.H.M. of amplitude 'A'. The displacement of the body from a point where kinetic energy is maximum to a point where potential energy is maximum, is

Show answer and solution

Answer: (b)

KE is maximum at the mean position $x = 0$ .
PE is maximum at the extreme positions $x = \pm A$ .
Displacement between them $= \pm A$ .
So the answer is (b) $\pm A$ .

Report a mistake

Q225JEE Main 2021 · 27 JUL · SHIFT 1Easy

Energy in SHM · KE and PE at a displacement

A particle starts executing simple harmonic motion (SHM) of amplitude 'a' and total energy E. At any instant, its kinetic energy is $\frac{3E}{4}$ then its displacement 'y' is given by:

Show answer and solution

Answer: (d)

$KE = \frac{3E}{4} \Rightarrow PE = \frac{E}{4}$ .
$\frac{y^2}{a^2} = \frac{1}{4} \Rightarrow y = \frac{a}{2}$
So the answer is (d) $y = \frac{a}{2}$ .

Report a mistake

Q226JEE Main 2021 · 25 JUL · SHIFT 2Easy

Energy in SHM · KE and PE at a displacement

In a simple harmonic oscillation, what fraction of total mechanical energy is in the form of kinetic energy, when the particle is midway between mean and extreme position.

Show answer and solution

Answer: (b)

Midway means $x = A/2$ , so $\frac{PE}{E} = \frac{x^2}{A^2} = \frac{1}{4}$ .
$\frac{KE}{E} = 1 - \frac{1}{4} = \frac{3}{4}$
So the answer is (b) $\frac{3}{4}$ .

Report a mistake

Q227JEE Main 2021 · 17 MAR · SHIFT 1Easy

Energy in SHM · KE and PE at a displacement

For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillation become equal?

Show answer and solution

Answer: (c)

$KE = PE \Rightarrow \frac{1}{2}k(A^2 - x^2) = \frac{1}{2}kx^2$
$2x^2 = A^2 \Rightarrow x = \pm\frac{A}{\sqrt{2}}$
So the answer is (c) $x = \pm\frac{A}{\sqrt{2}}$ .

Report a mistake

Q228NEET 2021Easy

Energy in SHM · Energy graphs and energy frequency

A body is executing simple harmonic motion with frequency 'n', the frequency of its potential energy is:

Show answer and solution

Answer: (c)

$PE = \frac{1}{2}kA^2\sin^2\omega t = \frac{1}{4}kA^2(1 - \cos 2\omega t)$
PE varies with angular frequency $2\omega$ , so frequency $2n$ .
So the answer is (c) $2n$ .

Report a mistake

Q229JEE Main 2021 · 27 AUG · SHIFT 1Easy

Energy in SHM · Energy graphs and energy frequency

The variation of displacement with time of a particle executing free simple harmonic motion is shown in the figure.

The potential energy U(x) versus time (t) plot of the particle is correctly shown in figure:

Show answer and solution

Answer: (d)

From the graph $x = A\sin\omega t$ , zero at O, A, B, C.
$U = \frac{1}{2}kx^2 = \frac{1}{2}kA^2\sin^2\omega t$ , never negative.
U is zero at O, A, B, C and maximum midway between them (humps of period $T/2$ ).
So the answer is (d).

Report a mistake

Q230MHT CET 2021 · 22 SEP · SHIFT 1Medium

Energy in SHM · Total energy

A mass $0.4\text{ kg}$ performs S.H.M. with a frequency $\frac{16}{\pi}\text{ Hz}$ . At a certain displacement it has kinetic energy $2\text{ J}$ and potential energy $1.2\text{ J}$ . The amplitude of oscillation is

Show answer and solution

Answer: (b)

$\omega = 2\pi \times \frac{16}{\pi} = 32$ rad/s
$E = 2 + 1.2 = 3.2$ J
$3.2 = \frac{1}{2}(0.4)(32)^2A^2 = 204.8A^2 \Rightarrow A^2 = 0.015625$
$A = 0.125$ m
So the answer is (b) 0.125 m.

Report a mistake

Q231MHT CET 2021 · 24 SEP · SHIFT 1Medium

Energy in SHM · KE and PE at a displacement

A body is executing S.H.M. under the action of force having maximum magntude $50\text{ N}$ . When its energy is half kinetic and half potential; the magnitude of the force acting on the particle is

Show answer and solution

Answer: (d)

$KE = PE \Rightarrow x = \frac{A}{\sqrt{2}}$ .
$F = kx$ , and $F_{max} = kA = 50$ N.
$F = \frac{50}{\sqrt{2}} = 25\sqrt{2}$ N
So the answer is (d) $25\sqrt{2}$ N.

Report a mistake

Q232MHT CET 2021 · 23 SEP · SHIFT 2Medium

Energy in SHM · KE and PE at a displacement

An object executes SHM along $x$ -axis with amplitude $0.06\text{ m}$ . At certain distance ' $x$ ' metre from mean position, it has kinetic energy $10\text{ J}$ and potential energy $8\text{ J}$ . the distance ' $x$ ' will be

Show answer and solution

Answer: (c)

$E = 10 + 8 = 18$ J
$\frac{x^2}{A^2} = \frac{PE}{E} = \frac{8}{18} = \frac{4}{9} \Rightarrow x = \frac{2}{3}A$
$x = \frac{2}{3} \times 0.06 = 0.04$ m
So the answer is (c) 0.04 m.

Report a mistake

Q233JEE Main 2021 · 27 JUL · SHIFT 2Medium

Energy in SHM · KE and PE at a given time

An object of mass 0.5 kg is executing simple harmonic motion. It amplitude is 5 cm and time period (T) is 0.2 s. What will be the potential energy of the object at an instant $t = \frac{T}{4}\text{ s}$ starting from mean position. Assume that the initial phase of the oscillation is zero.

Show answer and solution

Answer: (a)

At $t = \frac{T}{4}$ from the mean position the particle is at the extreme, $x = A$ , so all energy is potential.
$\omega = \frac{2\pi}{T} = \frac{2\pi}{0.2} = 10\pi$ rad/s
$U = \frac{1}{2}m\omega^2A^2 = \frac{1}{2}(0.5)(10\pi)^2(0.05)^2 \approx 0.62$ J
So the answer is (a) $0.62$ J.

Report a mistake

Q234JEE Main 2021 · 31 AUG · SHIFT 1Medium

Energy in SHM · KE and PE at a given time

A particle of mass 1 kg is hanging from a spring of force constant 100 Nm ${}^{-1}$ . The mass is pulled slightly downward and released so that it executes free simple harmonic motion with time period T. The time when the kinetic energy and potential energy of the system will become equal, is $\frac{T}{x}$ . The value of x is _____________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 8

Pulled down and released, so motion starts at the extreme: $x = A\cos\omega t$ .
$K = U$ when $x = \frac{A}{\sqrt{2}} \Rightarrow \cos\omega t = \frac{1}{\sqrt{2}} \Rightarrow \omega t = \frac{\pi}{4}$ .
$t = \frac{\pi/4}{2\pi/T} = \frac{T}{8}$
So $x = 8$ .

Report a mistake

Q235MHT CET 2020 · 16 OCT · SHIFT 1Easy

Energy in SHM · KE and PE at a displacement

For a particle performing SHM when displacement is $x$ , the potential energy and restoring force acting on it is denoted by $E$ and $F$ , respectively. The relation between $x, E$ and $F$ is

Show answer and solution

Answer: (b)

$E = \dfrac{1}{2}kx^2$ and restoring force $F = -kx$ .
$\dfrac{2E}{F} = \dfrac{kx^2}{-kx} = -x \Rightarrow \dfrac{2E}{F} + x = 0$ (option c).
If only the magnitude $F = kx$ is used, $\dfrac{2E}{F} - x = 0$ (option b).
The key takes magnitudes: (b).

Report a mistake

Q236MHT CET 2019 · 3 MAY · SHIFT 1Easy

Energy in SHM · Total energy

The total energy of a simple harmonic oscillaior is proportional to

Show answer and solution

Answer: (a)

$E = \dfrac{1}{2}m\omega^2A^2$
For a given oscillator, $E \propto A^2$ .
So the answer is (a) square of the amplitude.

Report a mistake

Q237MHT CET 2019 · 2 MAY · SHIFT 1Easy

Energy in SHM · Total energy

The quantity which does not vary periodically for a particle performing SHM is

Show answer and solution

Answer: (b)

Displacement, velocity and acceleration all vary sinusoidally with time.
Total energy $KE + PE = \dfrac{1}{2}kA^2$ stays constant.
So the answer is (b) total energy.

Report a mistake

Q238JEE Main 2019 · 9 JAN · SHIFT 2Easy

Energy in SHM · KE and PE at a displacement

A particle is executing simple harmonic motion (SHM) of amplitude A, along the x-axis, about x = 0. When its potential Energy (PE) equals kinetic energy (KE), the position of the particle will be:

Show answer and solution

Answer: (c)

$PE = KE \Rightarrow kx^2 = \frac{1}{2}kA^2$
$x = \frac{A}{\sqrt{2}}$ (on either side of $x = 0$ )
So the answer is (c) $\frac{A}{\sqrt{2}}$ .

Report a mistake

Q239JEE Main 2019 · 11 JAN · SHIFT 1Medium

Energy in SHM · KE and PE at a given time

A particle undergoing simple harmonic motion has time dependent displacement given by x(t) = Asin $\frac{\pi t}{90}$ . The ratio of kinetic to potential energy of this particle at t = 210 s will be:

Show answer and solution

Answer: (b)

Phase at $t = 210$ s: $\frac{\pi \times 210}{90} = \frac{7\pi}{3}$ , which is equivalent to $\frac{\pi}{3}$ .
$\frac{K}{U} = \frac{\cos^2\theta}{\sin^2\theta} = \cot^2 60^\circ = \frac{1}{3}$
So $K : U = 1 : 3$ , which matches option (b) read as $\frac{1}{3}$ .

Report a mistake

Q240JEE Main 2017Easy

Energy in SHM · Energy graphs and energy frequency

A particle is executing simple harmonic motion with a time period T. At time t = 0, it is at its position of equilibrium. The kinetic energy – time graph of the particle will look like:

Show answer and solution

Answer: (a)

At $t = 0$ the particle is at equilibrium: $x = A\sin\omega t$ , $KE = E\cos^2\omega t$ .
KE is maximum at $t = 0$ , zero at $t = T/4$ , maximum again at $T/2$ .
KE repeats every $T/2$ and is never negative.
So the answer is (a).

Report a mistake

Q241JEE Main 2015Easy

Energy in SHM · Energy graphs and energy frequency

For a simple pendulum, a graph is plotted between its kinetic energy $(KE)$ and potential energy $(PE)$ against its displacement $d.$ Which one of the following represents these correctly? $\left(\textit{graphs are schematic and not drawn to scale}\right)$

Show answer and solution

Answer: (d)

$PE = \frac{1}{2}kd^2$ : upward parabola, zero at $d = 0$ .
$KE = \frac{1}{2}k(A^2 - d^2)$ : downward parabola, maximum at $d = 0$ , zero at $d = \pm A$ .
Both are non-negative and cross where $KE = PE$ .
So the answer is (d).

Report a mistake

Q242NEET 2007 · AIPMTEasy

Energy in SHM · Total energy

The particle executing simple harmonic motion has a kinetic energy K0cos2 $\omega$ t. The maximum values of the potential energy and the total energy are respectively

Show answer and solution

Answer: (c)

Maximum KE $= K_0$ (when $\cos^2\omega t = 1$ ), and there PE $= 0$ , so total energy $= K_0$ .
Maximum PE occurs when KE $= 0$ : $PE_{max} = K_0$ .
So the answer is (c) $K_0$ and $K_0$ .

Report a mistake

Q243JEE Main 2007 · AIEEEMedium

Energy in SHM · Total energy

A particle of mass $m$ executes simple harmonic motion with amplitude a and frequency $v.$ The average kinetic energy during its motion from the position of equilibrium to the end is

Show answer and solution

Answer: (b)

$KE = E\cos^2\omega t$ ; its time average over a quarter period is $\frac{E}{2}$ .
$E = \frac{1}{2}m\omega^2a^2 = \frac{1}{2}m(2\pi v)^2a^2 = 2\pi^2ma^2v^2$
Average KE $= \frac{E}{2} = \pi^2ma^2v^2$
So the answer is (b) $\pi^2ma^2v^2$ .

Report a mistake

Q244JEE Main 2006 · AIEEEMedium

Energy in SHM · KE and PE at a given time

Starting from the origin a body oscillates simple harmonically with a period of $2\text{ s}$ After what time will its kinetic energy be $75\%$ of the total energy?

Show answer and solution

Answer: (a)

$K = 75\%$ of $E$ at $x = \frac{A}{2}$ , so $\sin\omega t = \frac{1}{2} \Rightarrow \omega t = \frac{\pi}{6}$ .
$t = \frac{T}{12} = \frac{2}{12} = \frac{1}{6}$ s
So the answer is (a) $\frac{1}{6}$ s.

Report a mistake

Q245JEE Main 2004 · AIEEEEasy

Energy in SHM · Total energy

The total energy of particle, executing simple harmonic motion is

Show answer and solution

Answer: (a)

$E = KE + PE = \dfrac{1}{2}k(A^2 - x^2) + \dfrac{1}{2}kx^2 = \dfrac{1}{2}kA^2$
No $x$ remains in the result.
So the answer is (a) independent of $x$ .

Report a mistake

Q246NEET 2003 · AIPMTEasy

Energy in SHM · KE and PE at a displacement

The potential energy of a simple harmonic oscillator when the particle is half way to its end point is

Show answer and solution

Answer: (c)

Halfway to the end point: $x = A/2$ .
$PE = \dfrac{1}{2}k\dfrac{A^2}{4} = \dfrac{1}{4}\cdot\dfrac{1}{2}kA^2 = \dfrac{E}{4}$
So the answer is (c) $\dfrac{1}{4}E$ .

Report a mistake

Q247JEE Main 2003 · AIEEEEasy

Energy in SHM · KE and PE at a displacement

A body executes simple harmonic motion. The potential energy $(P.E),$ the kinetic energy $(K.E)$ and total energy $(T.E)$ are measured as a function of displacement $x.$ Which of the following statements is true?

Show answer and solution

Answer: (a)

$KE = \frac{1}{2}k(A^2 - x^2)$ is maximum at $x = 0$ : (a) true, (c) false.
$PE = \frac{1}{2}kx^2$ is zero (minimum) at $x = 0$ : (d) false.
Total energy $\frac{1}{2}kA^2$ is constant, never zero: (b) false.
So the answer is (a).

Report a mistake

Q248NEET 2003 · AIPMTEasy

Energy in SHM · Energy graphs and energy frequency

A particle of mass m oscillates with simple harmonic motion between points x1 and x2, the equilibrium position being O. Its potential energy is plotted. It will be as given below in the graph

Show answer and solution

Answer: (a)

$U = \frac{1}{2}kx^2$ with $x$ measured from O.
This is an upward parabola, zero at O and maximum at the ends $x_1$ and $x_2$ .
So the answer is (a).

Report a mistake

Q249JEE Main 2002 · AIEEEEasy

Energy in SHM · Total energy

In a simple harmonic oscillator, at the mean position

Show answer and solution

Answer: (c)

At the mean position $x = 0$ : speed is maximum, so KE is maximum.
$PE = \dfrac{1}{2}kx^2 = 0$ , its minimum.
So the answer is (c).

Report a mistake

Q250NEET 2002 · AIPMTEasy

Energy in SHM · KE and PE at a displacement

Displacement between maximum potential energy position and maximum kinetic energy position for a particle executing simple harmonic motion is

Show answer and solution

Answer: (c)

Maximum KE at the mean position $x = 0$ .
Maximum PE at the extremes $x = \pm a$ .
Displacement between them $= \pm a$ .
So the answer is (c) $\pm a$ .

Report a mistake

Q251NEET 2001 · AIPMTEasy

Energy in SHM · Total energy

The total energy of particle performing SHM depend on

Show answer and solution

Answer: (b)

$E = \dfrac{1}{2}ka^2$
It depends only on $k$ and $a$ (not on $m$ or $x$ separately).
So the answer is (b) $k, a$ .

Report a mistake

Q252NEET 1996 · AIPMTEasy

Energy in SHM · Total energy

A linear harmonic oscillator of force constant $2 \times 10^{6}\text{ N/m}$ and amplitude $0.01\text{ m}$ has a total mechanical energy of $160\text{ J}$ . Its

Show answer and solution

Answer: (c)

Energy of oscillation $\dfrac{1}{2}kA^2 = \dfrac{1}{2}(2\times10^6)(0.01)^2 = 100\text{ J}$
Total energy is 160 J, so PE at the mean position is $160 - 100 = 60\text{ J}$ .
So maximum KE $= 100\text{ J}$ and PE varies from 60 J to 160 J.
The key takes the 100 J value: (c).

Report a mistake

Q253NEET 1995 · AIPMTEasy

Energy in SHM · KE and PE at a displacement

In a simple harmonic motion, when the displacement is one-half the amplitude, what fraction of the total energy is kinetic?

Show answer and solution

Answer: (b)

$PE = \frac{1}{2}kx^2$ , so $\frac{PE}{E} = \frac{x^2}{A^2}$ .
At $x = A/2$ : $PE = \frac{E}{4}$ .
$KE = E - \frac{E}{4} = \frac{3}{4}E$
So the answer is (b) $3/4$ .

Report a mistake

Q254NEET 1994 · AIPMTEasy

Energy in SHM · Energy graphs and energy frequency

A loaded vertical spring executes S.H.M. with a time period of $4\text{ sec}$ . The difference between the kinetic energy and potential energy of this system varies with a period of

Show answer and solution

Answer: (a)

$KE = E\cos^2\omega t$ , $PE = E\sin^2\omega t$
$KE - PE = E\cos 2\omega t$ , which has period $\frac{T}{2}$ .
$\frac{T}{2} = \frac{4}{2} = 2$ s
So the answer is (a) 2 sec.

Report a mistake

Q255NEET 1993 · AIPMTEasy

Energy in SHM · KE and PE at a displacement

A body executes simple harmonic motion with an amplitude $A$ . At what displacement from the mean position is the potential energy of the body is one fourth of its total energy?

Show answer and solution

Answer: (b)

$\frac{PE}{E} = \frac{x^2}{A^2}$
$\frac{x^2}{A^2} = \frac{1}{4} \Rightarrow x = \frac{A}{2}$
So the answer is (b) $A/2$ .

Report a mistake

Q256MHT CET 2026 · 20 MAY · SHIFT 1Easy

Simple Pendulum · Time period, length and mass

The bob of a simple pendulum is a hollow sphere with a small hole in the bottom. It is filled with sand and set into oscillation. Its periodic time (T) is measured at regular intervals. As the sand starts coming out we find that.

Show answer and solution

Answer: (d)

$T = 2\pi\sqrt{\frac{l}{g}}$ , where $l$ is measured to the centre of mass of the bob.
As sand flows out from the bottom, the centre of mass first moves down, so $l$ and $T$ increase.
When the sphere is nearly empty the centre of mass rises back to the centre, so $T$ decreases to its original value.
So the answer is (d) T first increases and then decreases.

Report a mistake

Q257MHT CET 2026 · 19 MAY · SHIFT 2Easy

Simple Pendulum · Time period, length and mass

A simple pendulum consists of a hollow sphere filled with mercury. As the mercury begins to leak out slowly through a small hole at the bottom, the time period of the pendulum will

Show answer and solution

Answer: (b)

Effective length $l$ is up to the centre of mass of the bob, and $T \propto \sqrt{l}$ .
As mercury leaks, the centre of mass drops first ( $T$ increases), then rises back to the sphere's centre when empty ( $T$ returns to original).
So the answer is (b) first increase and then decrease to the original value.

Report a mistake

Q258MHT CET 2026 · 15 APR · SHIFT 2Easy

Simple Pendulum · Time period, length and mass

A pendulum clock is running slow, In order to correct it, we should

Show answer and solution

Answer: (c)

A slow clock means its period is too large.
$T = 2\pi\sqrt{\frac{l}{g}}$ , so to decrease $T$ the length must be reduced (mass and small amplitude do not affect $T$ ).
So the answer is (c) reduce the length of pendulum.

Report a mistake

Q259JEE Main 2026 · 23 JAN · SHIFT 1Easy

Simple Pendulum · Time period, length and mass

A simple pendulum of string length 30 cm performs 20 oscillations in 10 s. The length of the string required for the pendulum to perform 40 oscillations in the same time duration is ______ cm. [Assume that the mass of the pendulum remains same.]

Show answer and solution

Answer: (b)

$T_1 = \frac{10}{20} = 0.5$ s and $T_2 = \frac{10}{40} = 0.25$ s.
$l \propto T^2 \Rightarrow l_2 = l_1\left(\frac{T_2}{T_1}\right)^2 = 30 \times \frac{1}{4}$
$l_2 = 7.5$ cm
So the answer is (b) 7.5 cm.

Report a mistake

Q260NEET 2026Easy

Simple Pendulum · Energy, speed and tension of a pendulum

The sum of kinetic energy and potential energy of a simple pendulum bob is 0.02 joule. The speed of the simple pendulum bob at equilibrium position is approximately:

(Consider mass of the bob $= 20\text{ g}$ )

Show answer and solution

Answer: (b)

At the equilibrium position the whole energy is kinetic: $\frac{1}{2}mv^2 = E$ .
$\frac{1}{2}(0.02)v^2 = 0.02 \Rightarrow v^2 = 2$
$v = \sqrt{2} \approx 1.41\text{ m/s}$
So the answer is (b) $1.41\text{ m/s}$ .

Report a mistake

Q261JEE Main 2026 · 22 JAN · SHIFT 2Easy

Simple Pendulum · Effect of g and seconds pendulum

Using a simple pendulum experiment $g$ is determind by measuring its time period $T$ . Which of the following plots represent the correct relation between the pendulum length $L$ and time period $T$ ?

Show answer and solution

Answer: (a)

$T = 2\pi\sqrt{\frac{L}{g}} \Rightarrow \frac{1}{T^2} = \frac{g}{4\pi^2}\cdot\frac{1}{L}$
So $\frac{1}{T^2} \propto \frac{1}{L}$ : a rectangular hyperbola, falling steeply and flattening out as $L$ grows.
So the answer is (a).

Report a mistake

Q262MHT CET 2026 · 13 MAY · SHIFT 2Easy

Simple Pendulum · Pendulum in a lift or vehicle

The time period of a simple pendulum inside a stationary lift is $\sqrt{5}$ s. The time period when the lift moves upwards with an acceleration $\frac{g}{4}$ is ( $g$ = acceleration due to gravity)

Show answer and solution

Answer: (d)

Lift accelerating up: $g_{eff} = g + \frac{g}{4} = \frac{5g}{4}$ .
$T' = T\sqrt{\frac{g}{g_{eff}}} = \sqrt{5}\cdot\frac{2}{\sqrt{5}}$
$T' = 2\text{ s}$
So the answer is (d) $2$ s.

Report a mistake

Q263MHT CET 2026 · 15 MAY · SHIFT 2Medium

Simple Pendulum · Time period, length and mass

Two simple pendulums having lengths $\ell_1$ and $\ell_2$ with negligible string mass undergo angular displacement $\theta_1$ and $\theta_2$ from their mean positions respectively. If the angular acceleration of both pendulums are same, then which equation is correct?

Show answer and solution

Answer: (d)

For a simple pendulum, angular acceleration $\alpha = -\frac{g}{\ell}\theta$ .
Equal $\alpha$ : $\frac{\theta_1}{\ell_1} = \frac{\theta_2}{\ell_2}$
$\theta_1\ell_2 = \theta_2\ell_1$
So the answer is (d).

Report a mistake

Q264MHT CET 2026 · 20 APR · SHIFT 1Medium

Simple Pendulum · Time period, length and mass

A simple pendulum of length $l_1$ has periodic time $2.4$ s. Another simple pendulum of length $l_2 < l_1$ , has periodic time $1.8$ s. The periodic time of simple pendulum of length $(l_1 - l_2)$ is nearly

Show answer and solution

Answer: (c)

$T^2 \propto l$ , so for length $(l_1 - l_2)$ : $T^2 = T_1^2 - T_2^2$ .
$T^2 = 2.4^2 - 1.8^2 = 5.76 - 3.24 = 2.52$
$T = \sqrt{2.52} \approx 1.59 \approx 1.6\text{ s}$
So the answer is (c) $1.6$ s.

Report a mistake

Q265MHT CET 2026 · 17 APR · SHIFT 1Medium

Simple Pendulum · Time period, length and mass

Two simple pendulums of lengths $L_1$ and $L_2$ have periodic time $T_1$ and $T_2$ respectively $(T_1 > T_2)$ . The time period of the pendulum of length $(L_1 - L_2)$ is

$[(L_1 - L_2) > 60$ cm $]$

Show answer and solution

Answer: (b)

$T^2 = \frac{4\pi^2}{g}L$ , so $T^2 \propto L$ .
For length $(L_1 - L_2)$ : $T^2 = T_1^2 - T_2^2$
So the answer is (b) $\sqrt{T_1^2 - T_2^2}$ .

Report a mistake

Q266MHT CET 2026 · 20 APR · SHIFT 1Medium

Simple Pendulum · Two pendulums in phase

Two oscillating simple pendulums with time periods $T$ and $\frac{4T}{3}$ are in phase at a given time. They will be again in phase after an elapse of time

Show answer and solution

Answer: (b)

They are in phase again when the faster one has made exactly one more oscillation: $(n+1)T = n\cdot\frac{4T}{3}$ .
$\frac{n}{3} = 1 \Rightarrow n = 3$
Time $= 3 \times \frac{4T}{3} = 4T$
So the answer is (b) $4T$ .

Report a mistake

Q267MHT CET 2026 · 15 MAY · SHIFT 1Medium

Simple Pendulum · Energy, speed and tension of a pendulum

Two simple pendulums, A of mass $M_1$ and length $L_1$ , and B of mass $M_2$ and length $L_2.\ M_1 = M_2$ and $L_1 = 2L_2$ . If vibrational energies of both are same then which one of the following is correct?

Show answer and solution

Answer: (b)

Energy of a pendulum: $E = \frac{1}{2}m\omega^2A^2 = \frac{mgA^2}{2L}$ .
Same $E$ and $m$ : $A^2 \propto L$ .
$L_1 = 2L_2 \Rightarrow A_A = \sqrt{2}A_B$ , so B has the smaller amplitude.
So the answer is (b).

Report a mistake

Q268MHT CET 2026 · 14 MAY · SHIFT 1Medium

Simple Pendulum · Energy, speed and tension of a pendulum

The path length of oscillation of simple pendulum of $1$ m is $8$ cm. Its maximum velocity is (Take $g = \pi^2$ m/s $^{2}$ )

Show answer and solution

Answer: (b)

Path length $= 2A = 8\text{ cm} \Rightarrow A = 4\text{ cm}$ .
$\omega = \sqrt{\frac{g}{L}} = \sqrt{\frac{\pi^2}{1}} = \pi\text{ rad/s}$
$v_{max} = A\omega = 4\pi\text{ cm/s}$
So the answer is (b) $4\pi$ cm/s.

Report a mistake

Q269MHT CET 2026 · 12 MAY · SHIFT 2Medium

Simple Pendulum · Energy, speed and tension of a pendulum

If length of oscillating simple pendulum is made $\left(\frac{1}{3}\right)$ times at a place keeping amplitude same then its total energy will be

Show answer and solution

Answer: (b)

$E = \frac{1}{2}m\omega^2A^2 = \frac{mgA^2}{2L}$ , so $E \propto \frac{1}{L}$ for the same $A$ .
$L \to \frac{L}{3} \Rightarrow E \to 3E$
So the answer is (b) 3 times.

Report a mistake

Q270MHT CET 2026 · 12 MAY · SHIFT 2Medium

Simple Pendulum · Energy, speed and tension of a pendulum

A simple pendulum of length $L$ and mass of the bob $m$ is oscillating with amplitude $A$ . The maximum tension in the string is ( $g$ = acceleration due to gravity)

Show answer and solution

Answer: (c)

Tension is maximum at the mean position: $T_{max} = mg + \frac{mv_{max}^2}{L}$ .
$v_{max} = A\omega = A\sqrt{\frac{g}{L}} \Rightarrow v_{max}^2 = \frac{gA^2}{L}$
$T_{max} = mg + \frac{mgA^2}{L^2} = mg\left[1 + \left(\frac{A}{L}\right)^2\right]$
So the answer is (c).

Report a mistake

Q271NEET 2026Medium

Simple Pendulum · Effect of g and seconds pendulum

Savitha, a XI standard student, while conducting an experiment to determine the effective length of a simple pendulum $L$ , notes down the data of time taken to complete 30 oscillations as 60 s and hence calculates the length of the simple pendulum as: (Take $\pi^{2} = 9.8$ , and $g = 9.8\text{ m/s}^{2}$ )

Show answer and solution

Answer: (d)

Period $T = \frac{60}{30} = 2\text{ s}$ .
$T = 2\pi\sqrt{L/g} \Rightarrow L = \frac{gT^2}{4\pi^2} = \frac{9.8 \times 4}{4 \times 9.8}$ .
So the answer is (d) $1\text{ m}$ .

Report a mistake

Q272MHT CET 2026 · 11 APR · SHIFT 2Medium

Simple Pendulum · Pendulum in a lift or vehicle

An oscillating pendulum suspended from the roof of a lift which is at rest has time period $T_1$ . When lift moves up with acceleration 'a' its time period is $T_2$ . When lift moves down with acceleration 'a' its time period is $T_3$ . The relation between $T_1,\ T_2$ and $T_3$ is

Show answer and solution

Answer: (b)

$T \propto \frac{1}{\sqrt{g_{eff}}} \Rightarrow \frac{1}{T^2} \propto g_{eff}$ .
Lift up: $g+a$ ; lift down: $g-a$ . Adding: $\frac{1}{T_2^2} + \frac{1}{T_3^2} = \frac{2}{T_1^2}$ .
$T_1^2 = \frac{2T_2^2T_3^2}{T_2^2+T_3^2}$ .
So the answer is (b) $T_1 = \frac{\sqrt{2}\,T_2T_3}{\sqrt{T_2^2+T_3^2}}$ .

Report a mistake

Q273MHT CET 2025 · 26 APR · SHIFT 1Easy

Simple Pendulum · Time period, length and mass

The ratio of the frequencies of two simple pendulums is $4 : 3$ at the same place. The ratio of their respective lengths is

Show answer and solution

Answer: (c)

$f = \frac{1}{2\pi}\sqrt{\frac{g}{l}} \Rightarrow l \propto \frac{1}{f^2}$
$\frac{l_1}{l_2} = \left(\frac{f_2}{f_1}\right)^2 = \left(\frac{3}{4}\right)^2 = \frac{9}{16}$
So the answer is (c) $9 : 16$ .

Report a mistake

Q274MHT CET 2025 · 21 APR · SHIFT 2Easy

Simple Pendulum · Time period, length and mass

The length of the simple pendulum is made 3 times the original length. If ' T ' is its original time period, then the new time period will be

Show answer and solution

Answer: (b)

$T \propto \sqrt{l}$
$l' = 3l \Rightarrow T' = \sqrt{3}\,T$
So the answer is (b) $\sqrt{3}\,T$ .

Report a mistake

Q275MHT CET 2025 · 26 APR · SHIFT 1Easy

Simple Pendulum · Energy, speed and tension of a pendulum

At a place, the length of the oscillating simple pendulum is made $\frac{1}{4}$ times keeping amplitude same then the total energy will be

Show answer and solution

Answer: (b)

$E = \frac{1}{2}m\omega^2A^2$ with $\omega^2 = \frac{g}{L}$ , so $E = \frac{mgA^2}{2L}$ .
For the same amplitude $A$ , $E \propto \frac{1}{L}$ .
$L' = \frac{L}{4} \Rightarrow E' = 4E$
So the answer is (b) 4 times.

Report a mistake

Q276MHT CET 2025 · 19 APR · SHIFT 2Easy

Simple Pendulum · Energy, speed and tension of a pendulum

If the length of the oscillating simple pendulum is made $\frac{1}{3}$ times the original keeping amplitude same then increase in its total energy at a place will be

Show answer and solution

Answer: (b)

$E = \frac{mgA^2}{2L}$ , so for the same amplitude $E \propto \frac{1}{L}$ .
$L' = \frac{L}{3} \Rightarrow E' = 3E$
Increase $= E' - E = 2E$ , i.e. 2 times the original energy.
So the answer is (b) 2 times.

Report a mistake

Q277JEE Main 2025 · 29 JAN · SHIFT 1Easy

Simple Pendulum · Effect of g and seconds pendulum

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A): Time period of a simple pendulum is longer at the top of a mountain than that at the base of the mountain.

Reason (R): Time period of a simple pendulum decreases with increasing value of acceleration due to gravity and vice-versa.

In the light of the above statements, choose the most appropriate answer from the options given below:

Show answer and solution

Answer: (c)

$g$ is smaller at the top of a mountain, and $T = 2\pi\sqrt{\frac{l}{g}}$ , so $T$ is longer there: (A) true.
$T \propto \frac{1}{\sqrt{g}}$ , so $T$ decreases when $g$ increases: (R) true.
(R) is exactly why (A) happens.
So the answer is (c).

Report a mistake

Q278JEE Main 2025 · 22 JAN · SHIFT 2Easy

Simple Pendulum · Effect of g and seconds pendulum

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A): A simple pendulum is taken to a planet of mass and radius, 4 times and 2 times, respectively, that of the Earth. The time period of the pendulum remains same on earth and the planet.

Reason (R): The mass of the pendulum remains unchanged at Earth and the other planet. In the light of the above statements, choose the correct answer from the options given below:

Show answer and solution

Answer: (d)

$g = \frac{GM}{R^2}$ ; on the planet $g' = \frac{G(4M)}{(2R)^2} = g$ .
Same $g$ gives the same $T$ : (A) true.
Mass of the bob does not change: (R) true, but $T$ does not depend on mass, so (R) does not explain (A).
So the answer is (d).

Report a mistake

Q279MHT CET 2025 · 19 APR · SHIFT 1Easy

Simple Pendulum · Pendulum in a lift or vehicle

The time period of a simple pendulum inside a stationary lift is $\sqrt{3}$ second. When the lift moves upwards with an acceleration $g/3$ , the time period will be ( $g =$ acceleration due to gravity)

Show answer and solution

Answer: (a)

Lift accelerating up: $g_{eff} = g + \frac{g}{3} = \frac{4g}{3}$ .
$T' = \sqrt{3}\cdot\sqrt{\frac{3}{4}} = \frac{3}{2}\text{ s}$
So the answer is (a) 1.5 s.

Report a mistake

Q280MHT CET 2025 · 19 APR · SHIFT 1Medium

Simple Pendulum · Time period, length and mass

' $P$ ' and ' $Q$ ' are fixed points in same plane and mass ' $m$ ' is tied by string as shown in figure. If the mass is displaced slightly out of this plane and released, it will oscillate with time period $(PQ = 2d, PR = QR = L)(g =$ gravitational acceleration)

Show answer and solution

Answer: (c)

Displaced out of the plane, the mass swings about the line $PQ$ .
Effective length = distance of $R$ from $PQ$ : $l_{eff} = \sqrt{L^2 - d^2}$ .
$T = 2\pi\sqrt{\frac{l_{eff}}{g}} = 2\pi\sqrt{\frac{(L^2 - d^2)^{1/2}}{g}}$
So the answer is (c).

Report a mistake

Q281JEE Main 2025 · 4 APR · SHIFT 1Medium

Simple Pendulum · Time period, length and mass

Two simple pendulums having lengths $l_1$ and $l_2$ with negligible string mass undergo angular displacements $\theta_1$ and $\theta_2$ , from their mean positions, respectively. If the angular accelerations of both pendulums are same, then which expression is correct?

Show answer and solution

Answer: (a)

For a simple pendulum, angular acceleration $\alpha = -\frac{g}{l}\theta$ .
Equal $\alpha$ : $\frac{\theta_1}{l_1} = \frac{\theta_2}{l_2}$
$\theta_1 l_2 = \theta_2 l_1$
So the answer is (a).

Report a mistake

Q282MHT CET 2025 · 25 APR · SHIFT 2Medium

Simple Pendulum · Energy, speed and tension of a pendulum

Two simple pendulums have first (A) bob of mass ' $M_1$ ' and length ' $L_1$ ', second (B) of mass ' $M_2$ ' and length ' $L_2$ '. $M_1 = M_2$ and $L_1 = 2L_2$ . If their total energies are same then the correct statement is

Show answer and solution

Answer: (b)

Energy of a pendulum with linear amplitude $A$ : $E = \frac{1}{2}m\omega^2A^2 = \frac{1}{2}m\frac{g}{L}A^2$ .
Same $E$ and same $m \Rightarrow A^2 \propto L$ .
$L_1 = 2L_2 \Rightarrow A_A^2 = 2A_B^2$ , so $A_B = A_A/\sqrt{2}$ .
So the answer is (b) amplitude of B is smaller than amplitude of A.

Report a mistake

Q283MHT CET 2025 · 26 APR · SHIFT 2Medium

Simple Pendulum · Pendulum in a lift or vehicle

A simple pendulum is suspended from ceiling of a lift when lift is at rest its period is ' T '. With what acceleration ' $a$ ' should lift be accelerated upward in order to reduce the period to ' $T/2$ '? (take ' g ' as acceleration due to gravity)

Show answer and solution

Answer: (b)

Lift accelerating up: $T' = 2\pi\sqrt{\frac{l}{g+a}}$ .
$\frac{T'}{T} = \sqrt{\frac{g}{g+a}} = \frac{1}{2} \Rightarrow g + a = 4g$ .
So the answer is (b) $a = 3g$ .

Report a mistake

Q284MHT CET 2025 · 21 APR · SHIFT 1Medium

Simple Pendulum · Pendulum in a lift or vehicle

A simple pendulum has time period ' $T_1$ '. The point of suspension is now moved upward according to equation $y = kt^{2}$ where $k = 1\text{ m/s}^{2}$ . If new time period is ' $T_2$ ' then $T_1^{2}/T_2^{2}$ will be ( $g = 10\text{ m/s}^{2}$ )

Show answer and solution

Answer: (c)

$y = kt^2 \Rightarrow a = \frac{d^2y}{dt^2} = 2k = 2\text{ m/s}^2$ upward.
$g_{eff} = g + a = 12\text{ m/s}^2$ .
$\frac{T_1^2}{T_2^2} = \frac{g_{eff}}{g} = \frac{12}{10}$ .
So the answer is (c) $6/5$ .

Report a mistake

Q285MHT CET 2024 · 10 MAY · SHIFT 2Easy

Simple Pendulum · Time period, length and mass

A simple pendulum of length ' $l$ ' has a brass bob attached at its lower end. It's period is ' T '. A steel bob of the same size, having density ' $x$ ' times that of brass, replaces the brass bob. Its length is then so changed that the period becomes ' 2 T '. What is the new length?

Show answer and solution

Answer: (c)

$T = 2\pi\sqrt{\frac{l}{g}}$ does not depend on the mass or density of the bob.
$T' = 2T \Rightarrow l' = 2^2 l = 4l$
So the answer is (c) $4l$ .

Report a mistake

Q286MHT CET 2024 · 3 MAY · SHIFT 2Easy

Simple Pendulum · Time period, length and mass

The period of a simple pendulum gets doubled when

Show answer and solution

Answer: (b)

$T \propto \sqrt{l}$ and is independent of mass.
$T' = 2T \Rightarrow l' = 4l$
So the answer is (b) its length is made four times.

Report a mistake

Q287NEET 2024Easy

Simple Pendulum · Time period, length and mass

If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is $\frac{x}{2}$ times its original time period. Then the value of $x$ is:

Show answer and solution

Answer: (b)

$T = 2\pi\sqrt{\frac{l}{g}}$ is independent of mass.
$l' = \frac{l}{2} \Rightarrow T' = \frac{T}{\sqrt{2}} = \frac{\sqrt{2}}{2}T$
Comparing with $\frac{x}{2}T$ : $x = \sqrt{2}$ .
So the answer is (b) $\sqrt{2}$ .

Report a mistake

Q288MHT CET 2024 · 11 MAY · SHIFT 1Easy

Simple Pendulum · Energy, speed and tension of a pendulum

A simple pendulum of length ' $L$ ' has mass ' $M$ ' and it oscillates freely with amplitude ' $A$ '. At extreme position, its potential energy is

Show answer and solution

Answer: (d)

At the extreme position all energy is potential: $U = E = \frac{1}{2}M\omega^2A^2$ .
For a simple pendulum $\omega^2 = \frac{g}{L}$ .
$U = \frac{1}{2}M\frac{g}{L}A^2 = \frac{MgA^2}{2L}$
So the answer is (d) $\frac{MgA^2}{2L}$ .

Report a mistake

Q289MHT CET 2024 · 2 MAY · SHIFT 1Easy

Simple Pendulum · Pendulum in a lift or vehicle

Choose the correct answer. When a point of suspension of pendulum is moved vertically upward with acceleration ' $a$ ', its period of oscillation

Show answer and solution

Answer: (a)

Suspension accelerating upward: $g_{eff} = g + a > g$ .
$T = 2\pi\sqrt{\frac{l}{g + a}}$ is smaller than $2\pi\sqrt{\frac{l}{g}}$ .
So the answer is (a) decreases.

Report a mistake

Q290MHT CET 2024 · 10 MAY · SHIFT 1Medium

Simple Pendulum · Time period, length and mass

Let ' $l_1$ ' be the length of simple pendulum. Its length changes to ' $l_2$ ' to increase the periodic time by $20\%$ . The ratio $\frac{l_2}{l_1} =$

Show answer and solution

Answer: (c)

$T \propto \sqrt{l} \Rightarrow \frac{l_2}{l_1} = \left(\frac{T_2}{T_1}\right)^2$
A 20% increase means $\frac{T_2}{T_1} = 1.2$ .
$\frac{l_2}{l_1} = 1.2^2 = 1.44$
So the answer is (c) 1.44.

Report a mistake

Q291MHT CET 2024 · 4 MAY · SHIFT 1Medium

Simple Pendulum · Time period, length and mass

A simple pendulum of length $l_1$ has time period $T_1$ . Another simple pendulum of length $l_2\ (l_1 > l_2)$ has time period $T_2$ . Then the time period of the pendulum of length $(l_1 - l_2)$ will be

Show answer and solution

Answer: (c)

$T^2 \propto l$ : $T_1^2 = kl_1$ , $T_2^2 = kl_2$ .
For length $(l_1 - l_2)$ : $T^2 = k(l_1 - l_2) = T_1^2 - T_2^2$
So the answer is (c) $\sqrt{T_1^2 - T_2^2}$ .

Report a mistake

Q292MHT CET 2024 · 16 MAY · SHIFT 2Medium

Simple Pendulum · Energy, speed and tension of a pendulum

The bob of a pendulum of length ' $l$ ' is pulled aside from its equilibrium position through an angle ' $\theta$ ' and then released. The bob will then pass through its equilibrium position with speed ' $v$ ', where ' $v$ ' equal to ( $g =$ acceleration due to gravity)

Show answer and solution

Answer: (a)

Height fallen from the extreme to the mean position: $h = l - l\cos\theta = l(1-\cos\theta)$ .
Energy conservation: $\frac{1}{2}mv^2 = mgl(1-\cos\theta)$ .
So the answer is (a) $v = \sqrt{2gl(1-\cos\theta)}$ .

Report a mistake

Q293JEE Main 2024 · 1 FEB · SHIFT 1Medium

Simple Pendulum · Energy, speed and tension of a pendulum

A simple pendulum of length $1\text{ m}$ has a wooden bob of mass $1\text{ kg}$ . It is struck by a bullet of mass $10^{-2}\text{ kg}$ moving with a speed of $2 \times 10^{2}\text{ m s}^{-1}$ . The bullet gets embedded into the bob. The height to which the bob rises before swinging back is. (use $g = 10\text{ m/s}^{2}$ )

Show answer and solution

Answer: (a)

Momentum conservation: $v = \frac{mu}{m+M} = \frac{10^{-2} \times 200}{1.01} \approx 1.98\text{ m/s}$ .
Energy conservation after collision: $h = \frac{v^2}{2g} = \frac{(1.98)^2}{20}$ .
$h \approx 0.196\text{ m} \approx 0.20\text{ m}$ .
So the answer is (a) $0.20\text{ m}$ .

Report a mistake

Q294JEE Main 2024 · 29 JAN · SHIFT 2Medium

Simple Pendulum · Energy, speed and tension of a pendulum

The bob of a pendulum was released from a horizontal position. The length of the pendulum is $10\text{ m}$ . If it dissipates $10\%$ of its initial energy against air resistance, the speed with which the bob arrives at the lowest point is:

[Use, $g : 10\text{ m s}^{-2}$ ]

Show answer and solution

Answer: (d)

Released from horizontal, so the drop to the lowest point is $L = 10\text{ m}$ .
Only $90\%$ of $mgL$ becomes KE: $\frac{1}{2}mv^2 = 0.9\,mgL$ .
$v^2 = 2 \times 0.9 \times 10 \times 10 = 180 \Rightarrow v = \sqrt{180}$ .
So the answer is (d) $6\sqrt{5}\text{ m s}^{-1}$ .

Report a mistake

Q295JEE Main 2024 · 5 APR · SHIFT 1Medium

Simple Pendulum · Effect of g and seconds pendulum

A simple pendulum doing small oscillations at a place $R$ height above earth surface has time period of $T_1 = 4\text{ s},\ T_2$ would be its time period if it is brought to a point which is at a height $2R$ from earth surface. Choose the correct relation [ $R =$ radius of earth]:

Show answer and solution

Answer: (a)

$g \propto \frac{1}{r^2}$ and $T \propto \frac{1}{\sqrt{g}} \Rightarrow T \propto r$ (distance from earth's centre).
At height $R$ : $r_1 = 2R$ ; at height $2R$ : $r_2 = 3R$ .
$\frac{T_2}{T_1} = \frac{3}{2} \Rightarrow 2T_2 = 3T_1$ .
So the answer is (a) $3T_1 = 2T_2$ .

Report a mistake

Q296MHT CET 2023 · 9 MAY · SHIFT 1Easy

Simple Pendulum · Time period, length and mass

A rubber ball filled with water, having a small hole is used as the bob of a simple pendulum. The time period of such a pendulum

Show answer and solution

Answer: (d)

Effective length is up to the centre of mass of the bob; $T \propto \sqrt{l}$ .
As water drains, the centre of mass first goes down ( $T$ increases), then returns to the centre when empty ( $T$ decreases back).
So the answer is (d) first increases and then decreases, finally having same value as at the beginning.

Report a mistake

Q297JEE Main 2023 · 1 FEB · SHIFT 2Easy

Simple Pendulum · Time period, length and mass

Choose the correct length (L) versus square of the time period ( $T^2$ ) graph for a simple pendulum executing simple harmonic motion.

Show answer and solution

Answer: (d)

$T = 2\pi\sqrt{\frac{L}{g}} \Rightarrow T^2 = \frac{4\pi^2}{g}L$
$T^2 \propto L$ : a straight line through the origin with positive slope $\frac{4\pi^2}{g}$ .
So the answer is (d).

Report a mistake

Q298JEE Main 2023 · 25 JAN · SHIFT 1Easy

Simple Pendulum · Effect of g and seconds pendulum

T is the time period of simple pendulum on the earth's surface. Its time period becomes $x$ T when taken to a height R (equal to earth's radius) above the earth's surface. Then, the value of $x$ will be:

Show answer and solution

Answer: (c)

At height $h = R$ : $g' = \frac{g}{(1 + h/R)^2} = \frac{g}{4}$ .
$T \propto \frac{1}{\sqrt{g}} \Rightarrow T' = T\sqrt{4} = 2T$
So the answer is (c) $x = 2$ .

Report a mistake

Q299MHT CET 2023 · 12 MAY · SHIFT 2Easy

Simple Pendulum · Pendulum in a lift or vehicle

In a stationary lift, time period of a simple pendulum is ' $T$ '. The lift starts accelerating downwards with acceleration $\left(\frac{g}{4}\right)$ , then the time period of the pendulum will be

Show answer and solution

Answer: (b)

Lift accelerating down: $g_{eff} = g - \frac{g}{4} = \frac{3g}{4}$ .
$T' = T\sqrt{\frac{g}{g_{eff}}} = T\sqrt{\frac{4}{3}}$
So the answer is (b) $\frac{2}{\sqrt{3}}T$ .

Report a mistake

Q300MHT CET 2023 · 11 MAY · SHIFT 1Easy

Simple Pendulum · Pendulum in a lift or vehicle

The time period of a simple pendulum inside a stationary lift is ' $T$ '. When the lift starts accelerating upwards with an acceleration $\left(\frac{g}{3}\right)$ , the time period of the pendulum will be

Show answer and solution

Answer: (b)

Lift accelerating up: $g_{eff} = g + \frac{g}{3} = \frac{4g}{3}$ .
$T' = T\sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}T$
So the answer is (b).

Report a mistake

Q301MHT CET 2023 · 13 MAY · SHIFT 1Easy

Simple Pendulum · Bob in a liquid

A simple pendulum has a time period ' $T$ ' in air. Its time period when it is completely immersed in a liquid of density one eighth the density of the material of bob is

Show answer and solution

Answer: (d)

Buoyancy reduces the effective weight: $g_{eff} = g\left(1 - \frac{\sigma}{\rho}\right)$ .
$\frac{\sigma}{\rho} = \frac{1}{8} \Rightarrow g_{eff} = \frac{7g}{8}$
$T' = T\sqrt{\frac{g}{g_{eff}}} = \sqrt{\frac{8}{7}}\,T$
So the answer is (d).

Report a mistake

Q302NEET 2023Easy

Simple Pendulum · Bob in a liquid

A simple pendulum oscillating in air has a period of $\sqrt{3}\text{ s}$ . If it is completely immersed in non- viscous liquid, having density $\left(\frac{1}{4}\right)$ th of the material of the bob, the new period will be :-

Show answer and solution

Answer: (c)

In a liquid: $g_{eff} = g\left(1 - \frac{\sigma}{\rho}\right) = g\left(1 - \frac{1}{4}\right) = \frac{3g}{4}$ .
$T' = \sqrt{3}\cdot\sqrt{\frac{4}{3}} = 2\text{ s}$
So the answer is (c) $2\text{ s}$ .

Report a mistake

Q303MHT CET 2023 · 14 MAY · SHIFT 1Medium

Simple Pendulum · Time period, length and mass

The bob of simple pendulum of length ' $L$ ' is released from a position of small angular displacement $\theta$ . Its linear displacement at time ' $t$ ' is ( $g =$ acceleration due to gravity)

Show answer and solution

Answer: (a)

Released from rest at the extreme, so the motion starts at maximum displacement: $x = A\cos\omega t$ .
Amplitude (arc) $A = L\theta$ and $\omega = \sqrt{\frac{g}{L}}$ .
$x = L\theta\cos\left[\sqrt{\frac{g}{L}}\,t\right]$
So the answer is (a).

Report a mistake

Q304MHT CET 2023 · 14 MAY · SHIFT 1Medium

Simple Pendulum · Energy, speed and tension of a pendulum

A simple pendulum of length ' $l$ ' and a bob of mass ' $\text{m}$ ' is executing S.H.M. of small amplitude ' $A$ '. The maximum tension in the string will be ( $g =$ acceleration due to gravity)

Show answer and solution

Answer: (b)

Tension is maximum at the mean position: $T_{max} = mg + \frac{mv_{max}^2}{l}$ .
$v_{max} = A\omega = A\sqrt{g/l} \Rightarrow v_{max}^2 = \frac{gA^2}{l}$ .
$T_{max} = mg + \frac{mgA^2}{l^2} = mg\left[1 + \left(\frac{A}{l}\right)^2\right]$ .
So the answer is (b).

Report a mistake

Q305JEE Main 2023 · 6 APR · SHIFT 2Medium

Simple Pendulum · Energy, speed and tension of a pendulum

A simple pendulum with length $100\text{ cm}$ and bob of mass $250\text{ g}$ is executing S.H.M. of amplitude $10\text{ cm}$ . The maximum tension in the string is found to be $\frac{x}{40}\text{ N}$ . The value of $x$ is ___________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 101

$T_{max} = mg\left[1 + \left(\frac{A}{l}\right)^2\right]$ (tension at mean position).
$m = 0.25\text{ kg},\ A/l = 10/100 = 0.1$ , take $g = 10\text{ m/s}^2$ .
$T_{max} = 0.25 \times 10 \times 1.01 = 2.525\text{ N} = \frac{101}{40}\text{ N}$ .
So $x = 101$ .

Report a mistake

Q306JEE Main 2022 · 29 JUN · SHIFT 2Easy

Simple Pendulum · Time period, length and mass

The motion of a simple pendulum executing S.H.M. is represented by the following equation. $y = A\sin(\pi t + \phi)$ , where time is measured in second. The length of pendulum is

Show answer and solution

Answer: (c)

Comparing with $y = A\sin(\omega t + \phi)$ : $\omega = \pi$ rad/s, so $T = 2$ s.
$\omega^2 = \frac{g}{l} \Rightarrow l = \frac{g}{\omega^2} = \frac{9.8}{\pi^2}$
$l \approx 0.994$ m $= 99.4$ cm
So the answer is (c) 99.4 cm.

Report a mistake

Q307JEE Main 2022 · 28 JUN · SHIFT 1Easy

Simple Pendulum · Energy, speed and tension of a pendulum

A pendulum is suspended by a string of length 250 cm. The mass of the bob of the pendulum is 200 g. The bob is pulled aside until the string is at 60 $^\circ$ with vertical as shown in the figure. After releasing the bob, the maximum velocity attained by the bob will be ____________ ms ${}^{-1}$ . (if g = 10 m/s2)

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 5

Maximum speed is at the lowest point: $v = \sqrt{2gl(1 - \cos\theta)}$ .
$l = 2.5\text{ m}$ , $\theta = 60^\circ$ , $1 - \cos 60^\circ = \frac{1}{2}$
$v = \sqrt{2 \times 10 \times 2.5 \times 0.5} = \sqrt{25}$
$v = 5\text{ m/s}$

Report a mistake

Q308JEE Main 2022 · 30 JUL · SHIFT 2Easy

Simple Pendulum · Pendulum in a lift or vehicle

An oscillating simple pendulum of time period $T_0$ is placed in a lift which is accelerating upwards with $2.5\text{ m/s}^2$ . The time period of pendulum in the lift will be (if $g = 10\text{ m/s}^2$ )

Show answer and solution

Answer: (b)

Lift accelerating up: $g_{eff} = 10 + 2.5 = 12.5\text{ m/s}^2$ .
$T = T_0\sqrt{\frac{10}{12.5}} = T_0\sqrt{\frac{4}{5}}$
So the answer is (b) $\frac{2}{\sqrt{5}}T_0$ .

Report a mistake

Q309JEE Main 2022 · 26 JUN · SHIFT 1Easy

Simple Pendulum · Pendulum in a lift or vehicle

Time period of a simple pendulum in a stationary lift is 'T'. If the lift accelerates with $\frac{g}{6}$ vertically upwards then the time period will be:

(Where g = acceleration due to gravity)

Show answer and solution

Answer: (c)

Lift accelerating up: $g_{eff} = g + \frac{g}{6} = \frac{7g}{6}$ .
$T' = T\sqrt{\frac{g}{g_{eff}}} = \sqrt{\frac{6}{7}}T$
So the answer is (c).

Report a mistake

Q310NEET 2022 · PHASE 1Medium

Simple Pendulum · Two pendulums in phase

Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is:

Show answer and solution

Answer: (a)

$T \propto \sqrt{l}$ : $\frac{T_{long}}{T_{short}} = \sqrt{\frac{121}{100}} = \frac{11}{10}$ .
They are in phase again when the shorter one makes one more vibration: $n\cdot 10 = (n-1)\cdot 11$ .
$n = 11$ vibrations of the shorter pendulum (10 of the longer).
So the answer is (a) 11.

Report a mistake

Q311JEE Main 2022 · 28 JUL · SHIFT 2Medium

Simple Pendulum · Effect of g and seconds pendulum

Assume there are two identical simple pendulum clocks. Clock - 1 is placed on the earth and Clock - 2 is placed on a space station located at a height h above the earth surface. Clock - 1 and Clock - 2 operate at time periods 4 s and 6 s respectively. Then the value of h is -

(consider radius of earth $R_E = 6400\text{ km}$ and $g$ on earth $10\text{ m/s}^{2}$ )

Show answer and solution

Answer: (c)

$T \propto \frac{1}{\sqrt{g}}$ and $g \propto \frac{1}{(R+h)^2}$ , so $T \propto (R+h)$ .
$\frac{T_2}{T_1} = \frac{6}{4} = \frac{R+h}{R} \Rightarrow h = 0.5R$ .
$h = 0.5 \times 6400 = 3200\text{ km}$ .
So the answer is (c) 3200 km.

Report a mistake

Q312JEE Main 2022 · 29 JUL · SHIFT 1Medium

Simple Pendulum · Pendulum in a lift or vehicle

The time period of oscillation of a simple pendulum of length L suspended from the roof of a vehicle, which moves without friction down an inclined plane of inclination $\alpha$ , is given by:

Show answer and solution

Answer: (a)

The vehicle slides freely with acceleration $g\sin\alpha$ down the incline.
In its frame the pseudo force cancels the along-incline part of gravity, leaving $g_{eff} = g\cos\alpha$ (perpendicular to the incline).
So the answer is (a) $2\pi\sqrt{L/(g\cos\alpha)}$ .

Report a mistake

Q313JEE Main 2022 · 29 JUL · SHIFT 2Medium

Simple Pendulum · Bob in a liquid

The metallic bob of simple pendulum has the relative density 5. The time period of this pendulum is $10\text{ s}$ . If the metallic bob is immersed in water, then the new time period becomes $5\sqrt{x}$ s. The value of $x$ will be ________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 5

Relative density 5: $g_{eff} = g\left(1 - \frac{1}{5}\right) = \frac{4g}{5}$ .
$T' = T\sqrt{\frac{g}{g_{eff}}} = 10\sqrt{\frac{5}{4}} = 5\sqrt{5}\text{ s}$ .
So $x = 5$ .

Report a mistake

Q314MHT CET 2021 · 21 SEP · SHIFT 2Easy

Simple Pendulum · Time period, length and mass

A pendulum clock is running fast. To correct its time, we should

Show answer and solution

Answer: (c)

A fast clock means its period is too small.
$T = 2\pi\sqrt{\frac{l}{g}}$ , so increasing $l$ increases $T$ .
So the answer is (c) increase the length of the pendulum.

Report a mistake

Q315JEE Main 2021 · 22 JUL · SHIFT 2Easy

Simple Pendulum · Time period, length and mass

T0 is the time period of a simple pendulum at a place. if the length of the pendulum is reduced to $\frac{1}{16}$ times of its initial value, the modified time period is:

Show answer and solution

Answer: (b)

$T \propto \sqrt{l}$
$l' = \frac{l}{16} \Rightarrow T' = \frac{T_0}{4}$
So the answer is (b) $\frac{1}{4}T_0$ .

Report a mistake

Q316MHT CET 2021 · 23 SEP · SHIFT 2Easy

Simple Pendulum · Energy, speed and tension of a pendulum

A bob of simple pendulum of mass 'm' perform $\text{SHM}$ with amplitude ' $A$ ' and period 'T'. Kinetic energy of pendulum of displacement $x = \frac{A}{2}$ will be

Show answer and solution

Answer: (d)

In SHM, $KE = \frac{1}{2}m\omega^2(A^2 - x^2)$ with $\omega = \frac{2\pi}{T}$ .
At $x = \frac{A}{2}$ : $A^2 - x^2 = \frac{3A^2}{4}$ .
$KE = \frac{1}{2}m\cdot\frac{4\pi^2}{T^2}\cdot\frac{3A^2}{4} = \frac{3m\pi^2A^2}{2T^2}$
So the answer is (d) $\frac{3m\pi^2A^2}{2T^2}$ .

Report a mistake

Q317JEE Main 2021 · 26 FEB · SHIFT 2Easy

Simple Pendulum · Effect of g and seconds pendulum

Given below are two statements:

Statement I: A second's pendulum has a time period of 1 second.

Statement II: It takes precisely one second to move between the two extreme positions. In the light of the above statements, choose the correct answer from the options given below:

Show answer and solution

Answer: (b)

A seconds pendulum has period $2\text{ s}$ , not $1\text{ s}$ : Statement I false.
Going from one extreme to the other takes $\frac{T}{2} = 1\text{ s}$ : Statement II true.
So the answer is (b).

Report a mistake

Q318JEE Main 2021 · 25 FEB · SHIFT 1Easy

Simple Pendulum · Effect of g and seconds pendulum

If the time period of a two meter long simple pendulum is 2s, the acceleration due to gravity at the place where pendulum is executing S.H.M. is:

Show answer and solution

Answer: (b)

$T = 2\pi\sqrt{\frac{L}{g}} \Rightarrow g = \frac{4\pi^2L}{T^2}$
$g = \frac{4\pi^2 \times 2}{2^2} = 2\pi^2\text{ m/s}^2$
So the answer is (b) $2\pi^2\text{ m s}^{-2}$ .

Report a mistake

Q319JEE Main 2021 · 16 MAR · SHIFT 1Easy

Simple Pendulum · Pendulum in a lift or vehicle

Time period of a simple pendulum is T inside a lift when the lift is stationary. If the lift moves upwards with an acceleration g/2, the time period of pendulum will be:

Show answer and solution

Answer: (b)

Lift accelerating up: $g_{eff} = g + \frac{g}{2} = \frac{3g}{2}$ .
$T' = T\sqrt{\frac{g}{g_{eff}}} = \sqrt{\frac{2}{3}}T$
So the answer is (b).

Report a mistake

Q320MHT CET 2021 · 21 SEP · SHIFT 2Medium

Simple Pendulum · Energy, speed and tension of a pendulum

A child is sitting on a swing which performs S.H.M. It has minimum and maximum heights from ground $0.75\text{ m}$ and $2\text{ m}$ respectively. Its maximum speed will be $\left[g = 10\,\frac{\text{m}}{\text{s}^{2}}\right]$

Show answer and solution

Answer: (c)

Height difference between extreme and lowest position: $h = 2 - 0.75 = 1.25\text{ m}$ .
$v_{max} = \sqrt{2gh} = \sqrt{2 \times 10 \times 1.25} = \sqrt{25}$ .
So the answer is (c) $5\text{ m/s}$ .

Report a mistake

Q321JEE Main 2021 · 31 AUG · SHIFT 2Medium

Simple Pendulum · Bob in a liquid

A bob of mass 'm' suspended by a thread of length l undergoes simple harmonic oscillations with time period T. If the bob is immersed in a liquid that has density $\frac{1}{4}$ times that of the bob and the length of the thread is increased by 1/3rd of the original length, then the time period of the simple harmonic oscillations will be :-

Show answer and solution

Answer: (d)

In a liquid, buoyancy reduces the effective gravity: $g_{eff} = g\left(1 - \frac{\sigma}{\rho}\right) = g\left(1-\frac{1}{4}\right) = \frac{3g}{4}$ .
New length $l' = l + \frac{l}{3} = \frac{4l}{3}$ .
$\frac{T'}{T} = \sqrt{\frac{l'/l}{g_{eff}/g}} = \sqrt{\frac{4/3}{3/4}} = \frac{4}{3}$ .
So the answer is (d) $\frac{4}{3}T$ .

Report a mistake

Q322MHT CET 2020 · 16 OCT · SHIFT 1Easy

Simple Pendulum · Energy, speed and tension of a pendulum

A simple pendulum of length $L$ has mass $m$ and it oscillates freely with amplitude $A$ . At extreme position, its potential energy is ( $g =$ acceleration due to gravity)

Show answer and solution

Answer: (d)

For a simple pendulum $\omega^2 = \frac{g}{L}$ .
At the extreme position all energy is potential: $PE = \frac{1}{2}m\omega^2A^2$ .
$PE = \frac{1}{2}m\cdot\frac{g}{L}\cdot A^2 = \frac{mgA^2}{2L}$
So the answer is (d) $\frac{mgA^2}{2L}$ .

Report a mistake

Q323MHT CET 2020 · 19 OCT · SHIFT 2Medium

Simple Pendulum · Energy, speed and tension of a pendulum

A bob of a simple pendulum has mass $m$ and is oscillating with an amplitude $a$ . If the length of the pendulum is $L$ , then the maximum tension in the string is $[\cos 0^\circ = 1,\ g =$ acceleration due to gravity]

Show answer and solution

Answer: (a)

At the mean position $T_{max} = mg + \frac{mv_{max}^2}{L}$ .
$v_{max} = a\omega = a\sqrt{g/L} \Rightarrow \frac{mv_{max}^2}{L} = mg\frac{a^2}{L^2}$ .
So the answer is (a) $mg\left[1+\left(\frac{a}{L}\right)^2\right]$ .

Report a mistake

Q324MHT CET 2020 · 16 OCT · SHIFT 2Medium

Simple Pendulum · Energy, speed and tension of a pendulum

A simple pendulum of length $l$ has a bob of mass $m$ . It executes SHM of small amplitude A. The maximum tension in the string is ( $g =$ acceleration due to gravity)

Show answer and solution

Answer: (b)

At the mean position $T_{max} = mg + \frac{mv_{max}^2}{l}$ , with $v_{max} = A\sqrt{g/l}$ .
$T_{max} = mg + mg\frac{A^2}{l^2}$ .
So the answer is (b) $mg\left(\frac{A^2}{l^2}+1\right)$ .

Report a mistake

Q325JEE Main 2020 · 8 JAN · SHIFT 2Medium

Simple Pendulum · Effect of g and seconds pendulum

A simple pendulum is being used to determine the value of gravitational acceleration g at a certain place. The length of the pendulum is 25.0 cm and a stop watch with 1s resolution measures the time taken for 40 oscillations to be 50 s. The accuracy in g is:

Show answer and solution

Answer: (a)

$g = \frac{4\pi^2 L}{T^2} \Rightarrow \frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta t}{t}$ .
$\frac{\Delta L}{L} = \frac{0.1}{25.0} = 0.4\%$ (least count 1 mm).
$2\frac{\Delta t}{t} = 2 \times \frac{1}{50} = 4\%$ .
So the answer is (a) $4.40\%$ .

Report a mistake

Q326JEE Main 2019 · 11 JAN · SHIFT 2Easy

Simple Pendulum · Effect of g and seconds pendulum

The mass and the diameter of a planet are three times the respective values for the Earth. The period of oscillation of simple pendulum on the Earth is 2 s. The period of oscillation of the same pendulum on the planet would be:

Show answer and solution

Answer: (d)

Diameter $3\times$ means radius $3\times$ : $g' = \frac{G(3M)}{(3R)^2} = \frac{g}{3}$ .
$T \propto \frac{1}{\sqrt{g}} \Rightarrow T' = 2\sqrt{3}\text{ s}$
So the answer is (d) $2\sqrt{3}$ s.

Report a mistake

Q327MHT CET 2019 · 3 MAY · SHIFT 1Easy

Simple Pendulum · Pendulum in a lift or vehicle

A person measures a time period of a simple pendulum inside a stationary lift and finds it to be $T$ . If the lift starts accelerating upwards with an acceleration $\left(\frac{g}{3}\right)$ , the time period of the pendulum will be

Show answer and solution

Answer: (b)

Lift accelerating up: $g_{eff} = g + \frac{g}{3} = \frac{4g}{3}$ .
$T' = T\sqrt{\frac{3}{4}} = \sqrt{3}\,\frac{T}{2}$
So the answer is (b).

Report a mistake

Q328MHT CET 2019 · 2 MAY · SHIFT 2Medium

Simple Pendulum · Time period, length and mass

Two pendulums begin to swing simultaneously. The first pendulum makes nine full oscillations when the other makes seven. The ratio of the lengths of the two pendulums is

Show answer and solution

Answer: (a)

Same time: $9T_1 = 7T_2 \Rightarrow \frac{T_1}{T_2} = \frac{7}{9}$ .
$l \propto T^2 \Rightarrow \frac{l_1}{l_2} = \frac{49}{81}$
So the answer is (a) $\frac{49}{81}$ .

Report a mistake

Q329JEE Main 2019 · 11 JAN · SHIFT 2Medium

Simple Pendulum · Energy, speed and tension of a pendulum

A pendulum is executing simple harmonic motion and its maximum kinetic energy is K1. If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is K2. Then:

Show answer and solution

Answer: (b)

Take amplitude as the angular amplitude $\theta_0$ (same in both cases).
Max KE = loss of PE from extreme to mean: $K = mgL(1-\cos\theta_0) \propto L$ .
Length doubled $\Rightarrow K_2 = 2K_1$ .
So the answer is (b) $K_2 = 2K_1$ .

Report a mistake

Q330JEE Main 2019 · 9 APR · SHIFT 1Medium

Simple Pendulum · Bob in a liquid

A simple pendulum oscillating in air has period T. The bob of the pendulum is completely immersed in a non-viscous liquid. The density of the liquid is 1/16 th of the material of the bob. If the bob is inside liquid all the time, its period of oscillation in this liquid is:

Show answer and solution

Answer: (c)

$g_{eff} = g\left(1 - \frac{\sigma}{\rho}\right) = g\left(1 - \frac{1}{16}\right) = \frac{15g}{16}$ .
$T' = T\sqrt{\frac{g}{g_{eff}}} = T\sqrt{\frac{16}{15}}$ .
So the answer is (c) $4T\sqrt{\frac{1}{15}}$ .

Report a mistake

Q331JEE Main 2019 · 11 JAN · SHIFT 2DifficultOld syllabus

Simple Pendulum · Pendulum in a lift or vehicle

A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and an amplitude of 10–2 m. The relative change in the angular frequency of the pendulum is best given by:

Show answer and solution

Answer: (b)

Support moves as $y = A\sin\Omega t$ , so its maximum acceleration is $\Omega^2A = 1^2 \times 10^{-2} = 10^{-2}\text{ m/s}^2$ .
$g_{eff} = g \pm \Omega^2A$ and $\omega \propto \sqrt{g_{eff}} \Rightarrow \frac{\Delta\omega}{\omega} = \frac{1}{2}\frac{\Delta g}{g}$ .
$\Delta\omega = \frac{1}{2} \times \frac{10^{-2}}{10} \times 10 = 5\times10^{-3}\text{ rad/s}$ , of order $10^{-3}$ rad/s.
So the answer is (b) $10^{-3}$ rad/s.

Report a mistake

Q332JEE Main 2015Medium

Simple Pendulum · Effect of g and seconds pendulum

The period of oscillation of a simple pendulum is $T = 2\pi\sqrt{\frac{L}{g}}$ . Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using wrist watch of 1 s resolution. The accuracy in the determination of g is:

Show answer and solution

Answer: (d)

$\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta t}{t}$ .
$\frac{\Delta L}{L} = \frac{0.1}{20.0} = 0.5\%$ ; $2\frac{\Delta t}{t} = 2 \times \frac{1}{90} \approx 2.2\%$ .
Total $\approx 2.7\% \approx 3\%$ .
So the answer is (d) $3\%$ .

Report a mistake

Q333JEE Main 2014 · 19 APR · SHIFT 1Medium

Simple Pendulum · Effect of g and seconds pendulum

In an experiment for determining the gravitational acceleration g of a place with the help of a simple pendulum, the measured time period square is plotted against the string length of the pendulum in the figure.

what is the value of g at the place?

Show answer and solution

Answer: (b)

$T^2 = \frac{4\pi^2}{g}L$ , so slope of the $T^2$ vs $L$ graph $= \frac{4\pi^2}{g}$ .
From the graph: slope $= \frac{8.0 - 4.0}{2.0 - 1.0} = 4\text{ s}^2/\text{m}$ .
$g = \frac{4\pi^2}{4} = \pi^2 \approx 9.87\text{ m/s}^2$ .
So the answer is (b) $9.87\text{ m/s}^2$ .

Report a mistake

Q334JEE Main 2013 · 9 APR · SHIFT 1Easy

Simple Pendulum · Two pendulums in phase

Two simple pendulums of length 1 m and 4 m respectively are both given small displacement in the same direction at the same instant. They will be again in phase after the shorter pendulum has completed number of oscillations equal to:

Show answer and solution

Answer: (a)

$\frac{T_{long}}{T_{short}} = \sqrt{\frac{4}{1}} = 2$
In the time the longer pendulum makes 1 oscillation, the shorter makes 2, and they are in phase again.
So the answer is (a) 2.

Report a mistake

Q335JEE Main 2005 · AIEEEEasy

Simple Pendulum · Time period, length and mass

The bob of a simple pendulum is a spherical hollow ball filled with water. A plugged hole near the bottom of the oscillating bob gets suddenly unplugged. During observation, till water is coming out, the time period of oscillation would

Show answer and solution

Answer: (b)

$T \propto \sqrt{l}$ , with $l$ measured to the centre of mass of the bob.
As water flows out, the centre of mass first moves down, so $T$ increases.
When the ball is nearly empty, the centre of mass returns to the ball's centre, so $T$ returns to its original value.
So the answer is (b) first increase and then decrease to the original value.

Report a mistake

Q336JEE Main 2004 · AIEEEMedium

Simple Pendulum · Bob in a liquid

The bob of a simple pendulum executes simple harmonic motion in water with a period $t,$ while the period of oscillation of the bob is $t_0$ in air. Neglecting frictional force of water and given that the density of the bob is $(4/3) \times 1000\text{ kg/m}^{3}.$ What relationship between $t$ and $t_0$ is true

Show answer and solution

Answer: (a)

$g_{eff} = g\left(1 - \frac{\rho_w}{\rho_b}\right) = g\left(1 - \frac{1000}{4000/3}\right) = g\left(1-\frac{3}{4}\right) = \frac{g}{4}$ .
$t = t_0\sqrt{\frac{g}{g_{eff}}} = t_0\sqrt{4}$ .
So the answer is (a) $t = 2t_0$ .

Report a mistake

Q337JEE Main 2003 · AIEEEEasy

Simple Pendulum · Time period, length and mass

The length of a simple pendulum executing simple harmonic motion is increased by $21\%$ . The percentage increase in the time period of the pendulum of increased length is

Show answer and solution

Answer: (d)

$T \propto \sqrt{l}$ ; $l' = 1.21\,l$
$T' = \sqrt{1.21}\,T = 1.1\,T$
Increase $= 10\%$ (exact, since the change is not small, the $\frac{1}{2}\times 21\%$ approximation is not used).
So the answer is (d) $10\%$ .

Report a mistake

Q338JEE Main 2002 · AIEEEEasy

Simple Pendulum · Time period, length and mass

A child swinging on a swing in sitting position, stands up, then the time period of the swing will

Show answer and solution

Answer: (b)

The swing acts like a pendulum whose length is up to the centre of mass of the child.
On standing up, the centre of mass rises, so the effective length decreases.
$T \propto \sqrt{l}$ , so $T$ decreases.
So the answer is (b) decrease.

Report a mistake

Q339NEET 2000 · AIPMTEasy

Simple Pendulum · Time period, length and mass

Two masses $M$ A and $M$ B are hung from two strings of length $l$ A and $l$ B respectively. They are executing SHM with frequency relation $f$ A = 2 $f$ B, then relation

Show answer and solution

Answer: (a)

$f = \frac{1}{2\pi}\sqrt{\frac{g}{l}}$ , independent of mass, so $l \propto \frac{1}{f^2}$ .
$f_A = 2f_B \Rightarrow \frac{l_A}{l_B} = \frac{1}{4}$
So the answer is (a) $l_A = \frac{l_B}{4}$ , does not depend on mass.

Report a mistake

Q340NEET 2000 · AIPMTEasy

Simple Pendulum · Energy, speed and tension of a pendulum

The bob of simple pendulum having length $l$ , is displaced from mean position to an angular position q with respect to vertical. If it is released, then velocity of bob at equilibrium position

Show answer and solution

Answer: (a)

The bob falls through height $h = l - l\cos\theta = l(1 - \cos\theta)$ (q in the question means $\theta$ ).
Energy conservation: $\frac{1}{2}mv^2 = mgh$ .
$v = \sqrt{2gl(1 - \cos\theta)}$
So the answer is (a).

Report a mistake

Q341NEET 1999 · AIPMTEasy

Simple Pendulum · Time period, length and mass

The time period of a simple pendulum is $2\text{ s}$ . If its length is increased by 4 times, then its period becomes

Show answer and solution

Answer: (d)

$T \propto \sqrt{l}$
$l' = 4l \Rightarrow T' = 2T = 2 \times 2 = 4$ s
So the answer is (d) $4$ s.

Report a mistake

Q342NEET 1998 · AIPMTEasy

Simple Pendulum · Two pendulums in phase

Two simple pendulums of length $0.5\text{ m}$ and $2.0\text{ m}$ respectively are given small linear displacement in one direction at the same time. They will again be in the same phase when the pendulum of shorter length has completed oscillations

Show answer and solution

Answer: (c)

$T \propto \sqrt{l}$ : $\frac{T_{long}}{T_{short}} = \sqrt{\frac{2.0}{0.5}} = 2$
They are in phase again when $n\,T_{short} = m\,T_{long}$ for the smallest integers: $n = 2$ , $m = 1$ .
So the answer is (c) 2.

Report a mistake

Q343NEET 1997 · AIPMTEasy

Simple Pendulum · Time period, length and mass

If the length of a simple pendulum is increased by $2\%$ , then the time period

Show answer and solution

Answer: (a)

$T \propto l^{1/2} \Rightarrow \frac{\Delta T}{T} = \frac{1}{2}\frac{\Delta l}{l}$
$\frac{\Delta T}{T} = \frac{1}{2} \times 2\% = 1\%$
So the answer is (a) increases by $1\%$ .

Report a mistake

Q344NEET 1995 · AIPMTMedium

Simple Pendulum · Energy, speed and tension of a pendulum

A simple pendulum with a bob of mass $m$ oscillates from $A$ to $C$ and back to $A$ such that $PB$ is $H$ . If the acceleration due to gravity is $g$ , then the velocity of the bob as it passes through $B$ is

Show answer and solution

Answer: (b)

At A the bob is momentarily at rest; B is lower than A by $H$ .
Energy conservation: $mgH = \frac{1}{2}mv_B^2$ .
So the answer is (b) $v_B = \sqrt{2gH}$ .

Report a mistake

Q345NEET 1994 · AIPMTEasy

Simple Pendulum · Pendulum in a lift or vehicle

A seconds pendulum is mounted in a rocket. Its period of oscillation will decrease when rocket is

Show answer and solution

Answer: (d)

$T = 2\pi\sqrt{\frac{l}{g_{eff}}}$ , so $T$ decreases only if $g_{eff} > g$ .
Down with acceleration: $g_{eff} = g - a$ (increases $T$ ). Orbit: $g_{eff} = 0$ (no oscillation). Uniform velocity: $g_{eff} = g$ (no change).
Up with acceleration: $g_{eff} = g + a > g$ , so $T$ decreases.
So the answer is (d).

Report a mistake

Q346NEET 1991 · AIPMTEasy

Simple Pendulum · Energy, speed and tension of a pendulum

The angular velocity and the amplitude of a simple pendulum is $\omega$ and $a$ respectively. At a displacement $x$ from the mean position if its kinetic energy is $T$ and potential energy is $V$ , then the ratio of $T$ to $V$ is

Show answer and solution

Answer: (c)

$T = \frac{1}{2}m\omega^2(a^2 - x^2)$ and $V = \frac{1}{2}m\omega^2x^2$ .
$\frac{T}{V} = \frac{a^2 - x^2}{x^2}$
So the answer is (c).

Report a mistake

Q347NEET 1991 · AIPMTMedium

Simple Pendulum · Pendulum in a lift or vehicle

A simple pendulum is suspended from the roof of a trolley which moves in a horizontal direction with an acceleration $a$ , then the time period is given by $T = 2\pi\sqrt{(l/g)}$ , where $g$ is equal to

Show answer and solution

Answer: (d)

In the trolley frame the bob feels gravity $g$ down and pseudo acceleration $a$ horizontally.
These are perpendicular, so $g_{eff} = \sqrt{g^2 + a^2}$ .
So the answer is (d) $\sqrt{g^2+a^2}$ .

Report a mistake

Q348JEE Main 2026 · 8 APR · SHIFT 2Easy

Spring Systems · Spring constant and cut springs

The frequency of oscillation of a mass $m$ suspended by a spring is $v_1$ . If the length of the spring is cut to half, the same mass oscillates with frequency $v_2$ . The value of $v_2/v_1$ is ______.

Show answer and solution

Answer: (c)

Half-length spring: $k_2 = 2k_1$ .
$\nu \propto \sqrt{k} \Rightarrow \frac{\nu_2}{\nu_1} = \sqrt{2}$ .
So the answer is (c) $\sqrt{2}$ .

Report a mistake

Q349JEE Main 2026 · 24 JAN · SHIFT 1Easy

Spring Systems · Spring constant and cut springs

A spring of force constant $15\text{ N/m}$ is cut into two pieces. If the ratio of their length is $1:3$ , then the force constant of smaller piece is $\rule{3em}{0.4pt}\text{ N/m}$ .

Show answer and solution

Answer: (c)

$kL = \text{constant}$ . Smaller piece has length $\frac{L}{4}$ .
$k_{small} = 4 \times 15 = 60\text{ N/m}$ .
So the answer is (c) 60 N/m.

Report a mistake

Q350MHT CET 2026 · 20 MAY · SHIFT 1Easy

Spring Systems · Time period of a spring-mass system

A spring of force constant $180$ N/m is loaded with a mass $0.20$ kg. The amplitude of oscillations is $4$ cm. When mass comes to equillibrium position, its velocity in m/s is

Show answer and solution

Answer: (b)

$\omega = \sqrt{k/m} = \sqrt{180/0.20} = 30\text{ rad/s}$ .
At equilibrium $v_{max} = A\omega = 0.04 \times 30 = 1.2\text{ m/s}$ .
So the answer is (b) $1.2$ .

Report a mistake

Q351MHT CET 2026 · 15 MAY · SHIFT 1Easy

Spring Systems · Time period of a spring-mass system

A body of mass $64$ g is made to oscillate turn by turn on two different springs A and B. Spring A and B has force constant $4$ N/m and $16$ N/m respectively. If $T_1$ and $T_2$ are the period of oscillations of springs A and B respectively, then $(T_1+T_2)/(T_1-T_2)$ will be

Show answer and solution

Answer: (c)

$T \propto \frac{1}{\sqrt{k}} \Rightarrow \frac{T_1}{T_2} = \sqrt{\frac{16}{4}} = 2$ , so $T_1 = 2T_2$ .
$\frac{T_1+T_2}{T_1-T_2} = \frac{3T_2}{T_2} = 3$ .
So the answer is (c) $3:1$ .

Report a mistake

Q352MHT CET 2026 · 19 APR · SHIFT 2Easy

Spring Systems · Time period of a spring-mass system

A spring force constant $180$ N/m is loaded with a mass $0.2$ kg. The amplitude of oscillations is $4$ cm. When mass comes to equilibrium position, its velocity is

Show answer and solution

Answer: (c)

$\omega = \sqrt{k/m} = \sqrt{180/0.2} = 30\text{ rad/s}$ .
$v_{max} = A\omega = 0.04 \times 30 = 1.2\text{ m/s}$ .
So the answer is (c) $1.2$ m/s.

Report a mistake

Q353MHT CET 2026 · 18 APR · SHIFT 2Easy

Spring Systems · Equal masses on different springs

Two particles A and B of equal masses are suspended from two massless springs of spring constants $K_A$ and $K_B$ respectively. The maximum velocities of the particles during oscillations are equal. The ratio of the amplitude of 'B' to that of 'A' is

Show answer and solution

Answer: (b)

$v_{max} = A\omega = A\sqrt{\frac{K}{m}}$ .
Equal $v_{max}$ and equal $m$ : $A\sqrt{K}$ is the same, so $A \propto \frac{1}{\sqrt{K}}$ .
$\frac{A_B}{A_A} = \sqrt{\frac{K_A}{K_B}}$
So the answer is (b) $\sqrt{K_A} : \sqrt{K_B}$ .

Report a mistake

Q354MHT CET 2026 · 20 APR · SHIFT 1Easy

Spring Systems · Series and parallel springs

Two springs of force constants ' $2k$ ' and ' $k$ ' are connected to a mass ' $m$ ' as shown. Mass is displaced slightly to one side and released. The frequency of oscillation of the two springs-mass system is

Show answer and solution

Answer: (d)

When $m$ moves by $x$ , one spring stretches and the other compresses by $x$ ; both forces act the same way, so the springs are in parallel.
$k_{eff} = 2k + k = 3k$
$f = \frac{1}{2\pi}\sqrt{\frac{3k}{m}}$
So the answer is (d).

Report a mistake

Q355MHT CET 2026 · 13 MAY · SHIFT 2Medium

Spring Systems · Spring constant and cut springs

A spring has length $l$ and force constant $k$ . It is cut into 2 springs of length $l_1$ and $l_2$ such that $l_1 = nl_2$ ( $n$ =an integer). The force constant of the spring of length $l_2$ is $\left(k \propto \frac{1}{l}\right)$

Show answer and solution

Answer: (a)

$k \propto \frac{1}{l}$ , so $k l$ is constant: $k_2 l_2 = k l$ .
$l = l_1 + l_2 = n l_2 + l_2 = (n+1)l_2$
$k_2 = \frac{k l}{l_2} = (n+1)k$
So the answer is (a) $k(n+1)$ .

Report a mistake

Q356MHT CET 2026 · 18 APR · SHIFT 2Medium

Spring Systems · Time period of a spring-mass system

Three masses 100g, 300g and 500g are suspended at the end of a spring as shown in figure and are in equilibrium. When the 500 g mass is removed, the system oscillates with a period of 3 second. When the 300 g mass is also removed, it will oscillate with a period of

Show answer and solution

Answer: (b)

$T \propto \sqrt{m}$ for the same spring.
After removing 500 g: $m = 100 + 300 = 400\text{ g}$ , $T = 3\text{ s}$ . After also removing 300 g: $m = 100\text{ g}$ .
$T' = 3\sqrt{\frac{100}{400}} = 3 \times \frac{1}{2} = 1.5\text{ s}$
So the answer is (b) $1.5$ s.

Report a mistake

Q357MHT CET 2026 · 17 APR · SHIFT 2Medium

Spring Systems · Time period of a spring-mass system

A light spring is suspended with mass ' $m_1$ ' at its lower end and its upper end is fixed to a rigid support. The mass is pulled down a short distance and then released. The period of oscillation is T second. When a mass ' ' is added to ' $m_1$ ' and the system is made to oscillate the period is found to be $\frac{3}{2}$ T. The ratio $m_2\ \left(\frac{m_1}{m_2}\right)$ is

Show answer and solution

Answer: (c)

$T \propto \sqrt{m}$ : $\left(\frac{3}{2}\right)^2 = \frac{m_1+m_2}{m_1}$ .
$\frac{9}{4} = 1 + \frac{m_2}{m_1} \Rightarrow \frac{m_2}{m_1} = \frac{5}{4}$
$m_1 : m_2 = 4 : 5$ , so the answer is (c).

Report a mistake

Q358MHT CET 2026 · 13 APR · SHIFT 1Medium

Spring Systems · Time period of a spring-mass system

A spring executes S.H.M. with mass 10 kg attached to it. The force constant of spring is 10 N/m. If at any instant its velocity is 40 cm/s, the displacement at that instant is

(Amplitude of S.H.M. is 0.5 m)

Show answer and solution

Answer: (b)

$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{10}{10}} = 1\text{ rad/s}$
$v = \omega\sqrt{A^2 - x^2}$ : $0.4 = 1 \times \sqrt{0.25 - x^2}$
$x^2 = 0.25 - 0.16 = 0.09 \Rightarrow x = 0.3\text{ m}$
So the answer is (b) $0.3$ m.

Report a mistake

Q359MHT CET 2026 · 14 MAY · SHIFT 1Medium

Spring Systems · Vertical spring and static extension

A mass is suspended from a vertical spring which is executing S.H.M. of frequency $10$ Hz. The spring is unstretched at the highest point of oscillation. Maximum speed of the mass is ( $g = 10$ m/s $^{2}$ )

Show answer and solution

Answer: (c)

Spring unstretched at the top, so amplitude = static extension: $A = \frac{mg}{k} = \frac{g}{\omega^2}$ .
$v_{max} = A\omega = \frac{g}{\omega}$ , with $\omega = 2\pi f = 20\pi$ rad/s.
$v_{max} = \frac{10}{20\pi} = \frac{1}{2\pi}$ m/s
So the answer is (c) $\frac{1}{2\pi}$ m/s.

Report a mistake

Q360MHT CET 2026 · 12 MAY · SHIFT 2Medium

Spring Systems · Vertical spring and static extension

A spring produces extension $'x'$ by applying a force $'F'$ N. A body of mass $'m'$ suspended from spring oscillates vertically with period $'T'$ . The mass of the suspended body is (Neglect mass of spring)

Show answer and solution

Answer: (a)

Spring constant $k = F/x$ .
$T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{mx}{F}}$
$m = \frac{T^2F}{4\pi^2x}$
So the answer is (a).

Report a mistake

Q361JEE Main 2026 · 6 APR · SHIFT 2Medium

Spring Systems · Vertical spring and static extension

A spring stretches by 2 mm when it is loaded with a mass of 200 g. From equilibrium position the mass is further pulled down by 2 mm and released. The frequency associated with the system and maximum energy in the spring are ______ Hz and ______ J, respectively.

(Take $g = 10\text{ m/s}^2$ )

Show answer and solution

Answer: (a)

$k = \frac{mg}{x} = \frac{0.2\times10}{2\times10^{-3}} = 1000$ N/m
$f = \frac{1}{2\pi}\sqrt{\frac{k}{m}} = \frac{\sqrt{5000}}{2\pi} = \frac{5\sqrt{50}}{\pi}$ Hz
Lowest point: total extension $= 2 + 2 = 4$ mm.
$U_{max} = \frac{1}{2}(1000)(4\times10^{-3})^2 = 8\times10^{-3}$ J
So the answer is (a).

Report a mistake

Q362MHT CET 2026 · 13 MAY · SHIFT 1Medium

Spring Systems · Series and parallel springs

$S_1$ and $S_2$ are two identical springs. The frequency of oscillation of mass m is F. If one spring is removed, the frequency will become

Show answer and solution

Answer: (d)

On displacement $x$ , one spring stretches and the other compresses; both push back, so $k_{eff} = 2k$ .
$F = \frac{1}{2\pi}\sqrt{\frac{2k}{m}}$ ; with one spring $F' = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$ .
$F' = F/\sqrt{2}$ , so the answer is (d).

Report a mistake

Q363MHT CET 2026 · 13 MAY · SHIFT 1Medium

Spring Systems · Series and parallel springs

A body has time period $T_1$ under the action of one force and $T_2$ under the action of another force. The square of the time period when both the forces are acting in the same direction is

Show answer and solution

Answer: (d)

Restoring forces $F_1 = -k_1x$ , $F_2 = -k_2x$ add: $k = k_1 + k_2$ .
$T^2 \propto \frac{1}{k}$ , so $\frac{1}{T^2} = \frac{1}{T_1^2} + \frac{1}{T_2^2}$ .
$T^2 = \frac{T_1^2T_2^2}{T_1^2+T_2^2}$ , so the answer is (d).

Report a mistake

Q364MHT CET 2026 · 12 MAY · SHIFT 1Medium

Spring Systems · Series and parallel springs

A simple harmonic oscillator having four identical springs (as shown in figure) has a time period $T_1$ . Another similar simple harmonic oscillator of the same mass $m$ but the four identical springs with spring constants $2k$ has a time period $T_2$ , then $\frac{T_1}{T_2}$ is

Show answer and solution

Answer: (c)

The arrangement is the same in both cases, so $k_{eff} \propto k$ of each spring.
$T \propto \frac{1}{\sqrt{k}}$ : $\frac{T_1}{T_2} = \sqrt{\frac{2k}{k}} = \sqrt{2}$
So the answer is (c) $\sqrt{2} : 1$ .

Report a mistake

Q365MHT CET 2026 · 18 APR · SHIFT 1Medium

Spring Systems · Series and parallel springs

All the springs in fig (a), (b) and (c) are identical, each one having force constant K. Mass m is attached to each system. If $T_a, T_b$ and $T_c$ are the periodic time of oscillations of the three systems in fig (a), (b) and (c) respectively, then

Show answer and solution

Answer: (d)

(a) single spring: $T_a = 2\pi\sqrt{m/K}$ .
(b) series, $K/2$ : $T_b = \sqrt{2}\,T_a$ . (c) parallel, $2K$ : $T_c = T_a/\sqrt{2}$ .
$\frac{T_b}{T_c} = \frac{\sqrt{2}}{1/\sqrt{2}} = 2$ , so $T_b = 2T_c$ .
So the answer is (d).

Report a mistake

Q366MHT CET 2026 · 11 APR · SHIFT 1Medium

Spring Systems · Series and parallel springs

Two massless springs of spring constants $K_1$ and $K_2$ are connected one after the other forming a single chain, suspended vertically and a certain mass is attached to the free end. If $x_1$ and $x_2$ are their respective extensions and 'F' is their stretching force, the total extension produced is

Show answer and solution

Answer: (d)

In series each spring carries the same force $F$ : $x_1 = F/K_1$ , $x_2 = F/K_2$ .
$x = x_1 + x_2 = F\left(\frac{1}{K_1} + \frac{1}{K_2}\right)$
So the answer is (d).

Report a mistake

Q367MHT CET 2026 · 15 MAY · SHIFT 2Medium

Spring Systems · Added mass and two-body systems

A horizontal spring executes S.H.M. with amplitude $A_1$ when mass $m$ is attached to it. When it passes through mean position another mass $m$ is placed on it. Both masses move together with amplitude $A_2$ . The ratio $\frac{A_1}{A_2}$ is

Show answer and solution

Answer: (b)

At the mean position momentum is conserved: $mv = 2mv' \Rightarrow v' = v/2$ .
$\omega' = \sqrt{\frac{k}{2m}} = \frac{\omega}{\sqrt{2}}$
$A_2 = \frac{v'}{\omega'} = \frac{v/2}{\omega/\sqrt{2}} = \frac{A_1}{\sqrt{2}}$
$\frac{A_1}{A_2} = \sqrt{2}$ , so the answer is (b).

Report a mistake

Q368MHT CET 2026 · 16 APR · SHIFT 2Medium

Spring Systems · Added mass and two-body systems

A mass $m_1$ performs S.H.M. with amplitude A which is connected to horizontal spring. While mass $m_1$ passing through mean position, another mass $m_2$ ( $m_2 < m_1$ ) is placed on it so that both masses move together with amplitude $A_1$ . The ratio of $A/A_1$ is

Show answer and solution

Answer: (a)

Momentum conserved: $m_1v = (m_1+m_2)v'$ .
$A = \frac{v}{\omega}$ with $\omega = \sqrt{k/m}$ , so $A \propto \frac{v}{\sqrt{k/m}} = \frac{mv}{\sqrt{km}} = \frac{p}{\sqrt{km}}$ .
Same $p$ : $\frac{A}{A_1} = \sqrt{\frac{m_1+m_2}{m_1}}$
So the answer is (a).

Report a mistake

Q369JEE Main 2026 · 28 JAN · SHIFT 2Difficult

Spring Systems · Added mass and two-body systems

As shown in the figure, a spring is kept in a stretched position with some extension by holding the masses $1$ kg and $0.2$ kg with a separation more than spring natural length and are released. Assuming the horizontal surface to be frictionless, the angular frequency (in SI unit) of the system is:

Show answer and solution

Answer: (c)

Two free masses joined by a spring oscillate with $\omega = \sqrt{\frac{k}{\mu}}$ , $\mu = \frac{m_1 m_2}{m_1+m_2}$ .
$\mu = \frac{1 \times 0.2}{1.2} = \frac{1}{6}\text{ kg}$ , $k = 150\text{ N/m}$ (from figure).
$\omega = \sqrt{150 \times 6} = \sqrt{900} = 30\text{ rad/s}$
So the answer is (c) 30.

Report a mistake

Q370MHT CET 2025 · 20 APR · SHIFT 1Easy

Spring Systems · Time period of a spring-mass system

If the period of a oscillation of mass ' m ' suspended from a spring is 2 s, then the period of suspended mass ' 4 m ' with the same spring will be

Show answer and solution

Answer: (d)

$T = 2\pi\sqrt{m/k} \Rightarrow T \propto \sqrt{m}$ .
$T' = 2\sqrt{4} = 4\text{ s}$ .
So the answer is (d) 4 s.

Report a mistake

Q371NEET 2025Easy

Spring Systems · Equal masses on different springs

Two identical point masses $P$ and $Q$ , suspended from two separate massless springs of spring constants $k_1$ and $k_2$ , respectively, oscillate vertically. If their maximum speeds are the same, the ratio ( $A_Q/A_P$ ) of the amplitude $A_Q$ of mass $Q$ to the amplitude $A_P$ of mass $P$ is

Show answer and solution

Answer: (b)

$v_{max} = A\sqrt{\frac{k}{m}}$ ; equal $v_{max}$ and $m$ give $A \propto \frac{1}{\sqrt{k}}$ .
$\frac{A_Q}{A_P} = \sqrt{\frac{k_1}{k_2}}$ (P has $k_1$ , Q has $k_2$ ).
So the answer is (b) $\sqrt{\frac{k_1}{k_2}}$ .

Report a mistake

Q372JEE Main 2025 · 29 JAN · SHIFT 2Easy

Spring Systems · Equal masses on different springs

Two bodies A and B of equal mass are suspended from two massless springs of spring constant k1 and k2, respectively. If the bodies oscillate vertically such that their amplitudes are equal, the ratio of the maximum velocity of A to the maximum velocity of B is

Show answer and solution

Answer: (b)

$v_{max} = A\sqrt{\frac{k}{m}}$ ; with equal $A$ and $m$ , $v_{max} \propto \sqrt{k}$ .
$\frac{v_A}{v_B} = \sqrt{\frac{k_1}{k_2}}$
So the answer is (b) $\sqrt{\frac{k_1}{k_2}}$ .

Report a mistake

Q373MHT CET 2025 · 26 APR · SHIFT 1Medium

Spring Systems · Time period of a spring-mass system

A spring executes S.H.M. with mass 1 kg attached to it. The force constant of the spring is $4\text{ N/m}$ . If at any instant its velocity is $20\text{ cm/s}$ , the displacement at that instant is (Amplitude of S.H.M. is 0.4 m)

Show answer and solution

Answer: (b)

$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{4}{1}} = 2\text{ rad/s}$
$v = \omega\sqrt{A^2 - x^2}$ : $0.2 = 2\sqrt{0.16 - x^2}$
$0.01 = 0.16 - x^2 \Rightarrow x^2 = 0.15$ , $x = \sqrt{0.15}\text{ m}$
So the answer is (b) $\sqrt{0.15}$ m.

Report a mistake

Q374MHT CET 2025 · 21 APR · SHIFT 1Medium

Spring Systems · Time period of a spring-mass system

A mass $m$ is suspended from a spring of negligible mass. The spring is pulled a little and then released, so that mass executes S.H.M. of time period $T$ . If the mass is increased by $m_0$ , the periodic time becomes $\frac{5T}{4}$ . The ratio $\frac{m_0}{M}$

Show answer and solution

Answer: (c)

$T \propto \sqrt{m}$ : $\left(\frac{5}{4}\right)^2 = \frac{m + m_0}{m}$ .
$\frac{25}{16} = 1 + \frac{m_0}{m} \Rightarrow \frac{m_0}{m} = \frac{9}{16}$
( $M$ in the question is the original mass $m$ .) So the answer is (c) $\frac{9}{16}$ .

Report a mistake

Q375MHT CET 2025 · 20 APR · SHIFT 2Medium

Spring Systems · Time period of a spring-mass system

A mass $x$ gram is suspended from a light spring. It is pulled in downward direction and released so that mass performs S.H.M. of period T. If mass is increased by Y gram, the period becomes $\frac{4T}{3}$ . The ratio of $Y/x$ is

Show answer and solution

Answer: (a)

$T \propto \sqrt{m}$ : $\left(\frac{4}{3}\right)^2 = \frac{x + Y}{x}$ .
$\frac{16}{9} = 1 + \frac{Y}{x} \Rightarrow \frac{Y}{x} = \frac{7}{9}$
So the answer is (a) $7 : 9$ .

Report a mistake

Q376NEET 2025Medium

Spring Systems · Time period of a spring-mass system

In an oscillating spring mass system, a spring is connected to a box filled with sand. As the box oscillates, sand leaks slowly out of the box vertically so that the average frequency $\omega(t)$ and average amplitude $A(t)$ of the system change with time $t$ . Which one of the following options schematically depicts these changes correctly?

Show answer and solution

Answer: (d)

Sand leaks out, so the mass $m$ decreases and $\omega = \sqrt{\frac{k}{m}}$ increases with time.
Sand leaving vertically takes its own share of energy; for the box, $\ddot{x} + \omega^2(t)\,x = 0$ with slowly changing $\omega$ .
For such slow change $\omega A^2$ stays constant (adiabatic invariant), so $A \propto \frac{1}{\sqrt{\omega}}$ decreases.
Graph with $\omega$ rising and $A$ falling: the answer is (d).

Report a mistake

Q377JEE Main 2025 · 8 APR · SHIFT 2Medium

Spring Systems · Time period of a spring-mass system

A block of mass 2 kg is attached to one end of a massless spring whose other end is fixed at a wall. The spring-mass system moves on a frictionless horizontal table. The spring's natural length is 2 m and spring constant is 200 N/m. The block is pushed such that the length of the spring becomes 1 m and then released. At distance x m (x < 2) from the wall, the speed of the block will be

Show answer and solution

Answer: (a)

Equilibrium is at the natural length, $x = 2$ m from the wall; compressed to 1 m, so amplitude $A = 1$ m.
$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{200}{2}} = 10\text{ rad/s}$
Displacement from equilibrium is $(2 - x)$ , so $v = \omega\sqrt{A^2 - (2-x)^2} = 10\left[1 - (2-x)^2\right]^{1/2}$ m/s.
So the answer is (a).

Report a mistake

Q378MHT CET 2025 · 19 APR · SHIFT 2Medium

Spring Systems · Vertical spring and static extension

A vertical spring oscillates with period 6 second with mass $m$ is suspended from it. When the mass is at rest, the spring is stretched through a distance of (Take, acceleration due to gravity, $g = \pi^2 = 10\text{ m/s}^2$ )

Show answer and solution

Answer: (d)

At rest $kx = mg$ , so $\frac{m}{k} = \frac{x}{g}$ and $T = 2\pi\sqrt{\frac{x}{g}}$ .
$x = \frac{gT^2}{4\pi^2} = \frac{10\times36}{4\times10} = 9$ m
So the answer is (d) 9 m.

Report a mistake

Q379MHT CET 2025 · 19 APR · SHIFT 1Medium

Spring Systems · Vertical spring and static extension

A mass suspended from a vertical spring performs S.H.M. of period 0.1 second. The spring is unstretched at the highest point of suspension. Maximum speed of the mass is (Gravitational acceleration $g = 10\text{ m/s}^2$ )

Show answer and solution

Answer: (a)

Spring unstretched at the top, so $A = \frac{g}{\omega^2}$ and $v_{max} = A\omega = \frac{g}{\omega}$ .
$\omega = \frac{2\pi}{T} = \frac{2\pi}{0.1} = 20\pi$ rad/s
$v_{max} = \frac{10}{20\pi} = \frac{1}{2\pi}$ m/s
So the answer is (a).

Report a mistake

Q380MHT CET 2025 · 5 MAY · SHIFT 2Medium

Spring Systems · Series and parallel springs

All the springs in fig. (a), (b) and (c) are identical, each having force constant K each. Mass m is attached to each system. If $T_a, T_b$ and $T_c$ are the time periods of oscillations of the three systems in fig. (a), (b) and (c) respectively, then

Show answer and solution

Answer: (d)

(a) single spring: $T_a = 2\pi\sqrt{m/K}$ .
(b) series, $K/2$ : $T_b = \sqrt{2}\,T_a$ . (c) parallel, $2K$ : $T_c = T_a/\sqrt{2}$ .
$T_b/T_c = 2$ , so $T_b = 2T_c$ .
So the answer is (d).

Report a mistake

Q381MHT CET 2025 · 25 APR · SHIFT 2Medium

Spring Systems · Series and parallel springs

As shown in the figure, $S_1$ and $S_2$ are identical springs with spring constant K each. The oscillation frequency of the mass ' $m$ ' is ' $f$ '. If the spring $S_2$ is removed, the oscillation frequency will become

Show answer and solution

Answer: (c)

Both springs act on $m$ (one stretched, one compressed), so $k_{eff} = 2K$ and $f = \frac{1}{2\pi}\sqrt{\frac{2K}{m}}$ .
With $S_2$ removed, $f' = \frac{1}{2\pi}\sqrt{\frac{K}{m}} = \frac{f}{\sqrt{2}}$
So the answer is (c).

Report a mistake

Q382MHT CET 2025 · 26 APR · SHIFT 2Medium

Spring Systems · Added mass and two-body systems

A mass ' $M$ ' attached to a horizontal spring executes S.H.M. of amplitude $A_1$ . When the mass M passes through its mean position, then a smaller mass ' $m$ ' is placed over it and both of them move together with amplitude $A_2$ . The ratio $\left(\frac{A_1}{A_2}\right)$ is

Show answer and solution

Answer: (c)

Momentum conserved at the mean position: $Mv = (M+m)v'$ .
$A = \frac{v}{\omega} = \frac{p}{\sqrt{k\,(\text{mass})}}$ , and $p$ is unchanged.
$\frac{A_1}{A_2} = \sqrt{\frac{M+m}{M}}$ , so the answer is (c).

Report a mistake

Q383JEE Main 2025 · 3 APR · SHIFT 1Medium

Spring Systems · Added mass and two-body systems

Two blocks of masses $m$ and $M, (M > m)$ , are placed on a frictionless table as shown in figure. A massless spring with spring constant k is attached with the lower block. If the system is slightly displaced and released, then ( $\mu =$ coefficient of friction between the two blocks) A. The time period of small oscillation of the two blocks is $T = 2\pi\sqrt{\frac{(m+M)}{k}}$

B. The acceleration of the blocks is $a = -\frac{kx}{M+m}$ ( $x =$ displacement of the blocks from the mean position)

C. The magnitude of the frictional force on the upper block is $\frac{m\mu|x|}{M+m}$

D. The maximum amplitude of the upper block, if it does not slip, is $\frac{\mu(M+m)g}{k}$

E. Maximum frictional force can be $\mu(M+m)g$ .

Choose the correct answer from the options given below:

Show answer and solution

Answer: (c)

A: both blocks move together with mass $M+m$ : $T = 2\pi\sqrt{\frac{M+m}{k}}$ . True.
B: $a = -\frac{kx}{M+m}$ . True.
C: friction on $m$ is $m|a| = \frac{mk|x|}{M+m}$ , not $\frac{m\mu|x|}{M+m}$ . False.
D: no slip needs $\frac{mkA}{M+m} \le \mu mg \Rightarrow A_{max} = \frac{\mu(M+m)g}{k}$ . True.
E: maximum static friction on $m$ is $\mu mg$ , not $\mu(M+m)g$ . False.
So the answer is (c) A, B, D only.

Report a mistake

Q384MHT CET 2024 · 2 MAY · SHIFT 2Easy

Spring Systems · Spring constant and cut springs

A spring has a certain mass suspended from it and its period of vertical oscillations is $T_1$ . The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillations is now $T_2$ . The ratio of $T_2/T_1$ is

Show answer and solution

Answer: (b)

$k \propto \frac{1}{\text{length}}$ , so each half has $k' = 2k$ .
$T = 2\pi\sqrt{m/k} \Rightarrow \frac{T_2}{T_1} = \sqrt{\frac{k}{2k}} = \frac{1}{\sqrt{2}}$ .
So the answer is (b) $1:\sqrt{2}$ .

Report a mistake

Q385JEE Main 2024 · 1 FEB · SHIFT 2Easy

Spring Systems · Time period of a spring-mass system

A mass $m$ is suspended from a spring of negligible mass and the system oscillates with a frequency $f_1$ . The frequency of oscillations if a mass $9m$ is suspended from the same spring is $f_2$ . The value of $\frac{f_1}{f_2}$ is ________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 3

$f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$ , so $f \propto \frac{1}{\sqrt{m}}$ .
$\frac{f_1}{f_2} = \sqrt{\frac{9m}{m}} = 3$
So $\frac{f_1}{f_2} = 3$ .

Report a mistake

Q386MHT CET 2024 · 4 MAY · SHIFT 1Easy

Spring Systems · Equal masses on different springs

Two bodies A and B of equal mass are suspended from two separate massless springs of spring constants $K_1$ and $K_2$ respectively. The two bodies oscillate vertically such that their maximum velocities are equal. The ratio of the amplitude of $B$ to that of $A$ is

Show answer and solution

Answer: (c)

$v_{max} = A\sqrt{\frac{K}{m}}$ ; equal $v_{max}$ and $m$ give $A \propto \frac{1}{\sqrt{K}}$ .
$\frac{A_B}{A_A} = \sqrt{\frac{K_1}{K_2}}$ (A has $K_1$ , B has $K_2$ ).
So the answer is (c) $\sqrt{\frac{K_1}{K_2}}$ .

Report a mistake

Q387MHT CET 2024 · 4 MAY · SHIFT 2Medium

Spring Systems · Time period of a spring-mass system

A mass ' $m$ ' attached to a spring oscillates with a period of 3 second. If the mass is increased by 0.6 kg, the period increases by 3 second. The initial mass ' $m$ ' is equal to

Show answer and solution

Answer: (b)

Period goes from 3 s to 6 s. $T \propto \sqrt{m}$ : $\left(\frac{6}{3}\right)^2 = \frac{m + 0.6}{m}$ .
$4m = m + 0.6 \Rightarrow m = 0.2\text{ kg}$
So the answer is (b) 0.2 kg.

Report a mistake

Q388MHT CET 2024 · 2 MAY · SHIFT 1Medium

Spring Systems · Time period of a spring-mass system

Three masses $500\text{ g}, 300\text{ g}$ and 100 g are suspended at the end of spring as shown in figure and are in equilibrium. When the 500 g mass is removed, the system oscillates with a period of 3 second. When the 300 g mass is also removed it will oscillate with a period of

Show answer and solution

Answer: (b)

$T \propto \sqrt{m}$ for the same spring.
After removing 500 g: $m = 400\text{ g}$ , $T = 3\text{ s}$ . After also removing 300 g: $m = 100\text{ g}$ .
$T' = 3\sqrt{\frac{100}{400}} = 1.5\text{ s}$
So the answer is (b) 1.5 s.

Report a mistake

Q389MHT CET 2024 · 3 MAY · SHIFT 2Medium

Spring Systems · Vertical spring and static extension

Frequency of a particle performing S.H.M. is 10 Hz. The particle is suspended from a vertical spring. At the highest point of its oscillation the spring is unstretched. Maximum speed of the particle is $(g = 10\text{ m/s}^2)$

Show answer and solution

Answer: (b)

Spring unstretched at the top, so $A = \frac{g}{\omega^2}$ and $v_{max} = \frac{g}{\omega}$ .
$\omega = 2\pi f = 20\pi$ rad/s
$v_{max} = \frac{10}{20\pi} = \frac{1}{2\pi}$ m/s
So the answer is (b).

Report a mistake

Q390MHT CET 2024 · 11 MAY · SHIFT 2Medium

Spring Systems · Series and parallel springs

All the springs in fig. (a), (b) and (c) are identical, each having force constant K. Mass attached to each system is ' $m$ '. If $T_a, T_b$ and $T_c$ are the time periods of oscillations of the three systems respectively, then

Show answer and solution

Answer: (d)

(a) single spring: $T_a = 2\pi\sqrt{m/K}$ .
(b) series, $K/2$ : $T_b = \sqrt{2}\,T_a$ . (c) parallel, $2K$ : $T_c = T_a/\sqrt{2}$ .
$T_b/T_c = 2$ , so $T_b = 2T_c$ .
So the answer is (d).

Report a mistake

Q391JEE Main 2024 · 31 JAN · SHIFT 2Medium

Spring Systems · Series and parallel springs

The time period of simple harmonic motion of mass $M$ in the given figure is $\pi\sqrt{\frac{\alpha M}{5k}}$ , where the value of $\alpha$ is _________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 12

Right branch: two $k$ in parallel give $2k$ , in series with $k$ : $\frac{2k\cdot k}{3k} = \frac{2k}{3}$ .
This branch is in parallel with the left spring $k$ : $k_{eff} = k + \frac{2k}{3} = \frac{5k}{3}$ .
$T = 2\pi\sqrt{\frac{3M}{5k}} = \pi\sqrt{\frac{12M}{5k}}$
$\alpha = 12$

Report a mistake

Q392MHT CET 2023 · 9 MAY · SHIFT 2Easy

Spring Systems · Spring constant and cut springs

A spring has a certain mass suspended from it and its period for vertical oscillations is ' $T_1$ '. The spring is now cut in to two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillations is now ' $T_2$ '. The ratio $T_1/T_2$ is

Show answer and solution

Answer: (b)

Half spring: $k' = 2k$ .
$\frac{T_1}{T_2} = \sqrt{\frac{k'}{k}} = \sqrt{2}$ .
So the answer is (b) $\sqrt{2}$ .

Report a mistake

Q393JEE Main 2023 · 30 JAN · SHIFT 2Easy

Spring Systems · Time period of a spring-mass system

For a simple harmonic motion in a mass spring system shown, the surface is frictionless. When the mass of the block is $1\text{ kg}$ , the angular frequency is $\omega_1$ . When the mass block is $2\text{ kg}$ the angular frequency is $\omega_2$ . The ratio $\omega_2/\omega_1$ is

Show answer and solution

Answer: (a)

Same spring, frictionless surface: $\omega = \sqrt{\frac{k}{m}}$ , so $\omega \propto \frac{1}{\sqrt{m}}$ .
$\frac{\omega_2}{\omega_1} = \sqrt{\frac{m_1}{m_2}} = \sqrt{\frac{1}{2}}$
So the answer is (a) $1/\sqrt{2}$ .

Report a mistake

Q394MHT CET 2023 · 12 MAY · SHIFT 1Easy

Spring Systems · Series and parallel springs

Four massless springs whose force constants are $2K,\ 2K,\ K$ and $2K$ respectively are attached to a mass $M$ kept on a frictionless plane as shown in figure. If mass $M$ is displaced in horizontal direction then frequency of oscillating system is

Show answer and solution

Answer: (b)

Left side: $2K$ and $2K$ in series, $k_L = \frac{2K \cdot 2K}{4K} = K$ .
Right side: $K$ and $2K$ side by side (parallel), $k_R = 3K$ .
Both sides act on $M$ together: $k_{eff} = K + 3K = 4K$ .
$f = \frac{1}{2\pi}\sqrt{\frac{4K}{M}}$ , so the answer is (b).

Report a mistake

Q395JEE Main 2023 · 6 APR · SHIFT 1Easy

Spring Systems · Series and parallel springs

A mass $m$ is attached to two springs as shown in figure. The spring constants of two springs are $K_1$ and $K_2$ . For the frictionless surface, the time period of oscillation of mass $m$ is:

Show answer and solution

Answer: (a)

Displacing $m$ stretches one spring and compresses the other; both restoring forces act together, so they are in parallel.
$k_{eff} = K_1 + K_2$
$T = 2\pi\sqrt{\frac{m}{K_1+K_2}}$ , so the answer is (a).

Report a mistake

Q396JEE Main 2023 · 24 JAN · SHIFT 1Easy

Spring Systems · Series and parallel springs

A block of mass 2 kg is attached with two identical springs of spring constant 20 N/m each. The block is placed on a frictionless surface and the ends of the springs are attached to rigid supports (see figure). When the mass is displaced from its equilibrium position, it executes a simple harmonic motion. The time period of oscillation is $\frac{\pi}{\sqrt{x}}$ in SI unit. The value of $x$ is ____________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 5

Springs on both sides act in parallel: $k_{eff} = 20 + 20 = 40\text{ N/m}$ .
$T = 2\pi\sqrt{\frac{2}{40}} = \frac{2\pi}{\sqrt{20}} = \frac{2\pi}{2\sqrt{5}} = \frac{\pi}{\sqrt{5}}$
Comparing with $\frac{\pi}{\sqrt{x}}$ : $x = 5$ .

Report a mistake

Q397MHT CET 2023 · 14 MAY · SHIFT 2Medium

Spring Systems · Time period of a spring-mass system

A light spring is suspended with mass $m_1$ at its lower end and its upper end fixed to a rigid support. The mass is pulled down a short distance and then released. The period of oscillation is $T$ second. When a mass $m_2$ is added to $m_1$ and the system is made to oscillate, the period is found to be $\frac{3}{2}T$ . The ratio $m_1 : m_2$ is

Show answer and solution

Answer: (c)

$T \propto \sqrt{m}$ : $\left(\frac{3}{2}\right)^2 = \frac{m_1+m_2}{m_1}$ .
$\frac{9}{4} = 1 + \frac{m_2}{m_1} \Rightarrow \frac{m_2}{m_1} = \frac{5}{4}$
So $m_1 : m_2 = 4 : 5$ , answer (c).

Report a mistake

Q398MHT CET 2023 · 11 MAY · SHIFT 2Medium

Spring Systems · Time period of a spring-mass system

The upper end of the spring is fixed and a mass ' $m$ ' is attached to its lower end. When mass is slightly pulled down and released, it oscillates with time period 3 second. If mass ' $m$ ' is increased by $1\text{ kg}$ , the time period becomes 5 second. The value of ' $m$ ' is (mass of spring is negligible)

Show answer and solution

Answer: (d)

$T \propto \sqrt{m}$ : $\left(\frac{5}{3}\right)^2 = \frac{m+1}{m}$ .
$25m = 9m + 9 \Rightarrow m = \frac{9}{16}\text{ kg}$
So the answer is (d) $\frac{9}{16}$ kg.

Report a mistake

Q399JEE Main 2023 · 24 JAN · SHIFT 2Medium

Spring Systems · Time period of a spring-mass system

A mass m attached to free end of a spring executes SHM with a period of 1s. If the mass is increased by 3 kg the period of the oscillation increases by one second, the value of mass m is ___________ kg.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 1

$T = 2\pi\sqrt{m/k}$ , so $T \propto \sqrt{m}$ .
$\frac{T_2}{T_1} = \frac{2}{1} = \sqrt{\frac{m+3}{m}} \Rightarrow m + 3 = 4m$
$m = 1$ kg

Report a mistake

Q400MHT CET 2023 · 13 MAY · SHIFT 2Medium

Spring Systems · Vertical spring and static extension

A mass $M$ is suspended from a light spring. An additional mass $M_1$ added extends the spring further by a distance $x$ . Now, the combined mass will oscillate on the spring with period $T =$

Show answer and solution

Answer: (b)

Extra mass $M_1$ gives extra extension $x$ : $k = \frac{M_1 g}{x}$ .
Total mass $M + M_1$ : $T = 2\pi\sqrt{\frac{M+M_1}{k}} = 2\pi\sqrt{\frac{(M+M_1)x}{M_1 g}}$
So the answer is (b).

Report a mistake

Q401MHT CET 2023 · 14 MAY · SHIFT 2Medium

Spring Systems · Series and parallel springs

A uniform circular disc of mass $12\text{ kg}$ is held by two identical springs. When the disc is slightly pressed down and released, it executes S.H.M. of period 2 second. The force constant of each spring is (nearly) (Take $\pi^2 = 10$ )

Show answer and solution

Answer: (c)

The two springs support the disc in parallel: $k_{eff} = 2k$ .
$T = 2\pi\sqrt{\frac{m}{2k}} \Rightarrow k = \frac{2\pi^2 m}{T^2}$
$k = \frac{2\times10\times12}{4} = 60$ N/m
So the answer is (c).

Report a mistake

Q402MHT CET 2023 · 13 MAY · SHIFT 2Medium

Spring Systems · Series and parallel springs

Under the influence of force $F_1$ the body oscillates with a period $T_1$ and due to another force $F_2$ body oscillates with period $T_2$ . If both forces acts simultaneously, then the resultant period is (consider displacement is same in all three cases)

Show answer and solution

Answer: (b)

Same displacement, so restoring forces add: $k = k_1 + k_2$ , and $k \propto \frac{1}{T^2}$ .
$\frac{1}{T^2} = \frac{1}{T_1^2} + \frac{1}{T_2^2}$
$T = \sqrt{\frac{T_1^2T_2^2}{T_1^2+T_2^2}}$ , so the answer is (b).

Report a mistake

Q403JEE Main 2023 · 31 JAN · SHIFT 1Medium

Spring Systems · Series and parallel springs

In the figure given below, a block of mass $M = 490\text{ g}$ placed on a frictionless table is connected with two springs having same spring constant $\left(K = 2\text{ N m}^{-1}\right)$ . If the block is horizontally displaced through ' $X$ ' $\text{m}$ then the number of complete oscillations it will make in $14\pi$ seconds will be _____________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 20

Both springs push back on the block: $k_{eff} = 2K = 4$ N/m.
$T = 2\pi\sqrt{\frac{0.49}{4}} = 2\pi(0.35) = 0.7\pi$ s
Number of oscillations $= \frac{14\pi}{0.7\pi} = 20$

Report a mistake

Q404JEE Main 2022 · 24 JUN · SHIFT 2Medium

Spring Systems · Equal masses on different springs

Two massless springs with spring constants 2 k and 9 k, carry 50 g and 100 g masses at their free ends. These two masses oscillate vertically such that their maximum velocities are equal. Then, the ratio of their respective amplitudes will be:

Show answer and solution

Answer: (b)

$v_{max} = A\omega$ , so equal $v_{max}$ gives $\frac{A_1}{A_2} = \frac{\omega_2}{\omega_1}$ .
$\omega_1^2 = \frac{2k}{0.05} = 40k$ , $\omega_2^2 = \frac{9k}{0.1} = 90k$
$\frac{A_1}{A_2} = \sqrt{\frac{90}{40}} = \frac{3}{2}$
So the answer is (b) 3 : 2.

Report a mistake

Q405JEE Main 2022 · 26 JUL · SHIFT 2Medium

Spring Systems · Series and parallel springs

As per given figures, two springs of spring constants $k$ and $2k$ are connected to mass $m$ . If the period of oscillation in figure (a) is $3\text{ s}$ , then the period of oscillation in figure (b) will be $\sqrt{x}\text{ s}$ . The value of $x$ is ___________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 2

(a) series: $k_a = \frac{k\cdot2k}{3k} = \frac{2k}{3}$ . (b) parallel: $k_b = 3k$ .
$\frac{T_b}{T_a} = \sqrt{\frac{k_a}{k_b}} = \sqrt{\frac{2}{9}} = \frac{\sqrt{2}}{3}$
$T_b = 3\times\frac{\sqrt{2}}{3} = \sqrt{2}$ s, so $x = 2$

Report a mistake

Q406JEE Main 2022 · 25 JUL · SHIFT 1Medium

Spring Systems · Added mass and two-body systems

In figure $(A)$ , mass ' $2m$ ' is fixed on mass ' $m$ ' which is attached to two springs of spring constant $k$ . In figure (B), mass ' $m$ ' is attached to two springs of spring constant ' $k$ ' and ' $2k$ . If mass ' $m$ ' in (A) and in (B) are displaced by distance ' $x$ horizontally and then released, then time period $T_1$ and $T_2$ corresponding to $(A)$ and (B) respectively follow the relation.

Show answer and solution

Answer: (a)

(A) mass $3m$ , both springs act: $k = 2k$ , so $T_1 = 2\pi\sqrt{\frac{3m}{2k}}$ .
(B) mass $m$ , $k_{eff} = k + 2k = 3k$ , so $T_2 = 2\pi\sqrt{\frac{m}{3k}}$ .
$\frac{T_1}{T_2} = \sqrt{\frac{3m}{2k}\cdot\frac{3k}{m}} = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}}$
So the answer is (a).

Report a mistake

Q407JEE Main 2022 · 27 JUL · SHIFT 1Medium

Spring Systems · Added mass and two-body systems

A mass $0.9\text{ kg}$ , attached to a horizontal spring, executes SHM with an amplitude $A_1$ . When this mass passes through its mean position, then a smaller mass of $124\text{ g}$ is placed over it and both masses move together with amplitude $A_2$ . If the ratio $\frac{A_1}{A_2}$ is $\frac{\alpha}{\alpha-1}$ , then the value of $\alpha$ will be ___________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 16

At the mean position momentum is conserved: $Mv_1 = (M+m)v_2 \Rightarrow \frac{v_1}{v_2} = \frac{M+m}{M}$ .
Amplitude $A = \frac{v_{max}}{\omega}$ with $\omega = \sqrt{k/\text{mass}}$ , so $\frac{\omega_2}{\omega_1} = \sqrt{\frac{M}{M+m}}$ .
$\frac{A_1}{A_2} = \frac{v_1}{v_2}\cdot\frac{\omega_2}{\omega_1} = \sqrt{\frac{M+m}{M}} = \sqrt{\frac{1.024}{0.9}} = \frac{16}{15}$
$\frac{\alpha}{\alpha-1} = \frac{16}{15} \Rightarrow \alpha = 16$

Report a mistake

Q408NEET 2021Easy

Spring Systems · Time period of a spring-mass system

A spring is stretched by 5 cm by a force 10 N. The time period of the oscillations when a mass of 2 kg is suspended by it is -

Show answer and solution

Answer: (a)

$k = \frac{F}{x} = \frac{10}{0.05} = 200\text{ N/m}$ .
$T = 2\pi\sqrt{\frac{2}{200}} = 2\pi \times 0.1 = 0.628\text{ s}$ .
So the answer is (a) 0.628 s.

Report a mistake

Q409JEE Main 2021 · 17 MAR · SHIFT 2Easy

Spring Systems · Equal masses on different springs

Two particles A and B of equal masses are suspended from two massless springs of spring constants K1 and K2 respectively. If the maximum velocities during oscillations are equal, the ratio of the amplitude of A and B is

Show answer and solution

Answer: (d)

$v_{max} = A\sqrt{\frac{K}{m}}$ ; equal $v_{max}$ and $m$ give $A \propto \frac{1}{\sqrt{K}}$ .
$\frac{A_A}{A_B} = \sqrt{\frac{K_2}{K_1}}$
So the answer is (d) $\sqrt{\frac{K_2}{K_1}}$ .

Report a mistake

Q410MHT CET 2021 · 24 SEP · SHIFT 1Easy

Spring Systems · Series and parallel springs

Two identical springs of constant ' $K$ ' are connected in series and parallel in shown in figure. A mass ' $M$ ' is suspended from them. The ratio of their frequencies is series to parallel combination will be

Show answer and solution

Answer: (a)

Series: $k_s = \frac{K \cdot K}{2K} = \frac{K}{2}$ . Parallel: $k_p = 2K$ .
$f \propto \sqrt{k}$ , so $\frac{f_s}{f_p} = \sqrt{\frac{K/2}{2K}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$
So the answer is (a) $1:2$ .

Report a mistake

Q411JEE Main 2021 · 26 FEB · SHIFT 1Easy

Spring Systems · Series and parallel springs

If two similar springs each of spring constant K1 are joined in series, the new spring constant and time period would be changed by a factor:

Show answer and solution

Answer: (c)

Two equal springs in series: $k' = \frac{K_1 \cdot K_1}{2K_1} = \frac{K_1}{2}$ , factor $\frac{1}{2}$ .
$T \propto \frac{1}{\sqrt{k}}$ , so $\frac{T'}{T} = \sqrt{2}$ .
So the answer is (c) $\frac{1}{2},\ \sqrt{2}$ .

Report a mistake

Q412JEE Main 2021 · 25 FEB · SHIFT 2Easy

Spring Systems · Series and parallel springs

Two identical springs of spring constant '2k' are attached to a block of mass m and to fixed support (see figure). When the mass is displaced from equilibrium position on either side, it executes simple harmonic motion. The time period of oscillations of this system is:

Show answer and solution

Answer: (b)

Springs on both sides of the block act in parallel: $k_{eff} = 2k + 2k = 4k$ .
$T = 2\pi\sqrt{\frac{m}{4k}} = \pi\sqrt{\frac{m}{k}}$
So the answer is (b) $\pi\sqrt{\frac{m}{k}}$ .

Report a mistake

Q413JEE Main 2021 · 17 MAR · SHIFT 1Easy

Spring Systems · Series and parallel springs

Consider two identical springs each of spring constant k and negligible mass compared to the mass M as shown. Fig. 1 shows one of them and Fig. 2 shows their series combination. The ratios of time period of oscillation of the two SHM is Tb/Ta = $\sqrt{x}$ , where value of x is ___________. (Round off to the Nearest Integer)

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 2

Fig. 1: $T_a = 2\pi\sqrt{\frac{M}{k}}$ .
Fig. 2 (series): $k_{eff} = \frac{k}{2}$ , $T_b = 2\pi\sqrt{\frac{2M}{k}}$ .
$\frac{T_b}{T_a} = \sqrt{2}$ , so $x = 2$ .

Report a mistake

Q414MHT CET 2021 · 20 SEP · SHIFT 1Medium

Spring Systems · Time period of a spring-mass system

A mass ' $m_1$ ' is suspended from a spring of negligible mass. A spring is pulled slightly in downward direction and released, mass performs S.H.M. of period ' $T_1$ '. If the mass is increased by ' $m_2$ ', the time period becomes ' $T_2$ '. The ratio $\frac{m_2}{m_1}$ is

Show answer and solution

Answer: (c)

$T \propto \sqrt{m}$ : $\frac{T_2^2}{T_1^2} = \frac{m_1 + m_2}{m_1} = 1 + \frac{m_2}{m_1}$ .
$\frac{m_2}{m_1} = \frac{T_2^2}{T_1^2} - 1 = \frac{T_2^2 - T_1^2}{T_1^2}$
So the answer is (c).

Report a mistake

Q415JEE Main 2021 · 1 SEP · SHIFT 2Medium

Spring Systems · Time period of a spring-mass system

A mass of 5 kg is connected to a spring. The potential energy curve of the simple harmonic motion executed by the system is shown in the figure. A simple pendulum of length 4 m has the same period of oscillation as the spring system. What is the value of acceleration due to gravity on the planet where these experiments are performed?

Show answer and solution

Answer: (c)

From the graph, equilibrium at $x = 2$ m and $U = 10$ J at the turning points, so $A = 2$ m.
$\frac{1}{2}kA^2 = 10 \Rightarrow \frac{1}{2}k(2)^2 = 10 \Rightarrow k = 5\text{ N/m}$
$T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{5}{5}} = 2\pi\text{ s}$
Pendulum: $2\pi\sqrt{\frac{4}{g}} = 2\pi \Rightarrow g = 4\text{ m/s}^2$
So the answer is (c) 4 m/s $^2$ .

Report a mistake

Q416JEE Main 2021 · 22 JUL · SHIFT 2Medium

Spring Systems · Time period of a spring-mass system

The motion of a mass on a spring, with spring constant K is as shown in figure.

The equation of motion is given by

x(t) = A sin $\omega$ t + B cos $\omega$ t with $\omega = \sqrt{\frac{K}{m}}$

Suppose that at time t = 0, the position of mass is x(0) and velocity v(0), then its displacement can also be represented as x(t) = C cos( $\omega$ t $-\phi$ ), where C and $\phi$ are:

Show answer and solution

Answer: (c)

$x(t) = C\cos(\omega t - \phi)$ , so $v(t) = -C\omega\sin(\omega t - \phi)$ .
At $t = 0$ : $x(0) = C\cos\phi$ and $v(0) = C\omega\sin\phi$ .
Square and add: $C = \sqrt{x(0)^2 + \frac{v(0)^2}{\omega^2}}$ . Divide: $\tan\phi = \frac{v(0)}{x(0)\omega}$ .
So the answer is (c).

Report a mistake

Q417MHT CET 2021 · 20 SEP · SHIFT 2Medium

Spring Systems · Vertical spring and static extension

A particle is suspended from a vertical spring which is executing S.H.M. of frequency $5\text{ Hz}$ . The spring is unstretched at the highest point of oscillation. Maximum speed of the particle is $(g = 10\text{ m/s}^2)$

Show answer and solution

Answer: (a)

Spring unstretched at the top, so $A = \frac{g}{\omega^2}$ and $v_{max} = \frac{g}{\omega}$ .
$\omega = 2\pi f = 10\pi$ rad/s
$v_{max} = \frac{10}{10\pi} = \frac{1}{\pi}$ m/s
So the answer is (a).

Report a mistake

Q418MHT CET 2021 · 23 SEP · SHIFT 2Medium

Spring Systems · Series and parallel springs

A body executes SHM under the action of force ' $F_1$ ' with time period ' $T_1$ '. If the force is changed to ' $F_2$ ', it executes SHM with period ' $T_2$ '. If both the forces ' $F_1$ ' and ' $F_2$ ' act simultaneously in the same direction on the body, its time period is

Show answer and solution

Answer: (d)

Forces add: $k = k_1 + k_2$ , and $k \propto \frac{1}{T^2}$ .
$\frac{1}{T^2} = \frac{1}{T_1^2} + \frac{1}{T_2^2}$
$T = \frac{T_1T_2}{\sqrt{T_1^2+T_2^2}}$ , so the answer is (d).

Report a mistake

Q419MHT CET 2021 · 22 SEP · SHIFT 2Medium

Spring Systems · Series and parallel springs

A particle connected to the end of a spring executes S.H.M. with period ' $T_1$ '. While the corresponding period for another spring is ' $T_2$ '. If the period of oscillation with two springs in series is 'T', then

Show answer and solution

Answer: (a)

Series: $\frac{1}{k} = \frac{1}{k_1} + \frac{1}{k_2}$ .
$T^2 = \frac{4\pi^2 m}{k} \propto \frac{1}{k}$ , so $T^2 = T_1^2 + T_2^2$ .
So the answer is (a) $T = \sqrt{T_1^2+T_2^2}$ .

Report a mistake

Q420MHT CET 2021 · 20 SEP · SHIFT 2Medium

Spring Systems · Series and parallel springs

A body performs S.H.M. under the action of force ' $F_1$ ' with period ' $T_1$ ' second. If the force is changed to ' $F_2$ ' it performs S.H.M. with period ' $T_2$ ' second. If both forces ' $F_1$ ' and ' $F_2$ ' act simultaneously in the same direction on the body, the period in second will be

Show answer and solution

Answer: (c)

Forces add: $k = k_1 + k_2$ , and $k \propto \frac{1}{T^2}$ .
$\frac{1}{T^2} = \frac{1}{T_1^2} + \frac{1}{T_2^2}$
$T = \frac{T_1T_2}{\sqrt{T_1^2+T_2^2}}$ , so the answer is (c).

Report a mistake

Q421JEE Main 2021 · 24 FEB · SHIFT 2Medium

Spring Systems · Series and parallel springs

In the given figure, a body of mass M is held between two massless springs, on a smooth inclined plane. The free ends of the springs are attached to firm supports. If each spring has spring constant k, the frequency of oscillation of given body is:

Show answer and solution

Answer: (c)

Gravity ( $Mg\sin\alpha$ ) only shifts the equilibrium; it does not change $k$ .
For displacement $x$ along the incline, one spring stretches and the other compresses: restoring force $2kx$ .
$f = \frac{1}{2\pi}\sqrt{\frac{2k}{M}}$ , so the answer is (c).

Report a mistake

Q422JEE Main 2021 · 24 FEB · SHIFT 1Medium

Spring Systems · Added mass and two-body systems

In the given figure, a mass M is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is k. The mass oscillates on a frictionless surface with time period T and amplitude A. When the mass is in equilibrium position, as shown in the figure, another mass m is gently fixed upon it. The new amplitude of oscillations will be:

Show answer and solution

Answer: (d)

Momentum conserved at equilibrium: $Mv = (M+m)v'$ .
$A = \frac{v}{\omega} = \frac{p}{\sqrt{k\,(\text{mass})}}$ with $p$ unchanged.
$A' = A\sqrt{\frac{M}{M+m}}$ , so the answer is (d).

Report a mistake

Q423JEE Main 2021 · 25 JUL · SHIFT 1Difficult

Spring Systems · Added mass and two-body systems

In the reported figure, two bodies A and B of masses 200 g and 800 g are attached with the system of springs. Springs are kept in a stretched position with some extension when the system is released. The horizontal surface is assumed to be frictionless. The angular frequency will be ____________ rad/s when k = 20 N/m.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 10

Springs $k$ and $4k$ in series: $k_{eq} = \frac{k \cdot 4k}{5k} = \frac{4k}{5} = 16\text{ N/m}$ .
Reduced mass $\mu = \frac{0.2 \times 0.8}{1.0} = 0.16\text{ kg}$ .
$\omega = \sqrt{\frac{k_{eq}}{\mu}} = \sqrt{\frac{16}{0.16}} = 10\text{ rad/s}$

Report a mistake

Q424MHT CET 2020 · 19 OCT · SHIFT 2Easy

Spring Systems · Time period of a spring-mass system

A body of mass 64 g is made to oscillate turn by turn on two different springs $A$ and $B$ . Spring $A$ and $B$ has force constant $4\,\frac{\text{N}}{\text{m}}$ and $16\,\frac{\text{N}}{\text{m}}$ respectively. If $T_1$ and $T_2$ are period of oscillations of springs $A$ and $B$ respectively, then $\frac{T_1+T_2}{T_1-T_2}$ will be

Show answer and solution

Answer: (c)

$T \propto \frac{1}{\sqrt{k}} \Rightarrow \frac{T_1}{T_2} = \sqrt{\frac{16}{4}} = 2$ .
$\frac{T_1+T_2}{T_1-T_2} = \frac{2+1}{2-1} = 3$ .
So the answer is (c) $3:1$ .

Report a mistake

Q425MHT CET 2020 · 16 OCT · SHIFT 2Easy

Spring Systems · Vertical spring and static extension

A block of mass $m$ attached to one end of the vertical spring produces extension $x$ . If the block is pulled and released, the periodic time of oscillation is

Show answer and solution

Answer: (d)

At equilibrium $kx = mg$ , so $\frac{m}{k} = \frac{x}{g}$ .
$T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{x}{g}}$
So the answer is (d) $2\pi\sqrt{\frac{x}{g}}$ .

Report a mistake

Q426MHT CET 2020 · 19 OCT · SHIFT 2Easy

Spring Systems · Equal masses on different springs

Two bodies $A$ and $B$ of equal mass are suspended from two separate massles springs of force constant $k_1$ and $k_2$ , respectively. The bodies oscillate vertically such that their maximum velocities are equal. The ratio of the amplitudes of body $A$ to that of body $B$ is

Show answer and solution

Answer: (a)

$v_{max} = A\sqrt{\frac{k}{m}}$ ; equal $v_{max}$ and $m$ give $A \propto \frac{1}{\sqrt{k}}$ .
$\frac{A_A}{A_B} = \sqrt{\frac{k_2}{k_1}}$
So the answer is (a) $\sqrt{\frac{k_2}{k_1}}$ .

Report a mistake

Q427JEE Main 2020 · 3 SEP · SHIFT 2Medium

Spring Systems · Added mass and two-body systems

A block of mass m attached to a massless spring is performing oscillatory motion of amplitude ‘A’ on a frictionless horizontal plane. If half of the mass of the block breaks off when it is passing through its equilibrium point, the amplitude of oscillation for the remaining system become fA. The value of f is:

Show answer and solution

Answer: (d)

When half the mass breaks off, the remaining part keeps the same speed $v$ at equilibrium.
$\omega' = \sqrt{\frac{k}{m/2}} = \sqrt{2}\,\omega$
$A' = \frac{v}{\omega'} = \frac{A}{\sqrt{2}}$ , so $f = \frac{1}{\sqrt{2}}$ .
So the answer is (d).

Report a mistake

Q428JEE Main 2020 · 6 SEP · SHIFT 2Difficult

Spring Systems · Vertical spring and static extension

When a particle of mass m is attached to a vertical spring of spring constant k and released, its motion is described by

y(t) = y0 sin2 $\omega$ t, where 'y' is measured from the lower end of unstretched spring. Then $\omega$ is:

Show answer and solution

Answer: (d)

$y = y_0\sin^2\omega t = \frac{y_0}{2}(1-\cos 2\omega t)$ : SHM about $y = \frac{y_0}{2}$ with angular frequency $2\omega$ .
Mean position is the equilibrium point: $k\frac{y_0}{2} = mg \Rightarrow \frac{k}{m} = \frac{2g}{y_0}$ .
$2\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{2g}{y_0}} \Rightarrow \omega = \sqrt{\frac{g}{2y_0}}$
So the answer is (d).

Report a mistake

Q429JEE Main 2020 · 9 JAN · SHIFT 2Difficult

Spring Systems · Added mass and two-body systems

A spring mass system (mass m, spring constant k and natural length $l$ ) rests in equilibrium on a horizontal disc. The free end of the spring is fixed at the centre of the disc. If the disc together with spring mass system, rotates about its axis with an angular velocity $\omega$ , (k $\gg m\omega^2$ ) the relative change in the length of the spring is best given by the option:

Show answer and solution

Answer: (b)

Spring force provides the centripetal force: $kx = m\omega^2(l + x)$ .
$x = \frac{m\omega^2 l}{k - m\omega^2} \approx \frac{m\omega^2 l}{k}$ since $k \gg m\omega^2$ .
Relative change $\frac{x}{l} = \frac{m\omega^2}{k}$ , answer (b).

Report a mistake

Q430JEE Main 2018Medium

Spring Systems · Time period of a spring-mass system

A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 1012/sec. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver = 108 and Avogadro number = 6.02 × 1023 gm mole–1)

Show answer and solution

Answer: (c)

For SHM $k = m\omega^2 = 4\pi^2 f^2 m$ , with $m$ the mass of one atom.
$m = \frac{108\times10^{-3}}{6.02\times10^{23}} = 1.79\times10^{-25}$ kg
$k = 4\pi^2(10^{12})^2(1.79\times10^{-25}) \approx 39.5\times0.179 \approx 7.1$ N/m
So the answer is (c) $7.1$ N/m.

Report a mistake

Q431JEE Main 2018 · 16 APR · SHIFT 1Difficult

Spring Systems · Added mass and two-body systems

An oscillator of mass M is at rest in its equilibrium position in a potential

V = $\frac{1}{2}$ k(x $-$ X)2. A particle of mass m comes from right with speed u and collides completely inelastically with M and sticks to it. This process repeats every time the oscillator crosses its equilibrium position. The amplitude of oscillations after 13 collisions is: (M = 10, m = 5, u = 1, k = 1)

Show answer and solution

Answer: (a)

Each collision happens at the equilibrium position, so momentum is conserved and the speed there gives $A = \frac{v}{\omega}$ .
1st collision: momentum $mu$ (leftward). The oscillator returns moving right with momentum $mu$ ; the 2nd particle brings $-mu$ , total $0$ . So after every even collision it stops, after every odd one momentum is $mu$ .
After 13 collisions: mass $= M + 13m = 75$ , $v = \frac{mu}{75} = \frac{5}{75} = \frac{1}{15}$ .
$\omega = \sqrt{\frac{k}{75}} = \frac{1}{\sqrt{75}}$ , $A = \frac{v}{\omega} = \frac{\sqrt{75}}{15} = \frac{1}{\sqrt{3}}$
So the answer is (a).

Report a mistake

Q432JEE Main 2017 · 8 APR · SHIFT 1Medium

Spring Systems · Series and parallel springs

A 1 kg block attached to a spring vibrates with a frequency of 1 Hz on a frictionless horizontal table. Two springs identical to the original spring are attached in parallel to an 8 kg block placed on the same table. So, the frequency of vibration of the 8 kg block is:

Show answer and solution

Answer: (c)

$f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$ : $1 = \frac{1}{2\pi}\sqrt{\frac{k}{1}}$ .
New system: $k' = 2k$ , $m' = 8$ kg, so $f' = \frac{1}{2\pi}\sqrt{\frac{2k}{8}} = \frac{1}{2}\cdot\frac{1}{2\pi}\sqrt{k}$ .
$f' = \frac{1}{2}$ Hz, so the answer is (c).

Report a mistake

Q433NEET 2017Difficult

Spring Systems · Series and parallel springs

A spring of force constant k is cut into lengths of ratio 1: 2: 3. They are connected in series and the new force constant is K'. Then they are connected in parallel and force constant is k''. Then k': k'' is

Show answer and solution

Answer: (b)

$k \propto \frac{1}{l}$ , so pieces of lengths $\frac{l}{6}, \frac{2l}{6}, \frac{3l}{6}$ have constants $6k, 3k, 2k$ .
In series they rebuild the original spring: $k' = k$ .
In parallel: $k'' = 6k + 3k + 2k = 11k$ .
So $k' : k'' = 1 : 11$ , answer (b).

Report a mistake

Q434NEET 2016 · PHASE 2Medium

Spring Systems · Time period of a spring-mass system

A body of mass m is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass m is slightly pulled down and released, it oscillates with a time period of 3s. When the mass m is increased by 1 kg, the time period of oscillations becomes 5 s. The value of m in kg is

Show answer and solution

Answer: (d)

$T \propto \sqrt{m}$ : $\left(\frac{5}{3}\right)^2 = \frac{m+1}{m}$ .
$25m = 9m + 9 \Rightarrow m = \frac{9}{16}\text{ kg}$
So the answer is (d) $\frac{9}{16}$ .

Report a mistake

Q435JEE Main 2014 · 9 APR · SHIFT 1Difficult

Spring Systems · Added mass and two-body systems

Two bodies of masses 1 kg and 4 kg are connected to a vertical spring, as shown in the figure. The smaller mass executes simple harmonic motion of angular frequency 25 rad /s, and amplitude 1.6 cm while the bigger mass remains stationary on the ground. The maximum force exerted by the system on the floor is (take g = 10 ms ______2).

Show answer and solution

Answer: (c)

Maximum spring compression is at the lowest point of the 1 kg mass: spring force $= mg + m\omega^2 A$ .
$mg + m\omega^2 A = 1 \times 10 + 1 \times 25^2 \times 0.016 = 10 + 10 = 20\text{ N}$
Force on floor $= Mg + 20 = 40 + 20 = 60\text{ N}$
So the answer is (c) 60 N.

Report a mistake

Q436JEE Main 2011 · AIEEEMedium

Spring Systems · Added mass and two-body systems

A mass $M,$ attached to a horizontal spring, executes $\text{S.H.M.}$ with amplitude $A_1.$ When the mass $M$ passes through its mean position then a smaller mass $m$ is placed over it and both of them move together with amplitude $A_2.$ The ratio of $\left(\frac{A_1}{A_2}\right)$ is:

Show answer and solution

Answer: (c)

Momentum conserved at the mean position: $Mv = (M+m)v'$ .
$A = \frac{v}{\omega} = \frac{p}{\sqrt{k\,(\text{mass})}}$ with $p$ unchanged.
$\frac{A_1}{A_2} = \left(\frac{M+m}{M}\right)^{1/2}$ , so the answer is (c).

Report a mistake

Q437NEET 2010 · AIPMT · PRELIMSEasy

Spring Systems · Time period of a spring-mass system

The period of oscillation of a mass M suspended from a spring of negligible mass is T. If along with it another mass M is also suspended, the period of oscillation will now be

Show answer and solution

Answer: (d)

$T \propto \sqrt{m}$ ; mass becomes $2M$ .
$T' = T\sqrt{2}$ .
So the answer is (d) $\sqrt{2}\,T$ .

Report a mistake

Q438JEE Main 2007 · AIEEEEasy

Spring Systems · Series and parallel springs

Two springs, of force constant $k_1$ and $k_2$ are connected to a mass $m$ as shown. The frequency of oscillation of the mass is $f.$ If both $k_1$ and $k_2$ are made four times their original values, the frequency of oscillation becomes

Show answer and solution

Answer: (a)

Springs on both sides act in parallel: $f = \frac{1}{2\pi}\sqrt{\frac{k_1+k_2}{m}}$ .
Both constants $\times 4$ : $k_{eff}$ becomes $4(k_1+k_2)$ , so $f' = \sqrt{4}\,f$ .
So the answer is (a) $2f$ .

Report a mistake

Q439NEET 2004 · AIPMTEasy

Spring Systems · Series and parallel springs

Two springs of spring constants k1 and k2 are joined in series. The effective spring constant of the combination is given by

Show answer and solution

Answer: (d)

In series the same force $F$ acts on both springs and the extensions add: $\frac{F}{k} = \frac{F}{k_1} + \frac{F}{k_2}$ .
$\frac{1}{k} = \frac{1}{k_1} + \frac{1}{k_2} \Rightarrow k = \frac{k_1k_2}{k_1+k_2}$
So the answer is (d).

Report a mistake

Q440JEE Main 2004 · AIEEEMedium

Spring Systems · Series and parallel springs

A particle at the end of a spring executes $\text{S.H.M.}$ with a period $t_1$ . While the corresponding period for another spring is $t_2$ . If the period of oscillation with the two springs in series is $T$ then

Show answer and solution

Answer: (b)

Series: $\frac{1}{k} = \frac{1}{k_1} + \frac{1}{k_2}$ .
$T^2 \propto \frac{1}{k}$ , so $T^2 = t_1^2 + t_2^2$ .
So the answer is (b).

Report a mistake

Q441NEET 2003 · AIPMTEasy

Spring Systems · Spring constant and cut springs

The time period of mass suspended from a spring is T. If the spring is cut into four equal parts and the same mass is suspended from one of the parts, then the new time period will be

Show answer and solution

Answer: (c)

Quarter of the spring: $k' = 4k$ .
$T' = 2\pi\sqrt{\frac{m}{4k}} = \frac{T}{2}$ .
So the answer is (c) $T/2$ .

Report a mistake

Q442JEE Main 2003 · AIEEEEasy

Spring Systems · Time period of a spring-mass system

A mass $M$ is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes $SHM$ of time period $T.$ If the mass is increased by $m.$ the time period becomes $\frac{5T}{3}$ . Then the ratio of $\frac{m}{M}$ is

Show answer and solution

Answer: (c)

$T = 2\pi\sqrt{\frac{M}{k}}$ , so $\left(\frac{T'}{T}\right)^2 = \frac{M+m}{M}$ .
$\left(\frac{5}{3}\right)^2 = \frac{25}{9} = 1 + \frac{m}{M}$
$\frac{m}{M} = \frac{25}{9} - 1 = \frac{16}{9}$
So the answer is (c) $\frac{16}{9}$ .

Report a mistake

Q443JEE Main 2003 · AIEEEEasy

Spring Systems · Equal masses on different springs

Two particles $A$ and $B$ of equal masses are suspended from two massless springs of spring constant $k_1$ and $k_2$ , respectively. If the maximum velocities, during oscillation, are equal, the ratio of amplitude of $A$ and $B$ is

Show answer and solution

Answer: (c)

$v_{max} = A\sqrt{\frac{k}{m}}$ ; equal $v_{max}$ and $m$ give $A \propto \frac{1}{\sqrt{k}}$ .
$\frac{A_A}{A_B} = \sqrt{\frac{k_2}{k_1}}$
So the answer is (c) $\sqrt{\frac{k_2}{k_1}}$ .

Report a mistake

Q444JEE Main 2002 · AIEEEEasy

Spring Systems · Spring constant and cut springs

If a spring has time period $T,$ and is cut into $n$ equal parts, then the time period of each part will be

Show answer and solution

Answer: (b)

Each of $n$ equal parts has $k' = nk$ .
$T' = 2\pi\sqrt{\frac{m}{nk}} = \frac{T}{\sqrt{n}}$ .
So the answer is (b) $T/\sqrt{n}$ .

Report a mistake

Q445NEET 2002 · AIPMTMedium

Spring Systems · Series and parallel springs

A mass is suspended separately by two different springs in (successive order then time periods is t1 and t2 respectively, If it is connected by both spring as shown in figure then time period is t0, the correct relation is

Show answer and solution

Answer: (b)

The figure shows the springs in parallel: $k_0 = k_1 + k_2$ .
$k \propto t^{-2}$ , so $t_0^{-2} = t_1^{-2} + t_2^{-2}$ .
So the answer is (b).

Report a mistake

Q446NEET 1998 · AIPMTEasy

Spring Systems · Time period of a spring-mass system

A mass $m$ is vertically suspended from a spring of negligible mass; the system oscillates with a frequency $n$ . What will be the frequency of the system, if a mass $4m$ is suspended from the same spring?

Show answer and solution

Answer: (c)

Spring-mass frequency $n = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$ , so $n \propto \frac{1}{\sqrt{m}}$ .
$\frac{n'}{n} = \sqrt{\frac{m}{4m}} = \frac{1}{2}$
So the answer is (c) $\frac{n}{2}$ .

Report a mistake

Q447NEET 1994 · AIPMTMedium

Spring Systems · Vertical spring and static extension

A body of mass $5\text{ kg}$ hangs from a spring and oscillates with a time period of $2\pi$ seconds. If the ball is removed, the length of the spring will decrease by

Show answer and solution

Answer: (d)

$T = 2\pi\sqrt{m/k} = 2\pi \Rightarrow \frac{m}{k} = 1$ s $^2$ .
Static extension $x = \frac{mg}{k} = g\times1 = g$ m
Removing the body removes this extension, so the answer is (d) $g$ metres.

Report a mistake

Q448NEET 1990 · AIPMTEasy

Spring Systems · Series and parallel springs

A mass $m$ is suspended from the two coupled springs connected in series. The force constant for springs are $k_1$ and $k_2$ . The time period of the suspended mass will be

Show answer and solution

Answer: (d)

Series combination: $k_{eff} = \frac{k_1k_2}{k_1+k_2}$ .
$T = 2\pi\sqrt{\frac{m}{k_{eff}}} = 2\pi\sqrt{\frac{m(k_1+k_2)}{k_1k_2}}$
So the answer is (d).

Report a mistake

Q449JEE Main 2026 · 24 JAN · SHIFT 1Easy

Other SHM Systems · Floating bodies and liquid columns

A cylindrical block of mass $M$ and area of cross section $A$ is floating in a liquid of density $\rho$ and with its axis vertical. When depressed a little and released the block starts oscillating. The period of oscillation is ______

Show answer and solution

Answer: (c)

Pushing down by $x$ adds buoyancy $\rho A g x$ , so $k_{eff} = \rho A g$ .
$T = 2\pi\sqrt{\frac{M}{\rho A g}}$ , answer (c).

Report a mistake

Q450MHT CET 2026 · 15 APR · SHIFT 1Easy

Other SHM Systems · Bowl, piston, tunnel and wire

A small sphere oscillates simple harmonically in a watch glass whose radius of curvature is $1.6$ m. The period of Oscillation of the sphere is (Acceleration due to gravity $g = 10$ m/s $^{2}$ )

Show answer and solution

Answer: (c)

A small sphere in a bowl moves like a pendulum of length $R$ : $T = 2\pi\sqrt{\frac{R}{g}}$ .
$T = 2\pi\sqrt{\frac{1.6}{10}} = 2\pi \times 0.4 = 0.8\pi\text{ s}$
So the answer is (c).

Report a mistake

Q451JEE Main 2026 · 4 APR · SHIFT 2Medium

Other SHM Systems · Physical and torsional pendulum

A uniform disc of radius $R$ and mass $M$ is free to oscillate about the axis $A$ as shown in the figure. For small oscillations the time period is ______.

(g is acceleration due to gravity)

Show answer and solution

Answer: (a)

Axis A is a tangent in the plane of the disc. About a diameter $I = \frac{MR^2}{4}$ ; parallel axes: $I_A = \frac{MR^2}{4} + MR^2 = \frac{5MR^2}{4}$ .
Physical pendulum: $T = 2\pi\sqrt{\frac{I_A}{Mgd}}$ with $d = R$ .
$T = 2\pi\sqrt{\frac{5MR^2/4}{MgR}} = 2\pi\sqrt{\frac{5R}{4g}}$ , answer (a).

Report a mistake

Q452MHT CET 2026 · 13 MAY · SHIFT 2Medium

Other SHM Systems · Floating bodies and liquid columns

A wooden cube (density $d$ ) of side $L$ floats on a liquid of density $\rho$ with its upper and lower surfaces horizontal. If the cube is pushed slightly down and released, it performs S.H.M. of period $T$ . The value of $T$ is ( $g$ = acceleration due to gravity)

Show answer and solution

Answer: (b)

$T = 2\pi\sqrt{\frac{m}{\rho A g}}$ with $m = dL^3$ , $A = L^2$ .
$T = 2\pi\sqrt{\frac{dL^3}{\rho L^2 g}} = 2\pi\sqrt{\frac{Ld}{\rho g}}$ , answer (b).

Report a mistake

Q453MHT CET 2026 · 16 APR · SHIFT 1Medium

Other SHM Systems · Floating bodies and liquid columns

A rectangular block of mass 'm' and cross-sectional area 'A' floats on a liquid of density ' $\varrho$ '. It is given a small vertical displacement from equilibrium, it starts oscillating with frequency (g=acceleration due to gravity)

Show answer and solution

Answer: (c)

Restoring force $A\varrho g x$ , so $\omega = \sqrt{\frac{A\varrho g}{m}}$ .
$f = \frac{\omega}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{A\varrho g}{m}}$ , answer (c).

Report a mistake

Q454NEET 2026 · RE-NEETMedium

Other SHM Systems · Floating bodies and liquid columns

A cylindrical cork of uniform density floats in a liquid of density $\rho_1$ . If the cork is depressed slightly and released, it oscillates harmonically with time period $T$ . If the same cork floats in another liquid of density $\rho_2$ , then the similar oscillation has time period $2T$ . The value of $\rho_2/\rho_1$ is:

Show answer and solution

Answer: (a)

$T = 2\pi\sqrt{\frac{m}{A\rho g}} \Rightarrow T \propto \frac{1}{\sqrt{\rho}}$
$\frac{T_2}{T_1} = 2 = \sqrt{\frac{\rho_1}{\rho_2}} \Rightarrow \frac{\rho_2}{\rho_1} = \frac{1}{4}$ , answer (a).

Report a mistake

Q455MHT CET 2025 · 25 APR · SHIFT 1Easy

Other SHM Systems · Bowl, piston, tunnel and wire

A small sphere oscillates simple harmonically in a watch glass whose radius of curvature is 1.6 m. The period of oscillation of the sphere is (acceleration due to gravity $g = 10\text{ m/s}^2$ )

Show answer and solution

Answer: (d)

A small sphere in a bowl moves like a pendulum of length $R$ : $T = 2\pi\sqrt{\frac{R}{g}}$ .
$T = 2\pi\sqrt{\frac{1.6}{10}} = 2\pi \times 0.4 = 0.8\pi\text{ s}$
So the answer is (d).

Report a mistake

Q456JEE Main 2025 · 23 JAN · SHIFT 1Medium

Other SHM Systems · Floating bodies and liquid columns

A light hollow cube of side length 10 cm and mass 10 g, is floating in water. It is pushed down and released to execute simple harmonic oscillations. The time period of oscillations is $y\pi \times 10^{-2}\text{ s}$ , where the value of $y$ is (Acceleration due to gravity, $g = 10\text{ m/s}^2$ , density of water $= 10^{3}\text{ kg/m}^3$ )

Show answer and solution

Answer: (a)

$T = 2\pi\sqrt{\frac{m}{\rho A g}}$ with $m = 0.01\text{ kg}$ , $A = 0.01\text{ m}^2$ .
$T = 2\pi\sqrt{\frac{0.01}{10^3 \times 0.01 \times 10}} = 2\pi \times 10^{-2}\text{ s}$
So $y = 2$ , answer (a).

Report a mistake

Q457MHT CET 2025 · 22 APR · SHIFT 2Medium

Other SHM Systems · Bowl, piston, tunnel and wire

A small spherical ball of radius ' $r$ ' is rolling on a curved surface which is frictionless and has a radius of curvature ' R '. Its motion is simple harmonic. Then its time period of oscillation is proportional to ( $g =$ acceleration due to gravity)

Show answer and solution

Answer: (c)

The centre of the ball moves on a circle of radius $R - r$ , so it acts like a pendulum of length $R - r$ .
$T = 2\pi\sqrt{\frac{R-r}{g}} \Rightarrow T \propto \sqrt{\frac{R-r}{g}}$ , answer (c).

Report a mistake

Q458MHT CET 2024 · 3 MAY · SHIFT 1Easy

Other SHM Systems · Bowl, piston, tunnel and wire

A small sphere oscillates simple harmonically in a watch glass whose radius of curvature is 1.6 m. The period of oscillation of the sphere in second is (acceleration due to gravity, $g = 10\text{ m/s}^2$ )

Show answer and solution

Answer: (a)

A small sphere in a bowl moves like a pendulum of length $R$ : $T = 2\pi\sqrt{\frac{R}{g}}$ .
$T = 2\pi\sqrt{\frac{1.6}{10}} = 2\pi \times 0.4 = 0.8\pi\text{ s}$
So the answer is (a).

Report a mistake

Q459MHT CET 2024 · 11 MAY · SHIFT 2Medium

Other SHM Systems · Floating bodies and liquid columns

A piece of wood has length, breadth and height, ' $a$ ', ' $b$ ' and ' $c$ ' respectively. Its relative density, is ' $d$ '. It is floating in water such that the side ' $a$ ' is vertical. It is pushed down a little and released. The time period of S.H.M. executed by it is ( $g =$ acceleration due to gravity)

Show answer and solution

Answer: (d)

Horizontal face area $bc$ , mass $m = d\rho_w abc$ ; restoring constant $= \rho_w (bc) g$ .
$T = 2\pi\sqrt{\frac{d\rho_w abc}{\rho_w bc\, g}} = 2\pi\sqrt{\frac{ad}{g}}$ , answer (d).

Report a mistake

Q460MHT CET 2024 · 3 MAY · SHIFT 1Medium

Other SHM Systems · Floating bodies and liquid columns

A tube of uniform bore of cross-sectional area ' $A$ ' has been set up vertically with open end facing up. Now ' $M$ ' gram of a liquid of density ' $d$ ' is poured into it. The column of liquid in this tube will oscillate with a period ' T ', which is equal to [ $g =$ acceleration due to gravity]

Show answer and solution

Answer: (b)

Treat it as a U-tube: displacing the liquid by $x$ makes a level difference $2x$ , restoring force $2Adgx$ .
$T = 2\pi\sqrt{\frac{M}{2Adg}}$ (equivalently $2\pi\sqrt{\frac{L}{2g}}$ with column length $L = \frac{M}{Ad}$ ).
So the answer is (b).

Report a mistake

Q461JEE Main 2021 · 26 FEB · SHIFT 1Medium

Other SHM Systems · Bowl, piston, tunnel and wire

Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance (R/2) from the earth's centre, where 'R' is the radius of the Earth. The wall of the tunnel is frictionless. If a particle is released in this tunnel, it will execute a simple harmonic motion with a time period:

Show answer and solution

Answer: (a)

Inside the earth $g' = g\frac{r}{R}$ ; its component along the chord is $\frac{g}{R}x$ , where $x$ is distance from the chord's midpoint.
$a = -\frac{g}{R}x \Rightarrow T = 2\pi\sqrt{\frac{R}{g}}$ , independent of the chord.
So the answer is (a).

Report a mistake

Q462JEE Main 2019 · 10 JAN · SHIFT 2Medium

Other SHM Systems · Floating bodies and liquid columns

A cylindrical plastic bottle of negligible mass is filled with 310 ml of water and left floating in a pond with still water. If pressed downward slightly and released, it starts performing simple harmonic motion at angular frequency $\omega$ . If the radius of the bottle is 2.5 cm then $\omega$ is close to – (density of water = 103 kg/m3).

Show answer and solution

Answer: (d)

$\omega = \sqrt{\frac{\rho A g}{m}}$ with $m = 0.31\text{ kg}$ , $A = \pi(0.025)^2 = 1.96 \times 10^{-3}\text{ m}^2$ .
$\omega = \sqrt{\frac{10^3 \times 1.96 \times 10^{-3} \times 9.8}{0.31}} = \sqrt{62} \approx 7.9\text{ rad/s}$
So the answer is (d).

Report a mistake

Q463JEE Main 2019 · 12 JAN · SHIFT 1Difficult

Other SHM Systems · Physical and torsional pendulum

Two light identical springs of spring constant k are attached horizontally at the two ends of a uniform horizontal rod AB of length $\ell$ and mass m. The rod is pivoted at its centre 'O' and can rotate freely in horizontal plane. The other ends of the two springs are fixed to rigid supports as shown in figure. The rod is gently pushed through a small angle and released. The frequency of resulting oscillation is:

Show answer and solution

Answer: (b)

Turn the rod by a small $\theta$ : each end moves $\frac{\ell}{2}\theta$ , each spring gives force $k\frac{\ell}{2}\theta$ at arm $\frac{\ell}{2}$ .
Restoring torque $\tau = 2k\left(\frac{\ell}{2}\right)^2\theta = \frac{k\ell^2}{2}\theta$ ; $I = \frac{m\ell^2}{12}$ .
$\omega^2 = \frac{k\ell^2/2}{m\ell^2/12} = \frac{6k}{m}$
$f = \frac{1}{2\pi}\sqrt{\frac{6k}{m}}$ , answer (b).

Report a mistake

Q464JEE Main 2019 · 9 JAN · SHIFT 2Difficult

Other SHM Systems · Physical and torsional pendulum

A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of 'm' are attached at distance 'L/2' from its centre on both sides, it reduces the oscillation frequency by 20%. The value of ratio m/M is close to:

Show answer and solution

Answer: (c)

Torsional pendulum: $f \propto \frac{1}{\sqrt{I}}$ . $I_0 = \frac{M(2L)^2}{12} = \frac{ML^2}{3}$ .
With the masses: $I_1 = \frac{ML^2}{3} + 2m\left(\frac{L}{2}\right)^2 = \frac{ML^2}{3} + \frac{mL^2}{2}$ .
$f_1 = 0.8f_0 \Rightarrow \frac{I_1}{I_0} = \frac{1}{0.64} = 1.5625$
$\frac{3m}{2M} = 0.5625 \Rightarrow \frac{m}{M} = 0.375 \approx 0.37$ , answer (c).

Report a mistake

Q465JEE Main 2015 · 11 APR · SHIFT 1Medium

Other SHM Systems · Floating bodies and liquid columns

A cylindrical block of wood (density = 650 kg m ${}^{-3}$ ), of base area 30 cm2 and height 54 cm, floats in a liquid of density 900 kg m ${}^{-3}$ . The block is depressed slightly and then released. The time period of the resulting oscillations of the block would be equal to that of a simple pendulum of length (nearly):

Show answer and solution

Answer: (c)

$T = 2\pi\sqrt{\frac{m}{\rho_l A g}}$ with $m = \rho_w A h$ , so $T = 2\pi\sqrt{\frac{\rho_w h}{\rho_l g}}$ .
Equivalent pendulum length $L = \frac{\rho_w h}{\rho_l} = \frac{650 \times 54}{900} = 39\text{ cm}$
So the answer is (c).

Report a mistake

Q466JEE Main 2015Difficult

Other SHM Systems · Bowl, piston, tunnel and wire

A pendulum made of a uniform wire of cross sectional area $A$ has time period $T.$ When an additional mass $M$ is added to its bob, the time period changes to $T_M.$ If the Young's modulus of the material of the wire is $Y$ then $\frac{1}{Y}$ is equal to:

( $g =$ gravitational acceleration)

Show answer and solution

Answer: (c)

Added mass stretches the wire: $\Delta L = \frac{MgL}{AY}$ , and $T \propto \sqrt{L}$ .
$\left(\frac{T_M}{T}\right)^2 = \frac{L + \Delta L}{L} = 1 + \frac{Mg}{AY}$
$\frac{1}{Y} = \left[\left(\frac{T_M}{T}\right)^2 - 1\right]\frac{A}{Mg}$ , answer (c).

Report a mistake

Q467JEE Main 2013 · 25 APR · SHIFT 1Difficult

Other SHM Systems · Floating bodies and liquid columns

A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half submerged in a liquid of density $\sigma$ at equilibrium position. When the cylinder is given a downward push and released, it starts oscillating vertically with a small amplitude. The time period T of the oscillations of the cylinder will be:

Show answer and solution

Answer: (d)

Extra downward displacement $x$ adds spring force $kx$ and extra buoyancy $A\sigma g x$ , both upward.
Net restoring constant $k + A\sigma g$ , so $T = 2\pi\sqrt{\frac{M}{k + A\sigma g}}$ , answer (d).

Report a mistake

Q468JEE Main 2013Difficult

Other SHM Systems · Bowl, piston, tunnel and wire

An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass $M.$ The piston and the cylinder have equal cross sectional area $A$ . When the piston is in equilibrium, the volume of the gas is $V_0$ and its pressure is $P_0.$ The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frequency

Show answer and solution

Answer: (c)

Isolated system: adiabatic, $PV^\gamma$ = const, so $\Delta P = -\frac{\gamma P_0}{V_0}\Delta V = -\frac{\gamma P_0 A x}{V_0}$ .
Restoring force $F = A\Delta P = -\frac{\gamma P_0 A^2}{V_0}x$ .
$f = \frac{1}{2\pi}\sqrt{\frac{\gamma P_0 A^2}{M V_0}}$ , answer (c).

Report a mistake

Q469NEET 2006 · AIPMTEasy

Other SHM Systems · Floating bodies and liquid columns

A rectangular block of mass m and area of cross-section A floats in a liquid of density $\rho$ . If it is given a small vertical displacement from equilibrium it undergoes with a time period T, then

Show answer and solution

Answer: (c)

Extra depth $x$ gives restoring force $A\rho g x$ , so $k_{eff} = A\rho g$ .
$T = 2\pi\sqrt{\frac{m}{A\rho g}}$ : $T \propto \sqrt{m}$ , $T \propto \frac{1}{\sqrt{\rho}}$ , $T \propto \frac{1}{\sqrt{A}}$ .
So the answer is (c).

Report a mistake

Q470MHT CET 2025 · 23 APR · SHIFT 1Medium

Damped, Forced Oscillations & Superposition · Damped oscillations

The amplitude of a damped oscillator becomes $\left(\frac{1}{3}\right)^{\text{rd}}$ of original amplitude in 2 seconds. If its amplitude after 6 second become $\left(\frac{1}{n}\right)$ times the original amplitude, the value of $n$ is ( $n$ is non zero integer)

Show answer and solution

Answer: (d)

Damped amplitude $A = A_0 e^{-\gamma t}$ , so in equal time intervals it falls by the same factor.
In 2 s: $A/A_0 = \frac{1}{3}$ .
6 s = three 2 s intervals: $A/A_0 = \left(\frac{1}{3}\right)^3 = \frac{1}{27}$ .
So the answer is (d) $n = 27$ .

Report a mistake

Q471JEE Main 2025 · 2 APR · SHIFT 1Medium

Damped, Forced Oscillations & Superposition · Superposition along one line

A particle is subjected to two simple harmonic motions as: $x_1 = \sqrt{7}\sin 5t\text{ cm}$ and $x_2 = 2\sqrt{7}\sin\left(5t+\frac{\pi}{3}\right)\text{ cm}$ where $x$ is displacement and $t$ is time in seconds. The maximum acceleration of the particle is $x \times 10^{-2}\text{ m s}^{-2}$ . The value of $x$ is:

Show answer and solution

Answer: (d)

Resultant amplitude: $A^2 = A_1^2 + A_2^2 + 2A_1A_2\cos\frac{\pi}{3}$ .
$A^2 = 7 + 28 + 2(\sqrt{7})(2\sqrt{7})\left(\frac{1}{2}\right) = 49 \Rightarrow A = 7\ \text{cm}$ .
$a_{max} = \omega^2 A = 25 \times 7 = 175\ \text{cm s}^{-2} = 175 \times 10^{-2}\ \text{m s}^{-2}$ .
So the answer is (d) 175.

Report a mistake

Q472MHT CET 2024 · 2 MAY · SHIFT 1Easy

Damped, Forced Oscillations & Superposition · Damped oscillations

A particle is performing simple harmonic motion and if the oscillations are damped oscillations then the angular frequency is given by

Show answer and solution

Answer: (c)

Damped motion $x = Ae^{-bt/2m}\cos(\omega' t + \phi)$ with $\omega'^2 = \omega_0^2 - \left(\frac{b}{2m}\right)^2$ .
$\omega' = \sqrt{\frac{k}{m} - \left(\frac{b}{2m}\right)^2}$ , answer (c).

Report a mistake

Q473NEET 2024Medium

Damped, Forced Oscillations & Superposition · Perpendicular SHMs and Lissajous figures

The two-dimensional motion of a particle, described by $\vec{r} = (\hat{i}+2\hat{j})A\cos\omega t$ is a/an: A. parabolic path

B. elliptical path

C. periodic motion

D. simple harmonic motion

Choose the correct answer from the options given below:

Show answer and solution

Answer: (d)

$x = A\cos\omega t$ , $y = 2A\cos\omega t \Rightarrow y = 2x$ : a straight line, not a parabola or ellipse.
Position repeats every $2\pi/\omega$ , so the motion is periodic.
$\vec{r} \propto \cos\omega t$ along a fixed line, so $\vec{a} = -\omega^2\vec{r}$ : SHM.
So the answer is (d) C and D only.

Report a mistake

Q474JEE Main 2022 · 28 JUN · SHIFT 1Easy

Damped, Forced Oscillations & Superposition · Perpendicular SHMs and Lissajous figures

Motion of a particle in x-y plane is described by a set of following equations $x = 4\sin\left(\frac{\pi}{2}-\omega t\right)\text{ m}$ and $y = 4\sin(\omega t)\text{ m}$ . The path of the particle will be:

Show answer and solution

Answer: (a)

$x = 4\sin\left(\frac{\pi}{2} - \omega t\right) = 4\cos\omega t$ , $y = 4\sin\omega t$ .
$x^2 + y^2 = 16$ : a circle of radius 4 m, answer (a).

Report a mistake

Q475JEE Main 2021 · 17 MAR · SHIFT 2MediumOld syllabus

Damped, Forced Oscillations & Superposition · Damped oscillations

A block of mass 1 kg attached to a spring is made to oscillate with an initial amplitude of 12 cm. After 2 minutes the amplitude decreases to 6 cm. Determine the value of the damping constant for this motion. (take ln 2 = 0.693)

Show answer and solution

Answer: (c)

$A = A_0 e^{-bt/2m}$ .
$\frac{6}{12} = e^{-b(120)/(2 \times 1)} \Rightarrow 60b = \ln 2 = 0.693$ .
$b = \frac{0.693}{60} = 1.16 \times 10^{-2}\ \text{kg s}^{-1}$ .
So the answer is (c).

Report a mistake

Q476JEE Main 2021 · 16 MAR · SHIFT 2MediumOld syllabus

Damped, Forced Oscillations & Superposition · Damped oscillations

Amplitude of a mass-spring system, which is executing simple harmonic motion decreases with time. If mass = 500g, Decay constant = 20 g/s then how much time is required for the amplitude of the system to drop to half of its initial value? (ln 2 = 0.693)

Show answer and solution

Answer: (b)

$A = A_0 e^{-bt/2m}$ ; for $A = A_0/2$ : $t = \frac{2m \ln 2}{b}$ .
$t = \frac{2 \times 500 \times 0.693}{20}$ (both in grams, units cancel).
$t = 34.65\ \text{s}$ .
So the answer is (b) 34.65 s.

Report a mistake

Q477MHT CET 2020 · 16 OCT · SHIFT 2Easy

Damped, Forced Oscillations & Superposition · Damped oscillations

The damping force of an oscillator is directly proportional to the velocity. The unit of constant of proportionality is

Show answer and solution

Answer: (b)

$F = bv \Rightarrow b = \frac{F}{v}$
Unit: $\frac{\text{kg m s}^{-2}}{\text{m s}^{-1}} = \text{kg s}^{-1}$ , answer (b).

Report a mistake

Q478MHT CET 2019 · 3 MAY · SHIFT 1Easy

Damped, Forced Oscillations & Superposition · Damped oscillations

In damped SHM, the SI unit of damping constant is

Show answer and solution

Answer: (b)

$b = \frac{F}{v}$ , unit $\frac{\text{N}}{\text{m/s}} = \frac{\text{kg m s}^{-2}}{\text{m s}^{-1}} = \frac{\text{kg}}{\text{s}}$
So the answer is (b).

Report a mistake

Q479JEE Main 2019 · 10 APR · SHIFT 1MediumOld syllabus

Damped, Forced Oscillations & Superposition · Damped oscillations

The displacement of a damped harmonic oscillator is given by

x(t) = e–0.1t cos (10 $\pi$ t + f).

Here t is in seconds. The time taken for its amplitude of vibration to drop to half of its initial value is close to:

Show answer and solution

Answer: (c)

Amplitude $= e^{-0.1t}$ (initial value 1).
$e^{-0.1t} = \frac{1}{2} \Rightarrow 0.1t = \ln 2 = 0.693$ .
$t = 6.93\ \text{s} \approx 7\ \text{s}$ .
So the answer is (c) 7 s.

Report a mistake

Q480JEE Main 2019 · 8 APR · SHIFT 2MediumOld syllabus

Damped, Forced Oscillations & Superposition · Damped oscillations

A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it will take to drop to 1/1000 of the original amplitude is close to :-

Show answer and solution

Answer: (c)

Period $= \frac{1}{5}\ \text{s}$ , so 10 oscillations take 2 s: amplitude halves every 2 s.
$\frac{1}{1000} \approx \frac{1}{1024} = \left(\frac{1}{2}\right)^{10}$ , so 10 halvings are needed.
$t = 10 \times 2 = 20\ \text{s}$ .
So the answer is (c) 20 s.

Report a mistake

Q481JEE Main 2018 · 15 APR · SHIFT 2Medium

Damped, Forced Oscillations & Superposition · Perpendicular SHMs and Lissajous figures

Two simple harmonic motions, as shown below, are at right angles. They are combined to form Lissajous figures.

x(t) = A sin (at + $\delta$ )

y(t) = B sin (bt)

Identify the correct match below.

Show answer and solution

Answer: (c)

(a) $a = b$ , $\delta = 0$ : $y/x = B/A$ , a straight line, not a parabola.
(b) $A = B$ , $a = b$ , $\delta = \frac{\pi}{2}$ : $x = A\cos bt$ , $y = A\sin bt$ , a circle, not a line.
(c) $A \neq B$ , $a = b$ , $\delta = \frac{\pi}{2}$ : $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$ , an ellipse. Correct.
(d) $a = 2b$ : $x = A\cos 2bt = A(1 - 2y^2/B^2)$ , a parabola, not a circle.
So the answer is (c).

Report a mistake

Q482JEE Main 2017 · 9 APR · SHIFT 1MediumOld syllabus

Damped, Forced Oscillations & Superposition · Damped oscillations

A block of mass 0.1 kg is connected to an elastic spring of spring constant 640 Nm−1 and oscillates in a damping medium of damping constant 10−2 kg s−1. The system dissipates its energy gradually. The time taken for its mechanical energy of vibration to drop to half of its initial value, is closest to:

Show answer and solution

Answer: (d)

$A = A_0 e^{-bt/2m}$ , so energy $E \propto A^2$ gives $E = E_0 e^{-bt/m}$ .
$E = E_0/2 \Rightarrow t = \frac{m \ln 2}{b}$ .
$t = \frac{0.1 \times 0.693}{10^{-2}} = 6.93\ \text{s} \approx 7\ \text{s}$ .
So the answer is (d) 7 s.

Report a mistake

Q483NEET 2015 · AIPMT · CANCELLED PAPEREasy

Damped, Forced Oscillations & Superposition · Superposition along one line

When two displacements represented by y1 = a sin $(\omega t)$ and y2 = b cos $(\omega t)$ aresuperimposed the motion is

Show answer and solution

Answer: (a)

$y = a\sin\omega t + b\cos\omega t = \sqrt{a^2+b^2}\,\sin(\omega t + \phi)$ with $\tan\phi = \frac{b}{a}$ .
Same $\omega$ , so the result is SHM with amplitude $\sqrt{a^2+b^2}$ , answer (a).

Report a mistake

Q484JEE Main 2015 · 10 APR · SHIFT 1MediumOld syllabus

Damped, Forced Oscillations & Superposition · Forced oscillations and resonance

A simple harmonic oscillator of angular frequency 2 rad s ${}^{-1}$ is acted upon by an external force F = sint N. If the oscillator is at rest in its equilibrium position at t = 0, its position at later times is proportional to:

Show answer and solution

Answer: (d)

Equation: $\ddot{x} + 4x = \frac{1}{m}\sin t$ (natural $\omega = 2$ ).
Particular solution: $x_p = \frac{\sin t}{m(4-1)} = \frac{\sin t}{3m}$ .
Add free part $C\sin 2t + D\cos 2t$ ; $x(0) = 0 \Rightarrow D = 0$ .
$\dot{x}(0) = 0 \Rightarrow \frac{1}{3m} + 2C = 0 \Rightarrow C = -\frac{1}{6m}$ .
$x = \frac{1}{3m}\left(\sin t - \frac{1}{2}\sin 2t\right)$ .
So the answer is (d) $\sin t - \frac{1}{2}\sin 2t$ .

Report a mistake

Q485JEE Main 2015 · 10 APR · SHIFT 1Medium

Damped, Forced Oscillations & Superposition · Perpendicular SHMs and Lissajous figures

x and y displacements of a particle are given as

x(t) = a sin $\omega$ t and y(t) = a sin2 $\omega$ t.

Its trajectory will look like:

Show answer and solution

Answer: (c)

$y = a\sin 2\omega t = 2a\sin\omega t\cos\omega t = 2x\sqrt{1 - x^2/a^2}$ .
$y = 0$ at $x = 0$ and $x = \pm a$ ; $|y|$ is maximum ( $= a$ ) at $x = \pm a/\sqrt{2}$ .
So the path is a figure of eight with one loop on each side of the y-axis, crossing at the origin.
So the answer is (c).

Report a mistake

Q486JEE Main 2015 · 11 APR · SHIFT 1DifficultOld syllabus

Damped, Forced Oscillations & Superposition · Damped oscillations

A pendulum with time period of 1s is losing energy due to damping. At certain time its energy is 45 J. If after completing 15 oscillations, its energy has become 15 J, its damping constant (in s ${}^{-1}$ ) is:

Show answer and solution

Answer: (a)

$A = A_0 e^{-bt}$ , so $E \propto A^2 = E_0 e^{-2bt}$ .
15 oscillations of 1 s: $t = 15\ \text{s}$ ; $\frac{15}{45} = e^{-2b(15)}$ .
$30b = \ln 3 \Rightarrow b = \frac{1}{30}\ln 3\ \text{s}^{-1}$ .
So the answer is (a) $\frac{1}{30}\ln 3$ .

Report a mistake

Q487JEE Main 2014 · 9 APR · SHIFT 1Medium

Damped, Forced Oscillations & Superposition · Perpendicular SHMs and Lissajous figures

A particle which is simultaneously subjected to two perpendicular simple harmonic motions represented by; x = a1 cos $\omega$ t and y = a2 cos 2 $\omega$ t traces a curve given by:

Show answer and solution

Answer: (a)

$y = a_2\cos 2\omega t = a_2(2\cos^2\omega t - 1)$ and $\cos\omega t = x/a_1$ .
$y = a_2\left(\frac{2x^2}{a_1^2} - 1\right)$ : a parabola opening upward.
Vertex at $(0, -a_2)$ ; at $x = \pm a_1$ , $y = +a_2$ .
So the answer is (a).

Report a mistake

Q488JEE Main 2014 · 11 APR · SHIFT 1DifficultOld syllabus

Damped, Forced Oscillations & Superposition · Damped oscillations

The angular frequency of the damped oscillator is given by,

$\omega = \sqrt{\left(\frac{k}{m}-\frac{r^2}{4m^2}\right)}$ where k is the spring constant, m is the mass of the oscillator and r is the damping constant. If the ratio $\frac{r^2}{mk}$ is 8%, the change in time period compared to the undamped oscillator is approximately as follows:

Show answer and solution

Answer: (a)

$\omega = \sqrt{\frac{k}{m}}\sqrt{1 - \frac{r^2}{4mk}} = \omega_0\sqrt{1 - \frac{0.08}{4}} = \omega_0\sqrt{1 - 0.02}$ .
$\omega \approx \omega_0(1 - 0.01)$ , so $\omega$ decreases by 1%.
$T \propto 1/\omega$ , so $T$ increases by about 1%.
So the answer is (a) increases by 1%.

Report a mistake

Q489JEE Main 2014 · 9 APR · SHIFT 1DifficultOld syllabus

Damped, Forced Oscillations & Superposition · Damped oscillations

The amplitude of a simple pendulum, oscillating in air with a small spherical bob, decreases from 10 cm to 8 cm in 40 seconds. Assuming that Stokes law is valid, and ratio of the coefficient of viscosity of air to that of carbon dioxide is 1.3, the time in which amplitude of this pendulum will reduce from 10 cm to 5 cm in carbon dioxide will be close to (ln 5 = 1.609, ln 2 = 0.693).

Show answer and solution

Answer: (c)

$A = A_0 e^{-bt/2m}$ with $b \propto \eta$ (Stokes law).
Air: $\ln\frac{10}{8} = \ln 5 - 2\ln 2 = 0.223 = \frac{b_{air}}{2m}(40) \Rightarrow \frac{b_{air}}{2m} = 5.575 \times 10^{-3}\ \text{s}^{-1}$ .
CO $_2$ : $\frac{b_{CO_2}}{2m} = \frac{5.575 \times 10^{-3}}{1.3} = 4.29 \times 10^{-3}\ \text{s}^{-1}$ .
10 cm to 5 cm: $t = \frac{\ln 2}{4.29 \times 10^{-3}} \approx 161\ \text{s}$ .
So the answer is (c) 161 s.

Report a mistake

Q490JEE Main 2013 · 23 APR · SHIFT 1Medium

Damped, Forced Oscillations & Superposition · Damped oscillations

Bob of a simple pendulum of length $l$ is made of iron. The pendulum is oscillating over a horizontal coil carrying direct current. If the time period of the pendulum is T then:

Show answer and solution

Answer: (d)

The DC coil acts as a magnet and pulls the iron bob down, so the effective $g$ increases.
$T = 2\pi\sqrt{l/g_{\text{eff}}}$ with $g_{\text{eff}} > g \Rightarrow T < 2\pi\sqrt{l/g}$ .
The moving iron bob in the coil's non-uniform field has eddy currents, which oppose motion, so damping is larger than in air alone.
So the answer is (d).

Report a mistake

Q491JEE Main 2013MediumOld syllabus

Damped, Forced Oscillations & Superposition · Damped oscillations

The amplitude of a damped oscillator decreases to $0.9$ times its original magnitude in $5\text{ s}$ . In another $10\text{ s}$ it will decrease to $\alpha$ times its original magnitude, where $\alpha$ equals

Show answer and solution

Answer: (c)

$A = A_0 e^{-\gamma t}$ : amplitude falls by factor 0.9 in every 5 s.
Total time $5 + 10 = 15\ \text{s}$ = three 5 s intervals.
$\alpha = (0.9)^3 = 0.729$ .
So the answer is (c) 0.729.

Report a mistake

Q492JEE Main 2012 · AIEEEDifficultOld syllabus

Damped, Forced Oscillations & Superposition · Damped oscillations

If a simple pendulum has significant amplitude (up to a factor of $1/e$ of original) only in the period between $t = 0\text{ s to } t = \tau\text{ s},$ then $\tau$ may be called the average life of the pendulum When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity with $b$ as the constant of proportionality, the average life time of the pendulum is (assuming damping is small) in seconds:

Show answer and solution

Answer: (d)

Retardation $= bv$ , so drag force $= mbv$ and $m\ddot{x} + mb\dot{x} + kx = 0$ .
Amplitude $A = A_0 e^{-bt/2}$ .
$A = A_0/e$ when $\frac{b\tau}{2} = 1 \Rightarrow \tau = \frac{2}{b}$ .
So the answer is (d) $\frac{2}{b}$ .

Report a mistake

Q493JEE Main 2004 · AIEEEEasyOld syllabus

Damped, Forced Oscillations & Superposition · Forced oscillations and resonance

In forced oscillation of a particle the amplitude is maximum for a frequency $\omega_1$ of the force while the energy is maximum for a frequency $\omega_2$ of the force; then

Show answer and solution

Answer: (c)

Amplitude resonance: $\omega_1 = \sqrt{\omega_0^2 - \frac{b^2}{2m^2}}$ ; energy (velocity) resonance: $\omega_2 = \omega_0$ .
For small (negligible) damping $\omega_1 \approx \omega_0 = \omega_2$ .
So $\omega_1 = \omega_2$ , answer (c) as per key.

Report a mistake

Q494JEE Main 2004 · AIEEEMedium

Damped, Forced Oscillations & Superposition · Forced oscillations and resonance

A particle of mass $m$ is attached to a spring (of spring constant $k$ ) and has a natural angular frequency $\omega_0.$ An external force $F(t)$ proportional to $\cos\omega t\ (\omega \neq \omega_0)$ is applied to the oscillator. The time displacement of the oscillator will be proportional to

Show answer and solution

Answer: (b)

Steady state: $m\ddot{x} + m\omega_0^2 x = F_0\cos\omega t$ , try $x = A\cos\omega t$ .
$-m\omega^2 A + m\omega_0^2 A = F_0 \Rightarrow A = \frac{F_0}{m(\omega_0^2 - \omega^2)}$ .
So the answer is (b) $\frac{1}{m(\omega_0^2 - \omega^2)}$ .

Report a mistake

Q495NEET 2003 · AIPMTEasy

Damped, Forced Oscillations & Superposition · Forced oscillations and resonance

In case of a forced vibration, the resonance peak becomes very sharp when the

Show answer and solution

Answer: (a)

Sharpness of resonance is measured by the quality factor $Q = \frac{m\omega_0}{b}$ .
Small damping gives large $Q$ and a sharp, tall peak, answer (a).

Report a mistake

Q496NEET 2002 · AIPMTMedium

Damped, Forced Oscillations & Superposition · Damped oscillations

When an oscillator completes 100 oscillations its amplitude is reduced to $\frac{1}{3}$ of initial value. What will be its amplitude, when it completes 200 oscillations?

Show answer and solution

Answer: (d)

Amplitude falls by the same factor in every equal number of oscillations: $A = A_0 e^{-\gamma t}$ .
After 100 oscillations: $A_0/3$ .
After 200 oscillations: $A_0 \times \frac{1}{3} \times \frac{1}{3} = \frac{A_0}{9}$ .
So the answer is (d) $\frac{1}{9}$ of the initial value.

Report a mistake

Q497NEET 1998 · AIPMTEasy

Damped, Forced Oscillations & Superposition · Forced oscillations and resonance

A particle, with restoring force proportional to displacement and resisting force proportional to velocity is subjected to a force $F\sin\omega t$ . If the amplitude of the particle is maximum for $\omega = \omega_1$ and the energy of the particle is maximum for $\omega = \omega_2$ , then

(where $\omega_0 \rightarrow$ natural angular frequency of oscillations of particle)

Show answer and solution

Answer: (c)

Amplitude resonance occurs at $\omega_1 = \sqrt{\omega_0^2 - \frac{b^2}{2m^2}}$ , slightly below $\omega_0$ , so $\omega_1 \neq \omega_0$ .
Velocity amplitude, and hence kinetic energy, is maximum exactly at $\omega_2 = \omega_0$ .
So the answer is (c).

Report a mistake

Q498NEET 1997 · AIPMTEasy

Damped, Forced Oscillations & Superposition · Perpendicular SHMs and Lissajous figures

Two SHM’s with same amplitude and time period, when acting together in perpendicular directions with a phase difference of $\pi/2$ , give rise to

Show answer and solution

Answer: (c)

$x = a\sin\omega t$ , $y = a\sin(\omega t + \frac{\pi}{2}) = a\cos\omega t$ .
$x^2 + y^2 = a^2$ : a circle, answer (c).

Report a mistake

Q499NEET 1990 · AIPMTEasy

Damped, Forced Oscillations & Superposition · Perpendicular SHMs and Lissajous figures

The composition of two simple harmonic motions of equal periods at right angle to each other and with a phase difference of $\pi$ results in the displacement of the particle along

Show answer and solution

Answer: (c)

$x = a\sin\omega t$ , $y = b\sin(\omega t + \pi) = -b\sin\omega t$ .
$y = -\frac{b}{a}x$ : a straight line, answer (c).

Report a mistake

Q500NEET 1989 · AIPMTEasy

Damped, Forced Oscillations & Superposition · Forced oscillations and resonance

A particle, with restoring force proportional to displacement and resisting force proportional to velocity is subjected to a force $F\sin\omega t$ . If the amplitude of the particle is maximum for $\omega = \omega_1$ and the energy of the particle is maximum for $\omega = \omega_2$ , then

Show answer and solution

Answer: (b)

Amplitude resonance occurs at $\omega_1 = \sqrt{\omega_0^2 - \frac{b^2}{2m^2}}$ , slightly below $\omega_0$ .
Velocity amplitude (energy) is maximum exactly at $\omega_2 = \omega_0$ .
For negligible damping $\omega_1 \to \omega_0$ , so both are taken equal to $\omega_0$ : answer (b) as per key.

Report a mistake