JEE Physics Waves MCQs Set 22

Answer: (a) \( \pi \)

Explanation:
1. At \( t = 0 \), \( y=A\sin(kx+\phi) \). We need \( y(0) = 0 \), so \( \phi=0 \) or \( \pi \).
2. For \( \phi=\pi \): \( y=-A\sin kx \), which is negative just to the right of the origin, as in the snapshot. So \( \phi=\pi \).

Teacher's Note:
1. Identify whether the situation is a travelling wave, a standing wave or a Doppler problem, and write the matching standard formula first.
2. Substitute carefully in SI units and test the result with a limiting case (for example, zero source speed gives no frequency change).

Answer: (b) 502 Hz

Explanation:
1. \( v_A=5 m/s \), \( v_B=7.5 m/s \). \( A \) receives \( f_1=500\dfrac{1500-5}{1500-7.5} \).
2. A re-emits \( f_1 \); \( B \) receives \( f_2=f_1\dfrac{1500+7.5}{1500+5}=500\times\dfrac{1495}{1492.5}\times\dfrac{1507.5}{1505}\approx501.7 \) Hz, i.e. about 502 Hz.

Teacher's Note:
1. In Doppler problems write \( f' = f (v +/- v_{observer})/(v -/+ v_{source}) \) with the upper signs for approach.
2. For an echo treat the wall first as an observer and then as a source of the received frequency.

Answer: (c) 20.0 cm

Explanation:
1. \( \mu=5\times10^{-3} \) kg/m, \( v=\sqrt{8/(5\times10^{-3})}=40 m/s \).
2. \( \lambda=v/f=0.4 m \), and the node separation is \( \lambda/2=20 \) cm.

Teacher's Note:
1. For stretched strings use \( v = \sqrt{T/\mu} \); both ends fixed give \( f = n v/(2L) \).
2. For pipes remember: open pipe all harmonics \( f = n v/(2L) \); closed pipe only odd harmonics \( f = n v/(4L) \).

Answer: (d) \( \dfrac{g}{5} \)

Explanation:
1. \( v=\sqrt{T/\mu} \) with \( T=M\sqrt{g^2+a^2} \), so \( v^2\propto\sqrt{g^2+a^2} \).
2. \( \left(\dfrac{60.5}{60}\right)^2=1.0167=\dfrac{\sqrt{g^2+a^2}}{g}\Rightarrow a\approx0.18g\approx\dfrac g5 \).

Teacher's Note:
1. Identify whether the situation is a travelling wave, a standing wave or a Doppler problem, and write the matching standard formula first.
2. Substitute carefully in SI units and test the result with a limiting case (for example, zero source speed gives no frequency change).

Answer: (a) \( 2\sqrt2\ \text{s} \)

Explanation:
1. At height \( x \) above the lower end the tension is \( \mu xg \), so \( v=\sqrt{gx} \).
2. \( t=\int_0^L\dfrac{dx}{\sqrt{gx}}=2\sqrt{L/g}=2\sqrt2 s \).

Teacher's Note:
1. Identify whether the situation is a travelling wave, a standing wave or a Doppler problem, and write the matching standard formula first.
2. Substitute carefully in SI units and test the result with a limiting case (for example, zero source speed gives no frequency change).

Answer: (b) \( \dfrac{\sqrt3\pi}{50}\hat j\ \text{m/s} \)

Explanation:
1. \( \omega=\dfrac{2\pi v}{\lambda}=0.4\pi \) rad/s and \( |v_y|=\omega\sqrt{A^2-y^2}=0.4\pi\times0.1\times\dfrac{\sqrt3}2=\dfrac{\sqrt3\pi}{50} m/s \).
2. For a wave moving in \( +x \), \( v_y=-v\,\dfrac{\partial y}{\partial x} \). The slope is negative at \( P \), so \( v_y \) is positive: along \( +j \).

Teacher's Note:
1. Identify whether the situation is a travelling wave, a standing wave or a Doppler problem, and write the matching standard formula first.
2. Substitute carefully in SI units and test the result with a limiting case (for example, zero source speed gives no frequency change).

Answer: (c) \( A\omega \)

Explanation:
1. \( v_y=\partial y/\partial t=-A\omega\cos(kx-\omega t) \).
2. Its maximum magnitude is \( A\omega \).

Teacher's Note:
1. Identify whether the situation is a travelling wave, a standing wave or a Doppler problem, and write the matching standard formula first.
2. Substitute carefully in SI units and test the result with a limiting case (for example, zero source speed gives no frequency change).

Answer: (d) \( \lambda=\dfrac{\pi y_0}{2} \)

Explanation:
1. Maximum particle speed \( =2\pi f y_0 \); wave speed \( =f\lambda \).
2. \( 2\pi fy_0=4f\lambda\Rightarrow\lambda=\dfrac{\pi y_0}2 \).

Teacher's Note:
1. Identify whether the situation is a travelling wave, a standing wave or a Doppler problem, and write the matching standard formula first.
2. Substitute carefully in SI units and test the result with a limiting case (for example, zero source speed gives no frequency change).

Answer: (a) (a), (c) and (d) only

Explanation:
1. Tension is largest at \( O \) and smallest at \( A \), so the speed \( \sqrt{T/\mu} \) depends only on position; the same path integral \( \int dx/v \) gives \( T_{AO}=T_{OA} \): \( (a) \) true.
2. Frequency is fixed and speed falls towards \( A \), so \( \lambda=v/f \) becomes shorter: \( (b) \) false. \( (c) \) and \( (d) \) are true because speed depends only on position.

Teacher's Note:
1. Identify whether the situation is a travelling wave, a standing wave or a Doppler problem, and write the matching standard formula first.
2. Substitute carefully in SI units and test the result with a limiting case (for example, zero source speed gives no frequency change).

Answer: (b) (b), (c) and (d) only

Explanation:
1. A function of \( (4x+5t) \) moves in the negative x-direction with speed \( 5/4=1.25 m/s \): \( (a) \) false, \( (b) \) true (\( 2.5 m \) in \( 2 s \)).
2. Maximum at \( 4x+5t=0 \): \( Y=0.8/5=0.16 m \): \( (c) \) true. \( Y \) depends on \( (4x+5t)^2 \), so it is symmetric: \( (d) \) true.

Teacher's Note:
1. Identify whether the situation is a travelling wave, a standing wave or a Doppler problem, and write the matching standard formula first.
2. Substitute carefully in SI units and test the result with a limiting case (for example, zero source speed gives no frequency change).

Answer: (c) \( \lambda=2\pi\times10^{-2}\ \text{m},\ f=\dfrac{10^3}{2\pi}\ \text{Hz} \)

Explanation:
1. \( a\omega=\dfrac v{10}=1 m/s \Rightarrow\omega=10^3 \) rad/s, so \( f=\dfrac{10^3}{2\pi} \) Hz.
2. \( \lambda=\dfrac vf=\dfrac{10\times2\pi}{10^3}=2\pi\times10^{-2} m \).

Teacher's Note:
1. Identify whether the situation is a travelling wave, a standing wave or a Doppler problem, and write the matching standard formula first.
2. Substitute carefully in SI units and test the result with a limiting case (for example, zero source speed gives no frequency change).

Answer: (d) a wave travelling in the negative x-direction with velocity 1.5 m/s and wavelength 0.2 m

Explanation:
1. The \( +\) sign between \( kx \) and \( \omega t \) means propagation in the negative x-direction.
2. \( k=10\pi\Rightarrow\lambda=0.2 m \); \( \omega=15\pi\Rightarrow v=\omega/k=1.5 m/s \).

Teacher's Note:
1. Identify whether the situation is a travelling wave, a standing wave or a Doppler problem, and write the matching standard formula first.
2. Substitute carefully in SI units and test the result with a limiting case (for example, zero source speed gives no frequency change).

Answer: (a) travelling with velocity 30 m/s in the negative x-direction, of wavelength \( \pi \) m, frequency \( 30/\pi \) Hz and amplitude \( 10^{-4} \) m (all four statements are true)

Explanation:
1. \( \omega=60 \), \( k=2 \): \( v=30 m/s \), direction negative \( x \) because of the + sign.
2. \( \lambda=2\pi/k=\pi m \), \( f=\omega/2\pi=30/\pi \) Hz, amplitude \( 10^{-4} m \).

Teacher's Note:
1. Identify whether the situation is a travelling wave, a standing wave or a Doppler problem, and write the matching standard formula first.
2. Substitute carefully in SI units and test the result with a limiting case (for example, zero source speed gives no frequency change).

Answer: (b) 0.5 m/s

Explanation:
1. The pulse shape \( \dfrac1{1+x^2} \) has its peak at the origin at \( t = 0 \) and at \( x = 1 m \) at \( t = 2 s \).
2. The shift is \( 1 m \) in \( 2 s \), so the speed is \( 0.5 m/s \).

Teacher's Note:
1. Identify whether the situation is a travelling wave, a standing wave or a Doppler problem, and write the matching standard formula first.
2. Substitute carefully in SI units and test the result with a limiting case (for example, zero source speed gives no frequency change).

Answer: (c) \( 2\pi\nu A,\ 4\pi^2\nu^2A \)

Explanation:
1. \( y=A\sin(kx-\omega t) \): velocity amplitude \( A\omega=2\pi\nu A \).
2. Acceleration amplitude \( A\omega^2=4\pi^2\nu^2A \).

Teacher's Note:
1. Identify whether the situation is a travelling wave, a standing wave or a Doppler problem, and write the matching standard formula first.
2. Substitute carefully in SI units and test the result with a limiting case (for example, zero source speed gives no frequency change).

Answer: (d) \( y=(0.1\ \text{m})\sin[(30\ \text{rad/s})t-(1.5\ \text{m}^{-1})x+\phi] \)

Explanation:
1. \( A\omega=3 \) and \( A\omega^2=90\Rightarrow\omega=30 \) rad/s, \( A=0.1 m \).
2. \( k=\omega/v=30/20=1.5 m ^{-1} \).

Teacher's Note:
1. Identify whether the situation is a travelling wave, a standing wave or a Doppler problem, and write the matching standard formula first.
2. Substitute carefully in SI units and test the result with a limiting case (for example, zero source speed gives no frequency change).

Answer: (a) 384 m/s

Explanation:
1. Successive resonances differ by \( \lambda/2 \): \( \lambda/2=40 \) cm, so \( \lambda=0.8 m \).
2. \( v=f\lambda=480\times0.8=384 m/s \).

Teacher's Note:
1. For stretched strings use \( v = \sqrt{T/\mu} \); both ends fixed give \( f = n v/(2L) \).
2. For pipes remember: open pipe all harmonics \( f = n v/(2L) \); closed pipe only odd harmonics \( f = n v/(4L) \).

Answer: (b) 320 m/s, 80 Hz

Explanation:
1. Third harmonic: \( L=\dfrac{3\lambda}2\Rightarrow\lambda=\dfrac{4}{3} m \), so \( v=240\times\dfrac43=320 m/s \).
2. Fundamental \( =240/3=80 \) Hz.

Teacher's Note:
1. For stretched strings use \( v = \sqrt{T/\mu} \); both ends fixed give \( f = n v/(2L) \).
2. For pipes remember: open pipe all harmonics \( f = n v/(2L) \); closed pipe only odd harmonics \( f = n v/(4L) \).

Answer: (c) 80 m

Explanation:
1. \( \lambda=\dfrac{2\pi}{0.157}=40 m \).
2. 4th harmonic: \( L=4\cdot\dfrac\lambda2=80 m \).

Teacher's Note:
1. For stretched strings use \( v = \sqrt{T/\mu} \); both ends fixed give \( f = n v/(2L) \).
2. For pipes remember: open pipe all harmonics \( f = n v/(2L) \); closed pipe only odd harmonics \( f = n v/(4L) \).

Answer: (d) 1 : 2

Explanation:
1. \( \mu\propto r^2 \): \( \mu_B=4\mu_A \), so \( v_A=2v_B \) at the same tension.
2. With the same frequency \( \lambda_A=2\lambda_B \), and the same length \( L \) holds \( p\lambda_A/2=L=q\lambda_B/2 \): \( p:q=\lambda_B:\lambda_A=1:2 \).

Teacher's Note:
1. For stretched strings use \( v = \sqrt{T/\mu} \); both ends fixed give \( f = n v/(2L) \).
2. For pipes remember: open pipe all harmonics \( f = n v/(2L) \); closed pipe only odd harmonics \( f = n v/(4L) \).

Answer: (a) 328 m/s

Explanation:
1. \( l_1+e=\dfrac{v}{4(512)} \) and \( l_2+e=\dfrac v{4(256)}=2(l_1+e) \). So \( 27+e=2(11+e)\Rightarrow e=5 \) cm.
2. \( \dfrac{v}{2048}=0.16 m \), so \( v\approx328 m/s \).

Teacher's Note:
1. For stretched strings use \( v = \sqrt{T/\mu} \); both ends fixed give \( f = n v/(2L) \).
2. For pipes remember: open pipe all harmonics \( f = n v/(2L) \); closed pipe only odd harmonics \( f = n v/(4L) \).

Answer: (b) 12.5 N

Explanation:
1. \( v=\omega/k=450/9=50 m/s \).
2. \( T=\mu v^2=5\times10^{-3}\times2500=12.5 N \).

Teacher's Note:
1. Identify whether the situation is a travelling wave, a standing wave or a Doppler problem, and write the matching standard formula first.
2. Substitute carefully in SI units and test the result with a limiting case (for example, zero source speed gives no frequency change).

Answer: (c) 6

Explanation:
1. Frequencies are odd multiples: 1.5, 4.5, 7.5, 10.5, 13.5, 16.5, 19.5 kHz (the next, 22.5 kHz, is inaudible).
2. Excluding the fundamental there are 6 overtones.

Teacher's Note:
1. For stretched strings use \( v = \sqrt{T/\mu} \); both ends fixed give \( f = n v/(2L) \).
2. For pipes remember: open pipe all harmonics \( f = n v/(2L) \); closed pipe only odd harmonics \( f = n v/(4L) \).

Answer: (d) 5 kHz

Explanation:
1. \( v=\sqrt{Y/\rho}=\sqrt{\dfrac{9.27\times10^{10}}{2.7\times10^3}}\approx5.86\times10^3 m/s \).
2. The middle is a node and the ends are antinodes: \( \lambda=2L=1.2 m \), so \( f=v/\lambda\approx4.9\approx5 \) kHz.

Teacher's Note:
1. For stretched strings use \( v = \sqrt{T/\mu} \); both ends fixed give \( f = n v/(2L) \).
2. For pipes remember: open pipe all harmonics \( f = n v/(2L) \); closed pipe only odd harmonics \( f = n v/(4L) \).

Answer: (a) f

Explanation:
1. Open pipe: \( f=\dfrac v{2L} \). Half in water gives a closed pipe of length \( L/2 \).
2. \( f'=\dfrac{v}{4(L/2)}=\dfrac v{2L}=f \).

Teacher's Note:
1. For stretched strings use \( v = \sqrt{T/\mu} \); both ends fixed give \( f = n v/(2L) \).
2. For pipes remember: open pipe all harmonics \( f = n v/(2L) \); closed pipe only odd harmonics \( f = n v/(4L) \).

Answer: (b) 6

Explanation:
1. Fundamental \( =\dfrac{340}{4\times0.85}=100 \) Hz; allowed frequencies are odd multiples: 100, 300, 500, 700, 900, 1100 Hz.
2. The next one (1300 Hz) exceeds 1250 Hz, so there are 6.

Teacher's Note:
1. For stretched strings use \( v = \sqrt{T/\mu} \); both ends fixed give \( f = n v/(2L) \).
2. For pipes remember: open pipe all harmonics \( f = n v/(2L) \); closed pipe only odd harmonics \( f = n v/(4L) \).

Answer: (c) 178.2 Hz

Explanation:
1. \( v=\sqrt{T/\mu}=\sqrt{\dfrac{Y\varepsilon}{\rho}}=\sqrt{\dfrac{2.2\times10^{11}\times0.01}{7.7\times10^3}}\approx534.5 m/s \).
2. \( f=\dfrac{v}{2L}=\dfrac{534.5}{3}\approx178.2 \) Hz.

Teacher's Note:
1. For stretched strings use \( v = \sqrt{T/\mu} \); both ends fixed give \( f = n v/(2L) \).
2. For pipes remember: open pipe all harmonics \( f = n v/(2L) \); closed pipe only odd harmonics \( f = n v/(4L) \).

Answer: (d) 15.2 cm

Explanation:
1. \( \lambda=\dfrac{336}{512}=0.656 m \), so \( \dfrac\lambda4=16.4 \) cm.
2. End correction \( e=0.3d=1.2 \) cm, so \( l=16.4-1.2=15.2 \) cm.

Teacher's Note:
1. For stretched strings use \( v = \sqrt{T/\mu} \); both ends fixed give \( f = n v/(2L) \).
2. For pipes remember: open pipe all harmonics \( f = n v/(2L) \); closed pipe only odd harmonics \( f = n v/(4L) \).

Answer: (a) 10 g

Explanation:
1. Pipe fundamental \( =\dfrac{320}{4(0.8)}=100 \) Hz.
2. String second harmonic: \( f=\dfrac{v_s}{L}\Rightarrow v_s=50 m/s \), \( \mu=\dfrac T{v_s^2}=0.02 \) kg/m, so \( m=\mu L=10 g \).

Teacher's Note:
1. For stretched strings use \( v = \sqrt{T/\mu} \); both ends fixed give \( f = n v/(2L) \).
2. For pipes remember: open pipe all harmonics \( f = n v/(2L) \); closed pipe only odd harmonics \( f = n v/(4L) \).

Answer: (b) \( \dfrac{l}{5} \)

Explanation:
1. Frequency \( \propto\sqrt T \) for identical strings, so \( T_{AB}=4T_{CD} \).
2. Moments about \( D \) give \( T_{AB}=mg\dfrac{l-x}l \) and \( T_{CD}=mg\dfrac xl \). Then \( \dfrac{l-x}{x}=4\Rightarrow x=\dfrac l5 \).

Teacher's Note:
1. For stretched strings use \( v = \sqrt{T/\mu} \); both ends fixed give \( f = n v/(2L) \).
2. For pipes remember: open pipe all harmonics \( f = n v/(2L) \); closed pipe only odd harmonics \( f = n v/(4L) \).