JEE Physics Sound Waves MCQs Set 06

Answer: (a) 16

Explanation:
1. Maxima occur where the path difference is \( n\lambda \) with \( |n|\le4 \).
2. The hyperbola branches \( n=\pm1,\pm2,\pm3 \) each meet the ellipse at 2 points: 12 points.
3. The perpendicular bisector (n = 0) meets it at 2 points, and the extensions of the axis (n = +4 and -4) each meet it at 1 point: 4 more points.
4. Total = 12 + 4 = 16.

Teacher's Note:
1. Count all the integer path differences that can occur, from -4 to +4.
2. Axis points beyond the sources have path difference exactly equal to the source separation, 4 lambda.

Answer: (d) \( N^2I_0 \)

Explanation:
1. In phase, the amplitudes add: \( A = NA_0 \).
2. Intensity is proportional to the square of the amplitude, so \( I = N^2I_0 \).

Teacher's Note:
1. Coherent addition of waves is by amplitude, not intensity.
2. Incoherent sources would give N times I0.

Answer: (b) \( 10,\ \tfrac{10}{2},\ \tfrac{10}{3},\ \tfrac{10}{4},\dots \)

Explanation:
1. Reflected path: \( 2\sqrt{60^2+25^2} = 130 \) m; direct path 120 m; path difference 10 m.
2. Constructive interference for \( n\lambda = 10 \): \( \lambda = 10/n \) with \( n = 1,2,3,\dots \)

Teacher's Note:
1. Use the Pythagoras theorem for the reflected ray (60, 25, 65).
2. Reflection at a rigid ceiling is taken as having no extra phase change here.

Answer: (a) The equation of the standing-wave part is \( y = 1.8A\sin ax\cos bt \)

Explanation:
1. \( \lambda = 2\pi/k = 2\pi/a \) and \( f = b/2\pi \), so the first statement is correct.
2. Intensity is proportional to amplitude squared: \( \sqrt{0.64} = 0.8 \), so the reflected amplitude is 0.8 A (correct).
3. Reflection from a denser medium inverts the displacement: \( y = A\cos(ax+bt)-0.8A\cos(ax-bt) \) (correct); the particle speed reaches \( 1.8bA \) at the antinodes (correct).
4. The standing part is \( 0.8A[\cos(ax+bt)-\cos(ax-bt)] = -1.6A\sin ax\sin bt \), not \( 1.8A\sin ax\cos bt \), so that statement is incorrect.

Teacher's Note:
1. Write the resultant as a standing wave plus a travelling wave.
2. The standing wave has amplitude twice the reflected amplitude, i.e. 1.6 A.

Answer: (d) \( \sqrt3/5 \)

Explanation:
1. \( v\propto\sqrt{\gamma/M} \): \( \dfrac{v_N}{v_{He}} = \sqrt{\dfrac{(7/5)/28}{(5/3)/4}} = \sqrt{\dfrac{0.05}{0.41667}} = \sqrt{0.12} = \dfrac{\sqrt3}{5} \).

Teacher's Note:
1. Nitrogen is diatomic (gamma 7/5), helium monatomic (gamma 5/3).
2. The comparison needs both gamma and molar mass.

Answer: (b) If the pressure rises (at constant temperature), the fundamental frequency increases

Explanation:
1. The speed of sound in an ideal gas is \( \sqrt{\gamma RT/M} \), which does not depend on pressure at constant temperature, so the frequency does not change.
2. Temperature increases the speed, hence the frequency (true); the first overtone of a closed pipe is 3 x 300 = 900 Hz (true); an open pipe needs L = v/2f, twice the closed length v/4f (true).

Teacher's Note:
1. Only temperature, gas type and pipe length matter for the pitch.
2. Density and pressure change together so the speed stays the same.

Answer: (d) 200 Hz

Explanation:
1. Open pipe fundamental: \( f = v/2L \). Closed pipe third harmonic: \( 3v/4L = 1.5f \).
2. \( 1.5f = f+100 \Rightarrow f = 200 \) Hz.

Teacher's Note:
1. Same length: the closed pipe fundamental is half the open pipe fundamental.
2. The third harmonic of the closed pipe is 1.5 times the open pipe fundamental.

Answer: (c) 551, 553, 558 Hz

Explanation:
1. Beats with the lowest fork give \( 550+\{1,2,3,5,7,8\} \), so the others are three of 551, 552, 553, 555, 557, 558 with the maximum 8 Hz.
2. Try 551, 553, 558: the pair differences are 1, 3, 8 (with 550), 2, 7 and 5 among themselves, giving exactly {1, 2, 3, 5, 7, 8}.
3. The other options give 10 Hz, or repeat 6 Hz, so they fail.

Teacher's Note:
1. List all six pairwise differences and match them with the given set.
2. The largest beat frequency is between the highest and the lowest fork.

Answer: (a) \( \dfrac{f_r+f_a}{2} \)

Explanation:
1. \( f_a = f\dfrac{c+v}{c} \), \( f_r = f\dfrac{c-v}{c} \).
2. Adding, \( f_a+f_r = 2f \), so \( f = (f_a+f_r)/2 \).

Teacher's Note:
1. For a moving observer the apparent frequency changes linearly with v.
2. The two shifts are equal and opposite, so the average is the true frequency.

Answer: (c) 22 m/s

Explanation:
1. Horn heard by the motorcyclist (both moving the same way, away from the police car's sound origin): \( f_1 = 176\dfrac{330-v}{330-22} \).
2. Siren heard by the motorcyclist (moving towards it): \( f_2 = 165\dfrac{330+v}{330} \).
3. No beats means \( f_1 = f_2 \): \( 0.5714(330-v) = 0.5(330+v) \), giving \( v = 22 \) m/s.

Teacher's Note:
1. Set the two apparent frequencies equal.
2. Be careful with the direction of the motion relative to each sound.

Answer: (a) 19/18

Explanation:
1. \( f_1 = f\dfrac{340}{340-34} = f\dfrac{340}{306} \) and \( f_2 = f\dfrac{340}{340-17} = f\dfrac{340}{323} \).
2. \( f_1/f_2 = 323/306 = 19/18 \).

Teacher's Note:
1. A faster source gives a higher frequency, so the ratio exceeds 1.
2. Reduce 323/306 by 17.

Answer: (d) 1.25

Explanation:
1. Moving source: \( f/f' = (c+v)/c = 1.2 \Rightarrow v = 0.2c \).
2. Moving observer: \( f' = f(c-v)/c = 0.8f \), so \( f/f' = 1/0.8 = 1.25 \).

Teacher's Note:
1. The two Doppler formulas are not symmetric when the speed is comparable to c.
2. Compute v/c from the first case, then reuse it.

Answer: (c) \( \left(\dfrac{V+U_w}{V}\right)\lambda_0 \)

Explanation:
1. The frequency is unchanged: \( f = V/\lambda_0 \).
2. Speed of the wave relative to the ground is \( V+U_w \), so \( \lambda = (V+U_w)/f = \dfrac{V+U_w}{V}\lambda_0 \).

Teacher's Note:
1. Wind changes the wave speed relative to the ground and hence the wavelength.
2. The frequency of the source is not affected by wind when both are at rest.

Answer: (b) \( 2u/\lambda \)

Explanation:
1. Between the sources, one wave is Doppler shifted up and the other down.
2. \( f_1 = f+\dfrac{u}{\lambda} \), \( f_2 = f-\dfrac{u}{\lambda} \), so the beat frequency is \( 2u/\lambda \).

Teacher's Note:
1. An approaching and a receding source give shifts of opposite sign.
2. For a moving observer the shift is u over lambda.

Answer: (a) \( \dfrac{c\nu}{c-\sqrt{2gd(1-\cos\theta)}},\ \dfrac{c\nu}{c+\sqrt{2gd(1-\cos\theta)}} \)

Explanation:
1. Maximum speed of the swing at the bottom: \( v_m = \sqrt{2gd(1-\cos\theta)} \).
2. At that point the source moves along the line of sight, so \( f_{max} = \dfrac{c\nu}{c-v_m} \) (approaching) and \( f_{min} = \dfrac{c\nu}{c+v_m} \) (receding).

Teacher's Note:
1. Energy conservation gives the speed at the lowest point of the swing.
2. Extreme Doppler shifts occur at the maximum speed along the line of sight.

Answer: (d) 630.5 Hz

Explanation:
1. Distance from the car to the man \( = \sqrt{800^2+600^2} = 1000 \) m.
2. Velocity of the car is 20 m/s; its component along the line of sight is \( 20\times\dfrac{800}{1000} = 16 \) m/s.
3. \( f' = 600\times\dfrac{330}{330-16} = 630.5 \) Hz.

Teacher's Note:
1. Only the component of the source's velocity along the line to the observer matters.
2. The 6-8-10 triangle gives cosine 4/5.

Answer: (c) Changing the amplitude of the signal voltage

Explanation:
1. The positions of maxima and minima depend on the path differences and on the wavelength.
2. Moving a speaker changes path differences; changing the frequency or the gas changes the wavelength; changing the amplitude only scales the intensities.

Teacher's Note:
1. Interference positions are set by geometry and wavelength.
2. Amplitude changes affect loudness, not the interference pattern.

Answer: (b) 9 : 1

Explanation:
1. Intensity \( I = \tfrac12\rho v\omega^2A^2 \propto f^2 \) at constant amplitude.
2. \( I_2/I_1 = (1200/400)^2 = 9 \).

Teacher's Note:
1. Intensity depends on the square of the frequency at a fixed displacement amplitude.
2. Tripling the frequency multiplies the intensity by nine.

Answer: (d) 2.32

Explanation:
1. The man hears the direct sound from a receding source: \( f_1 = 400\times\dfrac{345}{346} \).
2. He hears the echo, as from a source approaching the wall: \( f_2 = 400\times\dfrac{345}{344} \).
3. Beats = \( f_2-f_1 = 400\times345\left(\dfrac1{344}-\dfrac1{346}\right) = 2.32 \) per second.

Teacher's Note:
1. The same source is receding from the man and approaching the wall.
2. The two sound waves meet at the man with slightly different frequencies.

Answer: (c) 5

Explanation:
1. \( v = 360\sqrt{T/400} = 18\sqrt T \) m/s, and \( T = 100+\dfrac{800}{900}x \), so \( dx = \dfrac98dT \).
2. \( t = \int\dfrac{dx}{v} = \dfrac{9}{8\times18}\int_{100}^{900}T^{-1/2}dT = \dfrac1{16}\cdot2(30-10) = 2.5 \) s.
3. So K = 2.5 and 2K = 5.

Teacher's Note:
1. The speed changes along the tunnel, so integrate dx over v.
2. Substitute the temperature as a linear function of position.

Answer: (a) 6

Explanation:
1. Path difference \( = 2\lambda\sin\theta \) for a distant detector.
2. Maximum: \( 2\lambda\sin\theta = m\lambda \Rightarrow \sin\theta = m/2 \). Values m = 1 give \( \theta = \pi/6 \), while m = 2 gives \( \pi/2 \) (infinite distance) and m = 0 gives the centre O.
3. The only finite maximum away from O is at \( \theta = \pi/6 \), so n = 6.

Teacher's Note:
1. Far from the sources the path difference is d times the sine of the angle.
2. Exclude the central maximum and the one at infinity.

Answer: (d) 6

Explanation:
1. \( \lambda = 330/165 = 2 \) m. With a phase difference \( \pi \), minima occur where the path difference is \( n\lambda = 2n \) m.
2. On AB: point (0, y), path difference \( = \sqrt{16+(3-y)^2}-y \) varies from 5 m (at A) to 1 m (at B); the values 2 m and 4 m occur: 2 minima.
3. On BC: point (x, 3), path difference \( = \sqrt{x^2+9}-(4-x) \) varies from -1 m (at B) to 5 m (at C); the values 0, 2 and 4 occur: 3 minima.
4. The ratio is 2/3, so P x Q = 6.

Teacher's Note:
1. Opposite-phase sources give minima at whole-number path differences.
2. Count the integer multiples of the wavelength within the range of path difference on each side.

Answer: (b) 8

Explanation:
1. Maximum strain \( = \left|\dfrac{\partial y}{\partial x}\right|_{max} = 0.002\times\dfrac\pi3\left|\cos\dfrac{2\pi}3\right| = 0.002\times\dfrac\pi3\times0.5 \).
2. Stress \( = Y\times\text{strain} = \dfrac{3}{8\pi}\times0.002\times\dfrac\pi3\times0.5 = 1.25\times10^{-4} = \dfrac18\times10^{-3} \).
3. So n = 8.

Teacher's Note:
1. Tensile stress is Young's modulus times the strain dy/dx.
2. Take the maximum over time, which sets the time factor to 1.

Answer: (a) 4

Explanation:
1. At the moment when all elements pass through equilibrium the energy is purely kinetic: \( E = \int_0^{\pi/k}\tfrac12\rho S\omega^2a^2\sin^2kx\,dx \).
2. \( = \tfrac12\rho S\omega^2a^2\cdot\dfrac{\pi}{2k} = \dfrac{\pi S\rho\omega^2a^2}{4k} \), so p = 4.

Teacher's Note:
1. The total energy is constant, so evaluate it when the kinetic energy is greatest.
2. Adjacent nodes are pi over k apart.

Answer: (c) 12

Explanation:
1. A: second harmonic \( = v_A/L \). B: \( 3\times v_B/4L \). Equal gives \( v_A/v_B = 3/4 \).
2. B closed at both ends has fundamental \( v_B/2L \); A has \( v_A/2L \).
3. Ratio \( = v_A/v_B = 3/4 \), so pq = 12.

Teacher's Note:
1. The lengths are equal, so the frequency ratio equals the speed ratio.
2. A pipe closed at both ends behaves like an open pipe for its frequencies.

Answer: (b) 25

Explanation:
1. \( 0.1+e = \lambda/4 \) and \( 0.35+e = 3\lambda/4 \).
2. Subtracting, \( 0.25 = \lambda/2 \Rightarrow \lambda = 0.5 \) m, so \( e = 0.125-0.1 = 0.025 \) m = 25 mm.

Teacher's Note:
1. Two successive resonances differ by half a wavelength.
2. Use the first resonance to get e.

Answer: (a) 41

Explanation:
1. Frequency is inversely proportional to length: the 1.0 m wire has \( f_1 \) and the 1.05 m wire has \( f_2 = f_1/1.05 \).
2. The fork lies between them: \( f_1 - F = 5 \) and \( F - f_2 = 5 \), so \( f_1-f_2 = 10 \) and \( f_1(1-1/1.05) = 10 \Rightarrow f_1 = 210 \) Hz.
3. The fork frequency is \( 210-5 = 205 \) Hz \( = 5\times41 \), so f = 41.

Teacher's Note:
1. The same beat frequency with both lengths means the fork lies midway between the two wire frequencies.
2. Shorter wires vibrate faster.