Wave Motion

Answer: (a)

1. Frequency is set by the source and stays the same in both media: $n = \frac{v}{\lambda}$.
2. $n = \frac{v_P}{\lambda_P} = \frac{0.80}{0.150} = \frac{16}{3}$ Hz
3. $v_Q = n\lambda_Q = \frac{16}{3} \times 0.30 = 1.6$ m/s
4. So the answer is (a) $1.6$ m/s.

Answer: (d)

1. Frequency is the same in both media, so $\frac{v_Q}{v_P} = \frac{\lambda_Q}{\lambda_P}$.
2. $v_Q = 0.80 \times \frac{0.30}{0.150} = 1.6$ m/s
3. So the answer is (d) $1.6$ m s$^{-1}$.

Answer: (a)

1. Frequency $n = \frac{54}{60} = 0.9$ Hz
2. $v = n\lambda = 0.9 \times 5 = 4.5$ m/s
3. So the answer is (a) $4.5$ m/s.

Answer: (c)

1. Frequency does not change on entering a new medium, so $\lambda = \frac{v}{n} \propto v$.
2. $\frac{\lambda_{brass}}{\lambda_{air}} = \frac{3500}{350} = 10$
3. So the answer is (c) it increases by a factor $10$.

Answer: (c)

1. Distance between consecutive crests is the wavelength: $\lambda = 40$ m.
2. Time between consecutive waves is the period: $T = \frac{\lambda}{v} = \frac{40}{10} = 4$ s
3. So the answer is (c) $4$ s.

Answer: (a)

1. Loudness is the ear's response to intensity (energy per unit area per second).
2. Pitch is the ear's response to frequency.
3. So the answer is (a) intensity of sound and frequency of sound.

Answer: (a)

1. (A) True: pitch is the shrillness or sharpness of sound.
2. (B) True: a tone is a sound of a single frequency.
3. (C) True: pitch depends on frequency, loudness on intensity, so a high pitch sound can be soft.
4. (D) Wrong: a female voice usually has higher frequency, so it is sharper than a male voice.
5. So the answer is (a) D.

Answer: (b)

1. Percussion instruments produce sound when struck: drum, daphali, xylophone.
2. The saxophone is a wind instrument (vibrating air column).
3. So the answer is (b) Saxophone.

Answer: (b)

1. For a point source, $I = \frac{P}{4\pi r^2} \propto \frac{1}{r^2}$.
2. Also $I \propto A^2$, so $A \propto \frac{1}{r}$.
3. At $2r$: $A' = \frac{A}{2}$
4. So the answer is (b) $\frac{A}{2}$.

Answer: (b)

1. $V = \frac{4}{3}\pi r^3$, so $V \to 8V$ means $r \to 2r$.
2. $I = \frac{P}{4\pi r^2} \propto \frac{1}{r^2}$, so $I \to \frac{I}{4}$
3. So the answer is (b) decrease by 4 times.

Answer: (c)

1. Wave energy $E \propto A^2 n^2$.
2. Equal energies: $A_Q n_Q = A n \Rightarrow A_Q = A \frac{n}{n/4}$
3. $A_Q = 4A$
4. So the answer is (c) $4A$.

Answer: (b)

1. Frequency $n = \frac{45}{60} = 0.75$ Hz
2. $v = n\lambda = 0.75 \times 7 = 5.25$ m/s
3. So the answer is (b) $5.25$ m/s.

Answer: (b)

1. Wave energy $E \propto A^2 n^2$, so equal energies give $A n =$ constant.
2. $A_Q = A_p \frac{n}{n/4} = 4A_p$
3. So the answer is (b) $4A_p$.

Answer: (c)

1. Sound is a mechanical wave: it needs a material medium (solid, liquid or gas) to travel.
2. In vacuum there are no particles to vibrate, so sound cannot travel there.
3. Statements (a), (b), (d) are true.
4. So the answer is (c) Sound can travel through vacuum.

Answer: (b)

1. Five consecutive crests have four gaps: $4\lambda = 0.8$ m, so $\lambda = 0.2$ m.
2. $n = \frac{v}{\lambda} = \frac{56}{0.2} = 280$ Hz
3. So the answer is (b) 280 Hz.

Answer: (d)

1. Daphali, sambal and cymbals are struck, so they are percussion instruments.
2. The clarinet is a wind instrument (vibrating air column).
3. So the answer is (d) Clarinet.

Answer: (c)

1. $L_2 - L_1 = 10\log_{10}\frac{I_2}{I_1}$
2. $60 - 30 = 30 = 10\log_{10}\frac{I_2}{I_1} \Rightarrow \frac{I_2}{I_1} = 10^{3}$
3. So the answer is (c) 1000.

Answer: 3

1. Take the given $16 \times 10^{-8}$ W m$^{-2}$ as the intensity at $1$ m from the source; $I \propto \frac{1}{r^2}$.
2. $I_2 = \frac{16}{2^2} \times 10^{-8} = 4 \times 10^{-8}$ W m$^{-2}$
3. $I_4 = \frac{16}{4^2} \times 10^{-8} = 1 \times 10^{-8}$ W m$^{-2}$
4. Difference $= 3 \times 10^{-8}$ W m$^{-2}$, so the answer is $3$.

Answer: (a)

1. Wave energy $E \propto A^2 n^2$, so equal energies give $A n =$ constant.
2. $A_Y = A \frac{n}{n/3} = 3A$
3. So the answer is (a) 3A.

Answer: (b)

1. Wave energy $E \propto A^2 n^2$, so equal energies give $A n =$ constant.
2. $A_Q = A_P \frac{n}{n/4} = 4A_P$
3. So the answer is (b) $4A_P$.

Answer: (c)

1. In each vibration the wave advances one wavelength, so distance $= N\lambda = N\frac{v}{n}$.
2. $\lambda = \frac{320}{480} = \frac{2}{3}$ m
3. Distance $= 180 \times \frac{2}{3} = 120$ m
4. So the answer is (c) 120 m.

Answer: (a)

1. Pushing the block down by $x$ displaces extra liquid volume $Ax$, giving extra upthrust $F = -A\rho g\,x$.
2. This is SHM with force constant $k = A\rho g$.
3. $n = \frac{1}{2\pi}\sqrt{\frac{k}{m}} = \frac{1}{2\pi}\sqrt{\frac{A\rho g}{m}}$
4. So the answer is (a).

Answer: (c)

1. Frequency is fixed by the source, so it does not change on refraction.
2. In a denser medium the wave speed decreases, and $\lambda = \frac{v}{n}$ decreases with it.
3. So the answer is (c) wavelength and speed decrease, frequency remains constant.

Answer: (a)

1. Sound in air travels as compressions and rarefactions, so particles vibrate along the direction of travel: longitudinal.
2. Light is an electromagnetic wave with $\vec E$ and $\vec B$ perpendicular to the direction of travel: transverse.
3. So the answer is (a).

Answer: (c)

1. $\lambda = \frac{v}{n}$ for each fork.
2. $\lambda_1 = \frac{320}{320} = 1$ m, $\lambda_2 = \frac{320}{480} = 0.667$ m
3. $\Delta\lambda = 1 - 0.667 = 0.333$ m $\approx 33$ cm
4. So the answer is (c) 33 cm.

Answer: (a)

1. Distance $= N\lambda = N\frac{v}{n}$.
2. $\lambda = \frac{330}{220} = 1.5$ m
3. Distance $= 80 \times 1.5 = 120$ m
4. So the answer is (a) 120 m.

Answer: (c)

1. One vibration moves the wave forward by one wavelength $\lambda = \frac{V}{n}$.
2. After $x$ vibrations: distance $= x\lambda = \frac{xV}{n}$
3. So the answer is (c) $\frac{xV}{n}$.

Answer: (b)

1. Two crests are a whole number of wavelengths apart: $5 = n\lambda$.
2. A crest and a trough are an odd number of half wavelengths apart: $1.5 = (2m+1)\frac{\lambda}{2}$, so $\lambda = \frac{3}{2m+1}$.
3. Both hold for $\lambda = 1, \frac{1}{3}, \frac{1}{5}, \dots$ e.g. $\lambda = 1$: $5 = 5\lambda$, $1.5 = 3 \cdot \frac{\lambda}{2}$.
4. So the answer is (b) $\frac{1}{1}, \frac{1}{3}, \frac{1}{5}, \dots$

Answer: (b)

1. Write $s = \Delta p\, v^a \rho^b f^c$; $[s] =$ L and $[\Delta p] =$ M L$^{-1}$ T$^{-2}$.
2. Matching M, L, T gives $b = -1$, $a = -1$, $c = -1$: $s = \frac{\Delta p}{\rho v f}$.
3. $s = \frac{10}{1 \times 300 \times 1000} = 3.3 \times 10^{-5}$ m $\approx \frac{3}{100}$ mm
4. So the answer is (b) $\frac{3}{100}$ mm.

Answer: (c)

1. Sound level $L = 10\log_{10}(I/I_0)$, so $120 = 10\log_{10}(I/10^{-12})$ gives $I = 1\ \text{W/m}^2$.
2. Point source: $I = \dfrac{P}{4\pi r^2} \Rightarrow r = \sqrt{\dfrac{P}{4\pi I}} = \sqrt{\dfrac{2}{4\pi \times 1}}$
3. $r = \sqrt{0.159} \approx 0.40\ \text{m}$
4. So the answer is (c) $40\ \text{cm}$.

Answer: (b)

1. Effective length $\approx L + r$, so $T = 2\pi\sqrt{(L+r)/g}$ and $\dfrac{\Delta T}{T} = \dfrac{1}{2}\dfrac{\Delta r}{L}$.
2. $\Delta r = 2L\dfrac{\Delta T}{T} = 2 \times 1 \times 5\times10^{-4} = 10^{-3}\ \text{m}$
3. So the answer is (b) $0.1\ \text{cm}$.

Answer: (b)

1. Crest to neighbouring trough at one instant is $\frac{\lambda}{2} = 4$ cm, so $\lambda = 8$ cm.
2. Crest to trough at the same place (vertical) is $2A = 1$ cm, so $A = 0.5$ cm.
3. Next crest at the same place after one period: $T = 0.4$ s, so $\omega = \frac{2\pi}{T} = 5\pi$ rad/s.
4. $v_{max} = A\omega = 0.5 \times 5\pi = \frac{5\pi}{2}$ cm/s
5. So the answer is (b) $\frac{5\pi}{2}$ cm/s.

Answer: (c)

1. Frequency stays $700$ Hz in both media, so $\lambda \propto v$.
2. $\frac{\lambda_{brass}}{\lambda_{air}} = \frac{3500}{350} = 10$
3. So the answer is (c) increases by a factor 10.

Answer: (b)

1. $v = a\omega\cos(\omega t + \frac{\pi}{6})$, $v_{max} = a\omega$.
2. $v = \frac{v_{max}}{2} \Rightarrow \cos(\omega t + \frac{\pi}{6}) = \frac{1}{2} \Rightarrow \omega t + \frac{\pi}{6} = \frac{\pi}{3}$
3. $\omega t = \frac{\pi}{6} \Rightarrow \frac{2\pi}{T}t = \frac{\pi}{6} \Rightarrow t = \frac{T}{12}$
4. So the answer is (b) $T/12$.

Answer: (a)

1. $\Delta L = 10\log_{10}(I_1/I_2)$
2. $20 = 10\log_{10}(I_1/I_2) \Rightarrow I_1/I_2 = 10^2$
3. So the answer is (a) $100$.

Answer: (c)

1. Light (electromagnetic) travels in vacuum; sound (mechanical) does not, so (a) is false.
2. Sound in air is longitudinal; light is transverse.
3. So the answer is (c).

Answer: (b)

1. Sabine's formula: $T = \frac{0.161\,V}{aS}$, so $T \propto \frac{V}{S}$.
2. Doubling every dimension: $V \to 8V$, $S \to 4S$.
3. $T' = \frac{8}{4} \times 1 = 2$ s
4. So the answer is (b) 2.

Answer: (c)

1. For a point source, $I \propto \frac{1}{r^2}$.
2. $\frac{I_P}{I_Q} = \frac{r_Q^2}{r_P^2} = \frac{9}{4}$
3. So the answer is (c) 9 : 4.

Answer: (b)

1. With $\delta = \alpha + \frac{\pi}{2}$: $x = a\cos(\omega t + \alpha + \frac{\pi}{2}) = -a\sin(\omega t + \alpha)$.
2. $y = a\cos(\omega t + \alpha)$
3. $x^2 + y^2 = a^2$: equal amplitudes with $\frac{\pi}{2}$ phase difference.
4. So the answer is (b) a circle.

Answer: (a)

1. Going from maximum displacement to zero (mean position) takes a quarter period: $\frac{T}{4} = 0.170$ s.
2. $T = 0.680$ s
3. $n = \frac{1}{T} = \frac{1}{0.680} \approx 1.47$ Hz
4. So the answer is (a) $1.47$ Hz.

Answer: (c)

1. $\lambda = \frac{v}{n}$
2. $\lambda = \frac{1.7 \times 10^{3}}{4.2 \times 10^{6}} \approx 4.0 \times 10^{-4}$ m
3. So the answer is (c) $4 \times 10^{-4}$ m.

Answer: (b)

1. In a wave, the particles of the medium only oscillate about their mean positions; they do not move along with the wave.
2. So no mass is carried forward; the disturbance carries energy.
3. So the answer is (b) energy.

Answer: (c)

1. Intensity $I \propto A^2 n^2$.
2. $\frac{I'}{I} = (2)^2 \times \left(\frac{1}{4}\right)^2 = 4 \times \frac{1}{16} = \frac{1}{4}$
3. So the answer is (c) decreasing by a factor of 4.

Answer: (a)

1. $v = \sqrt{T/\mu} \Rightarrow T \propto v^2$
2. $T'/T = (1.2)^2 = 1.44$
3. So the answer is (a) $44\%$.

Answer: (c)

1. $T \propto v^2$, so tripling the speed makes $T' = 9T = 9 \times 2 = 18$ kg wt.
2. Increase $= 18 - 2 = 16$ kg wt.
3. So the answer is (c) $16$.

Answer: (a)

1. Water vapour ($M = 18$) is lighter than dry air ($M \approx 29$), so moist air has lower density.
2. $v = \sqrt{\gamma P/\rho}$ increases as $\rho$ decreases.
3. So the answer is (a) Speed of sound waves is more.

Answer: 819

1. $v \propto \sqrt{T}$, so doubling $v$ needs $T' = 4 \times 273 = 1092$ K.
2. $\alpha = 1092 - 273 = 819$
3. Answer: $819^\circ$C.

Answer: (d)

1. $\mu = 10^{-3}/0.5 = 2\times10^{-3}$ kg/m, $v = \sqrt{0.8/(2\times10^{-3})} = 20$ m/s.
2. A pulse reflects at both fixed ends (two inversions), returning upright after a round trip $2L$.
3. Minimum $\Delta t = \dfrac{2L}{v} = \dfrac{2\times0.5}{20} = 0.05\ \text{s}$
4. So the answer is (d) $0.05$ s.

Answer: (b)

1. $v = \sqrt{\dfrac{\gamma RT}{M}}$; both monoatomic (same $\gamma$) at the same $T$, so $v \propto 1/\sqrt{M}$.
2. $\dfrac{v_1}{v_2} = \sqrt{\dfrac{m_2}{m_1}}$
3. So the answer is (b).

Answer: (d)

1. $\dfrac{v_{He}}{v_{N_2}} = \sqrt{\dfrac{\gamma_{He}}{\gamma_{N_2}}\cdot\dfrac{M_{N_2}}{M_{He}}}$
2. $= \sqrt{\dfrac{5/3}{7/5}\times\dfrac{28}{4}} = \sqrt{\dfrac{25}{21}\times 7} = \sqrt{\dfrac{25}{3}}$
3. So the answer is (d) $\dfrac{5}{\sqrt{3}}$.

Answer: (b)

1. $v = \sqrt{\gamma P/\rho}$ with $\rho$ constant, so $v \propto \sqrt{P}$.
2. $\dfrac{P_2}{P_1} = (1.4)^2 = 1.96$
3. So the answer is (b) $1.96$.

Answer: (b)

1. $t = L/v = L\sqrt{\mu/T}$, same $T$ for both.
2. $\dfrac{t_1}{t_2} = \dfrac{L_A}{L_B}\sqrt{\dfrac{\mu_A}{\mu_B}} = \dfrac{2.5}{1.5}\sqrt{\dfrac{1}{2}} = 1.667 \times 0.707$
3. $t_1/t_2 \approx 1.18$
4. So the answer is (b) 1.18.

Answer: (a)

1. For a hanging rope, at height $y$ above the lower end $v = \sqrt{gy}$.
2. Upper end: $y = 2.5$ m, $v = \sqrt{25} = 5$ m/s.
3. At $y = 0.5$ m: $v = \sqrt{5}$ m/s.
4. So the answer is (a) $5$ m/s, $\sqrt{5}$ m/s.

Answer: (c)

1. $\mu = \rho \pi R^2$, so with equal $T$: $v = \sqrt{T/\mu} \propto \dfrac{1}{R}$.
2. $\dfrac{v_2}{v_1} = \dfrac{R}{R/2} = 2$
3. So the answer is (c) $2$.

Answer: (c)

1. A: true, $v = \sqrt{E/\rho}$ and solids have far larger elastic moduli, so sound is faster in solids.
2. R: false, gases are highly compressible, so their bulk modulus is much smaller than that of solids.
3. So the answer is (c) A is true but R is false.

Answer: (b)

1. $\mu = 0.2/2 = 0.1\ \text{kg/m}$
2. $v = \sqrt{T/\mu} = \sqrt{2.5/0.1} = 5\ \text{m/s}$
3. $t = L/v = 2/5 = 0.4\ \text{s}$
4. So the answer is (b) $0.4$ s.

Answer: (d)

1. $v = \sqrt{\dfrac{\gamma RT}{M}}$ with $M = 0.032$ kg/mol, $T = 273$ K.
2. $v = \sqrt{\dfrac{1.4 \times 8.3 \times 273}{0.032}} = \sqrt{9.91\times10^4}$
3. $v \approx 315\ \text{m/s}$
4. So the answer is (d) 315 m/s.

Answer: (c)

1. $v = \sqrt{T/\mu} = \sqrt{\dfrac{70}{7\times10^{-3}}} = \sqrt{10^4}$
2. So the answer is (c) $100$ m/s.

Answer: (d)

1. Both diatomic (same $\gamma$), same $T$: $v \propto \dfrac{1}{\sqrt{M}}$.
2. $\dfrac{v_{H_2}}{v_{O_2}} = \sqrt{\dfrac{32}{2}} = 4$
3. So the answer is (d) $4:1$.

Answer: (c)

1. At height $y$ above the lower end, tension $= \mu g y$, so $v = \sqrt{gy}$.
2. Midpoint: $y = 10$ m, $v = \sqrt{10 \times 10} = 10$ m/s.
3. So the answer is (c) $10\ \text{ms}^{-1}$.