JEE Physics Sound Waves MCQs Set 09

Answer: (c) 1, 2 and 4 only

Explanation:
1. Frequency is fixed by the source and does not change at a boundary (1 is true).
2. \( \lambda=v/f \); v is larger in B, so the wavelength increases (2 is true, 3 is false).
3. With no absorption or reflection, all the energy is transmitted, so the intensity is unchanged (4 is true).

Teacher's Note:
1. Frequency stays constant when a wave changes medium; speed and wavelength change together.

Answer: (d) All four, 1, 2, 3 and 4

Explanation:
1. Every particle oscillates as \( A(x)\cos\omega t \) with the same time factor.
2. So all reach zero displacement (and maximum velocity) together at \( \omega t=\pi/2 \), and all reach maximum displacement (and are momentarily at rest) together at \( \omega t=0 \).
3. All four statements are true.

Teacher's Note:
1. Every particle in a standing wave shares the same time factor, so they all pass through the same phase of oscillation simultaneously.

Answer: (a) 3 and 4 only

Explanation:
1. Between two successive nodes the factor \( \sin kx \) has a single sign, so those particles are in phase.
2. Across a node \( \sin kx \) changes sign, so particles on the two sides are in opposite phase.
3. Statements 3 and 4 are correct.

Teacher's Note:
1. The sign of the spatial factor sin kx decides the phase.
2. It flips at every node.

Answer: (b) \( y=A\sin\dfrac{n\pi x}{L}\cos\omega t \)

Explanation:
1. Both ends are fixed, so y must be zero at x = 0 and x = L for all t.
2. \( \sin\dfrac{n\pi x}{L} \) vanishes at both ends, whereas the cosine form does not vanish at x = 0.
3. The valid shape is \( y=A\sin\dfrac{n\pi x}{L}\cos\omega t \) (a sine in time works equally well).

Teacher's Note:
1. A clamped end must be a node, so the spatial factor must be a sine that vanishes at both ends.

Answer: (a) 2 and 3 only

Explanation:
1. A standing wave has amplitude \( 2A \), so \( 2A=2 \) mm and \( A=1 \) mm.
2. \( k=6.28=2\pi/\lambda \), so \( \lambda=1 \) m.
3. The smallest string length (fixed ends) is \( \lambda/2=50 \) cm.

Teacher's Note:
1. The maximum amplitude of a standing wave is twice the amplitude of each travelling wave.
2. The shortest string holds half a wavelength.

Answer: (b) 4V

Explanation:
1. The string frequency in the fundamental mode is \( f=v/2L \).
2. In air \( V=f\lambda=f\cdot\dfrac L2 \), so \( f=\dfrac{2V}{L} \).
3. \( v=2Lf=2L\cdot\dfrac{2V}{L}=4V \).

Teacher's Note:
1. The string and the sound share the same frequency.
2. Use it to link the two speeds.

Answer: (a) \( L=(n+1)\,l \)

Explanation:
1. The nth overtone is the (n+1)th harmonic, so the string has n+1 loops.
2. The distance between nodes is the length of one loop: \( l=\dfrac{L}{n+1} \).
3. \( L=(n+1)l \).

Teacher's Note:
1. The nth overtone is the (n+1)th harmonic.
2. The loop length is the node-to-node distance.

Answer: (a) \( 2\pi\times10^{-2}\ \text{m} \)

Explanation:
1. Maximum particle speed \( =\omega A=2\pi fA=\dfrac v{10} \).
2. With \( v=f\lambda \): \( 2\pi A=\dfrac\lambda{10} \).
3. \( \lambda=20\pi A=20\pi\times10^{-3}=2\pi\times10^{-2} \) m.

Teacher's Note:
1. Relate the maximum particle speed 2 pi f A to the wave speed f lambda; the frequency cancels.

Answer: (a) \( v\propto\sqrt{x} \)

Explanation:
1. The tension at distance x from the lower end is \( T=\mu xg \).
2. \( v=\sqrt{T/\mu}=\sqrt{gx} \).
3. So \( v\propto\sqrt x \).

Teacher's Note:
1. Tension in a hanging rope grows linearly with the length below the point, so the wave speed grows as the square root of x.

Answer: (b) 2 and 4 only

Explanation:
1. An open pipe has displacement antinodes at both ends and a displacement node at the centre in the fundamental mode.
2. A displacement node is a pressure antinode.
3. So the molecules are at rest at the centre (2) and the pressure variation is maximum there (4).

Teacher's Note:
1. Displacement nodes and pressure antinodes coincide in a pipe.

Answer: (a) 3 cm

Explanation:
1. \( \lambda/2=74.1-24.1=50 \) cm, so \( \lambda/4=25 \) cm.
2. First resonance with end correction: \( 24.1+0.3d=25 \).
3. \( d=3 \) cm.

Teacher's Note:
1. The difference between successive resonance lengths is half a wavelength.
2. End correction is 0.3 times the diameter.

Answer: (a) \( \dfrac{2c}{3} \)

Explanation:
1. \( \gamma=1+\dfrac2f=1+\dfrac26=\dfrac43 \).
2. \( c=\sqrt{3RT/M} \) and \( v=\sqrt{\gamma RT/M} \), so \( v/c=\sqrt{\gamma/3}=\sqrt{4/9} \).
3. \( v=\dfrac{2c}{3} \).

Teacher's Note:
1. Speed of sound is root of gamma P by rho, while rms speed is root of 3P by rho.
2. Their ratio involves only gamma.

Answer: (c) \( V_d\\lt V_m\ \text{because moist air has lower density than dry air} \)

Explanation:
1. Water vapour (molar mass 18) replaces air (mean molar mass 29), so moist air is less dense at the same pressure and temperature.
2. \( v=\sqrt{B/\rho} \) with almost the same B, so a lower density gives a higher speed.
3. \( V_m>V_d \).

Teacher's Note:
1. Humid air is lighter than dry air, so sound travels slightly faster in it.

Answer: (b) the overtones present in the sound

Explanation:
1. Different sources emit the same fundamental frequency with different mixes of overtones.
2. This quality (timbre) lets us distinguish the source.

Teacher's Note:
1. Timbre depends on the overtone structure, not on pitch or loudness.

Answer: (a) 16

Explanation:
1. Initially \( I_0=(\sqrt I+\sqrt I)^2=4I \).
2. Now \( I_1=0.36I \) and \( I_2=I \), so amplitudes are in the ratio 0.6 : 1.
3. \( \dfrac{I_{max}}{I_{min}}=\dfrac{(1+0.6)^2}{(1-0.6)^2}=\dfrac{2.56}{0.16}=16 \).

Teacher's Note:
1. Intensities add through amplitudes.
2. The amplitude ratio is the square root of the intensity ratio.

Answer: (b) 258 Hz

Explanation:
1. The string frequency is 254 Hz or 258 Hz initially (2 beats).
2. Increasing tension increases the frequency. If it were 254 Hz it would move closer to 256 Hz and the beats would decrease.
3. Beats rose to 3, so the frequency moved away from 256 Hz: initial 258 Hz.

Teacher's Note:
1. Raising tension raises frequency.
2. The change in beat count tells you which side of the fork the string started.

Answer: (a) 500 Hz

Explanation:
1. \( f'=f\dfrac{v}{v-v_s}=450\times\dfrac{330}{330-33} \).
2. \( f'=450\times\dfrac{330}{297}=500 \) Hz.

Teacher's Note:
1. For an approaching source the observed frequency is increased by the factor v over v minus the source speed.

Answer: (a) 5

Explanation:
1. Frequency received by the wall: \( f_1=165\times\dfrac{335}{335-5}=167.5 \) Hz.
2. Frequency of the reflection heard by the moving passenger: \( f_2=167.5\times\dfrac{335+5}{335}=170 \) Hz.
3. Beats \( =170-165=5 \) per second.

Teacher's Note:
1. Apply the Doppler shift twice: wall as an observer, then as a source.
2. The passenger hears the horn and its echo together.