JEE Physics Sound Waves MCQs Set 01

Answer: (b) 1 and 3 only

Explanation:
1. \( \frac\lambda2=83.9-50.7=33.2\ \text{cm}\Rightarrow\lambda=66.4\ \text{cm} \) (3 true); \( v=500\times0.664=332\ \text{m/s} \) (1 true).
2. Fundamental would be at about \( \lambda/4=16.6\ \text{cm} \); 50.7 cm is near \( \frac{3\lambda}{4}=49.8\ \text{cm} \), the second resonance, so 4 is false.
3. This gives an end correction of \( 49.8-50.7=-0.9\ \text{cm} \) (negative, not physical), so 2 is false.

Teacher's Note:
1. The difference of two successive resonance lengths is half a wavelength.
2. End correction is the amount by which the real column length falls short of the quarter-wave multiples.

Answer: (a) 0.2 m

Explanation:
1. \( \lambda_r=\frac{v-v_w}{f'}=\frac{332-64}{1320}=0.203\ \text{m}\approx0.2\ \text{m} \) (the reflected waves move at 332 m/s in air).

Teacher's Note:
1. The reflecting wall emits 1320 waves per second while moving in the direction of the reflected waves.
2. A moving source towards the direction of propagation shortens the wavelength.

Answer: (a) 12 Hz

Explanation:
1. At t = 1 s the body moves down at 10 m/s. Before crossing: both approach, \( f_1=150\,\frac{300+2}{300-10}=156.2\ \text{Hz} \).
2. After crossing: both recede, \( f_2=150\,\frac{300-2}{300+10}=144.2\ \text{Hz} \).
3. Difference \( \approx12\ \text{Hz} \).

Teacher's Note:
1. Before crossing the observer moves towards the source and the source towards the observer; after crossing both move away.
2. The source speed at t = 1 s is gt = 10 m/s.

Answer: (d) \( \frac{V+V_w-V_s}{f} \)

Explanation:
1. Towards the wall the sound moves at \( V+V_w \) relative to the ground; the source follows its own waves at \( V_s \), reducing the spacing.
2. \( \lambda=\frac{V+V_w-V_s}{f} \).

Teacher's Note:
1. A source moving towards the direction of propagation compresses the wavelength.
2. The wind speed adds to the sound speed in its own direction.

Answer: (c) 30 m/s

Explanation:
1. \( \frac{f_a}{f_r}=\frac{v+u}{v-u}=\frac{2.2}{1.8}=\frac{11}{9}\Rightarrow9v+9u=11v-11u \).
2. \( u=\frac{v}{10}=30\ \text{m/s} \).

Teacher's Note:
1. The ratio of approach to recede frequencies eliminates the source frequency.
2. Solve the linear equation for u.

Answer: (b) \( f'=f\left(1+\frac{at}{v}\right):\text{ a straight line rising with }t \)

Explanation:
1. Observer speed \( u=at \).
2. \( f'=\frac{v+at}{v}f \), which is linear in t with positive slope.

Teacher's Note:
1. Frequency heard is linear in the observer's speed when the source is stationary.
2. Uniform acceleration makes speed linear in time.

Answer: (b) the same wavelength and velocity but a different amplitude

Explanation:
1. Superposition of waves of the same frequency moving at the same speed gives a wave of the same frequency, wavelength and speed.
2. The amplitude becomes \( 2A\cos45^\circ=\sqrt2A \).

Teacher's Note:
1. Only the amplitude and phase of the resultant change.
2. Speed is a property of the medium.

Answer: (a) 2 and 4 only

Explanation:
1. At an open end the pressure stays at atmospheric (pressure node): the pulse reflects with opposite pressure, a low-pressure pulse.
2. At a closed end the air cannot move (pressure antinode): the pulse reflects with the same pressure, a high-pressure pulse.

Teacher's Note:
1. Pressure reflects like displacement does at the opposite kind of end.
2. Open end: pressure node; closed end: pressure antinode.

Answer: (d) 178.2 Hz

Explanation:
1. Stress \( =Y\times0.01=2.2\times10^9\ \text{N/m}^2 \).
2. \( v=\sqrt{\frac{\text{stress}}{\rho}}=\sqrt{\frac{2.2\times10^9}{7.7\times10^3}}=534.5\ \text{m/s} \).
3. \( f_1=\frac{v}{2L}=\frac{534.5}{3}=178.2\ \text{Hz} \).

Teacher's Note:
1. Tension per unit area is stress, and mu = rho times area.
2. Fundamental of a string fixed at both ends is v/2L.

Answer: (c) 330 Hz

Explanation:
1. \( L_2D=\sqrt{40^2+9^2}=41\ \text{m} \), path difference \( =1\ \text{m} \).
2. First minimum at \( \Delta=\lambda/2 \) (\( f=165 \) Hz); first maximum at \( \Delta=\lambda\Rightarrow f=\frac{330}{1}=330\ \text{Hz} \).

Teacher's Note:
1. As frequency rises from zero the detector goes max, min, max, so the first minimum comes at lambda = 2 m and the next maximum at lambda = 1 m.
2. A 9-40-41 right triangle gives the longer path.

Answer: (c) 20 m/s

Explanation:
1. \( \frac{f'}{f}=\frac{v+u}{v-u}=\frac98\Rightarrow8v+8u=9v-9u\Rightarrow u=\frac{v}{17}=20\ \text{m/s} \).

Teacher's Note:
1. Both the source and observer approach at speed u, so the numerator has v+u and the denominator v-u.
2. 340/17 = 20.

Answer: (b) \( \frac{2c}{3} \)

Explanation:
1. \( \gamma=1+\frac2f=\frac43 \).
2. \( v=\sqrt{\frac{\gamma RT}{M}},\ c=\sqrt{\frac{3RT}{M}}\Rightarrow v=c\sqrt{\frac{\gamma}{3}}=c\sqrt{\frac49}=\frac{2c}{3} \).

Teacher's Note:
1. Relate gamma to the degrees of freedom: gamma = 1 + 2/f.
2. Compare v and c, both depend on RT/M.

Answer: (a) the displacement is zero

Explanation:
1. The closed end holds the air particles still: a displacement node.
2. It is simultaneously a pressure antinode, so the pressure variation is maximum there.

Teacher's Note:
1. A closed end is a displacement node and a pressure antinode.
2. Reflection at a closed end is without phase change for pressure.

Answer: (d) 33 cm and 13.2 cm

Explanation:
1. \( L_{open}=\frac{330}{2\times500}=0.33\ \text{m} \).
2. \( L_{closed}=\frac{264}{4\times500}=0.132\ \text{m}=13.2\ \text{cm} \).

Teacher's Note:
1. Open pipe: L = v/2f; closed pipe: L = v/4f.
2. Use the speed that belongs to each gas.

Answer: (d) 10 cm

Explanation:
1. \( \lambda=\frac{350}{1750}=0.2\ \text{m}=20\ \text{cm} \).
2. Successive resonances of a closed column differ in length by \( \lambda/2=10\ \text{cm} \).

Teacher's Note:
1. Raising the pipe increases the air column length.
2. Successive resonance lengths are half a wavelength apart.

Answer: (c) \( \frac{vx}{2l^2} \)

Explanation:
1. \( f_1-f_2=\frac{v}{2l}-\frac{v}{2(l+x)}=\frac{v\,x}{2l(l+x)} \).
2. For \( x\ll l \) this is \( \frac{vx}{2l^2} \).

Teacher's Note:
1. Subtract the two fundamental frequencies and simplify using x much smaller than l.
2. The beat frequency is the difference between the two frequencies.

Answer: (b) \( y_1,\ y_2\text{ and }y_4\text{ only} \)

Explanation:
1. A steady interference pattern needs equal frequencies and constant phase difference.
2. y1, y2 and y4 have angular frequency \( \pi \); y3 has \( 2\pi \) and does not interfere steadily with them.

Teacher's Note:
1. Different frequencies give a phase difference that changes with time.
2. Only the angular frequency matters, not the phase constants.

Answer: (a) \( \frac{\pi}{2}\ \text{rad} \)

Explanation:
1. \( \Delta\phi=\omega\Delta t=2\pi(100)(2.5\times10^{-3})=\frac\pi2 \).

Teacher's Note:
1. At a fixed point the phase advances by omega times the elapsed time.
2. 2.5 ms is a quarter of the 10 ms period.

Answer: (a) 4I

Explanation:
1. \( I_A=I+4I+2\sqrt{4}I\cos\frac\pi2=5I \).
2. \( I_B=I+4I+4I\cos\pi=I \), so the difference is \( 4I \).

Teacher's Note:
1. Resultant intensity is I1 + I2 + 2 sqrt(I1 I2) cos(phi).
2. At a phase difference of pi the amplitudes subtract.

Answer: (d) \( 20\log3 \)

Explanation:
1. Amplitudes are in the ratio 2 : 1, so \( I_{max}\propto(2+1)^2=9 \) and \( I_{min}\propto(2-1)^2=1 \).
2. Difference \( =10\log\frac{9}{1}=20\log3 \) dB.

Teacher's Note:
1. Intensity is proportional to amplitude squared; with intensity ratio 4:1 the amplitude ratio is 2:1.
2. Loudness difference = 10 log of the intensity ratio.

Answer: (c) 250 Hz

Explanation:
1. With 250 Hz: beats with A = 6 Hz and with B = 12 Hz (twice as large).
2. Check others: 268 Hz gives 12 and 6 (wrong order); 260 Hz gives 4 and 2 (wrong order).

Teacher's Note:
1. Test each option against both forks.
2. The unknown note must be farther from B than from A.

Answer: (b) 0.4 m from the open end

Explanation:
1. First overtone: \( \lambda=\frac{4L}{3}=1.6\ \text{m} \).
2. Pressure antinodes are displacement nodes: at the closed end and at \( \frac\lambda2=0.8\ \text{m} \) from it, i.e. 0.4 m from the open end.

Teacher's Note:
1. Measure positions of nodes from the closed end first and convert them.
2. The open end is a pressure node.

Answer: (b) 344

Explanation:
1. Closed column first overtone: \( f=\frac{3v}{4L}=\frac{3\times340}{4\times0.75}=340\ \text{Hz} \) (string frequency).
2. Beats with the fork are 4; increasing tension raises the string frequency and reduces the beats, so the fork is above the string: \( n=340+4=344\ \text{Hz} \).

Teacher's Note:
1. If the fork were at 336 Hz, raising the string frequency would increase the beats.
2. The first overtone of a closed pipe is its third harmonic.

Answer: (a) 5 Hz

Explanation:
1. Each walking man is \( \sqrt{5^2+12^2}=13\ \text{m} \) from the stationary man, with \( \cos\theta=\frac{5}{13} \) between the line of motion and the line of sight.
2. Front man recedes: component \( 1\times\frac5{13} \); rear man approaches: component \( 2\times\frac5{13} \).
3. \( \Delta f\approx\frac{1430}{330}\times\frac{15}{13}=5\ \text{Hz} \).

Teacher's Note:
1. Only the velocity component along the line of sight matters.
2. The two shifts have opposite signs, so the components add.

Answer: (d) a

Explanation:
1. Third overtone of a closed pipe is the 7th harmonic: \( l=\frac{7\lambda}{4}\Rightarrow\lambda=\frac{4l}{7} \).
2. Distance from the closed end \( \frac l7=\frac\lambda4 \): a displacement antinode, so the amplitude is a.

Teacher's Note:
1. Closed pipe overtones are 3f, 5f, 7f; the third overtone is 7f.
2. A quarter wavelength from a node is an antinode.

Answer: (c) 12%

Explanation:
1. Approaching: \( 1000\times\frac{320}{300}=1066.7\ \text{Hz} \). Receding: \( 1000\times\frac{320}{340}=941.2\ \text{Hz} \).
2. Change \( =\frac{1066.7-941.2}{1066.7}\approx12\% \).

Teacher's Note:
1. Calculate both Doppler frequencies and express the drop relative to the higher value.
2. The shift is not symmetric because the denominators differ.

Answer: (c) 2 and 4 only

Explanation:
1. \( f\propto\sqrt{\gamma/M} \).
2. 1: M rises from 2 to 32, so f falls (statement false). 2: M falls from 32 to 28, same gamma, so f rises (true). 3: He has higher gamma and much lower M than N2, so f rises (false). 4: He (\( \gamma=1.67,\ M=4 \)) to CH4 (\( \gamma=1.33,\ M=16 \)): f falls (true).

Teacher's Note:
1. Speed of sound in a gas is sqrt(gamma R T / M); compare gamma/M for the two gases.
2. The pipe length is fixed, so frequency is proportional to speed.

Answer: (b) \( \frac{a}{v_s-v_2} \)

Explanation:
1. Sound travels at \( v_s \) relative to air; the observer is moving away with speed \( v_2 \), so the pulse closes the gap a at the relative rate \( v_s-v_2 \).
2. \( t_1=\frac{a}{v_s-v_2} \) (the source speed does not matter once the pulse has been emitted).

Teacher's Note:
1. The speed of a pulse in still air does not depend on the speed of its source.
2. The observer moving away reduces the closing speed.