JEE Physics Waves MCQs Set 23

Answer: (c) \( n=5,\ f_2=\dfrac54f_1 \)

Explanation:
1. Open pipe, 2nd harmonic: \( f_1=\dfrac vL \). Closed pipe harmonics: \( \dfrac{nv}{4L}=\dfrac n4f_1 \) with \( n \) odd.
2. Since \( f_2>f_1 \), the smallest odd \( n \) is 5, giving \( f_2=\dfrac54f_1 \).

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) \( \dfrac{4L}3\sqrt{\dfrac{\rho_1}{\rho_2}} \)

Explanation:
1. \( v=\sqrt{B/\rho}\propto\rho^{-1/2} \), so \( v_2/v_1=\sqrt{\rho_1/\rho_2} \).
2. Closed first overtone \( \dfrac{3v_1}{4L} \) equals open first overtone \( \dfrac{v_2}{L'} \): \( L'=\dfrac{4L}{3}\dfrac{v_2}{v_1} \).

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) 25 kg

Explanation:
1. \( f=\dfrac p{2l}\sqrt{\dfrac T\mu} \), so \( p\sqrt{M}=\) constant.
2. \( 5\sqrt9=3\sqrt M\Rightarrow\sqrt M=5\Rightarrow M=25 \) kg.

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) 1

Explanation:
1. \( \nu=\dfrac1{2l}\sqrt{\dfrac T\mu} \) and \( \mu\propto r^2 \), so \( \nu\propto\dfrac1{lr} \).
2. \( \dfrac{\nu_1}{\nu_2}=\dfrac{(2L)(r)}{(L)(2r)}=1 \).

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) 200 Hz

Explanation:
1. Open pipe fundamental \( f_0=\dfrac v{2L} \); third harmonic of the closed pipe \( =\dfrac{3v}{4L} \).
2. \( \dfrac{3v}{4L}-\dfrac v{2L}=\dfrac v{4L}=100\Rightarrow\dfrac v{2L}=200 \) 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) \( 300\left(\dfrac{2\rho-1}{2\rho}\right)^{1/2} \)

Explanation:
1. Tension \( T=V\rho g \); with half submerged \( T'=V\rho g-\dfrac12Vg=Vg\dfrac{2\rho-1}2 \).
2. \( f\propto\sqrt T\Rightarrow f'=300\sqrt{\dfrac{2\rho-1}{2\rho}} \).

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) \( \dfrac{1}{6} \)

Explanation:
1. \( \dfrac{v}{4L_1}=\dfrac{3v}{2L_2} \).
2. \( \dfrac{L_1}{L_2}=\dfrac{2}{12}=\dfrac16 \).

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) \( -a\cos(kx+\omega t) \)

Explanation:
1. A stationary wave needs a wave of equal amplitude travelling in the opposite direction, \( \pm a\cos(kx+\omega t) \).
2. At \( x = 0 \) the sum must vanish: \( a\cos\omega t+c\cos\omega t=0\Rightarrow c=-a \).

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) 1024

Explanation:
1. Closed: \( \dfrac{v}{4L}=512 \). Open: \( \dfrac v{2L}=2\times512 \).
2. Fundamental of the open tube is 1024 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) A: p, t; B: p, s; C: q, s; D: q, r

Explanation:
1. A: closed pipe, longitudinal, \( L=\lambda/4\Rightarrow\lambda_f=4L (p, t) \). \( B \): open pipe, longitudinal, \( \lambda_f=2L (p, s) \).
2. \( C \): wire fixed at both ends, transverse, \( L=\lambda/2\Rightarrow\lambda_f=2L (q, s) \). \( D \): extra node at the middle gives \( L=\lambda (q, r) \).

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) (a) and (c) only

Explanation:
1. \( \lambda/2=83.9-50.7=33.2 \) cm \( \Rightarrow\lambda=66.4 \) cm, \( v=500\times0.664=332 m/s \): \( (a), (c) \) true.
2. \( \lambda/4=16.6 \) cm, so 50.7 cm is nearly \( 3\lambda/4=49.8 \) cm (third harmonic), not the fundamental; the end correction is not \( +0.9 \) cm. \( (b), (d) \) false.

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

Explanation:
1. All four satisfy \( v=\omega/k=100 m/s \).
2. At \( x = 3 m \) we need an antinode, \( |\sin3k|=1 \): \( (a) \sin\dfrac\pi2 \), \( (c) \sin\dfrac{5\pi}2 \), \( (d) \sin\dfrac{15\pi}2 \) are all \( \pm1 \), but \( (b) \) gives \( \sin\pi=0 \), a node. So \( (a), (c), (d) \).

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) (b) and (c) only

Explanation:
1. \( \lambda=\dfrac{2\pi}{62.8}=0.1 m \); fifth harmonic \( L=\dfrac{5\lambda}2=0.25 m \): \( (b) \) true.
2. Mid-point \( x=0.125 \): \( \sin(7.85)=1 \), so the displacement is \( 0.01 m \): \( (c) \) true. The frequency is \( 628/2\pi=100 \) Hz, which is the fifth harmonic, so the fundamental is 20 Hz and \( (d) \) is false. There are 6 nodes including the ends: \( (a) \) false.

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) (a), (b) and (c) only

Explanation:
1. Reflection at fixed or free ends gives waves travelling in opposite directions, which form standing waves in \( (a), (b), (c) \).
2. In \( (d) \) the two waves travel the same way and cancel; no stationary pattern forms.

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) (b) and (c) only

Explanation:
1. Clamped edges require \( u=0 \) at \( x=0,L \) and \( y=0,L \).
2. \( (b) \) and \( (c) \) vanish on all four edges. \( (a) \) and \( (d) \) contain \( \cos \) factors that are non-zero at \( x = 0 \), so they are not allowed.

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) (a), (b), (c) and (d)

Explanation:
1. Amplitude factor \( \cos10\pi x \): nodes where \( 10\pi x=\dfrac\pi2,\dfrac{3\pi}2 \), i.e. \( x=0.05,0.15 m \): \( (a) \) true. Antinodes at \( x=0.1,0.2,0.3 m \): \( (b) \) true.
2. \( \lambda=\dfrac{2\pi}{10\pi}=0.2 m \) and \( v=\dfrac\omega k=\dfrac{50\pi}{10\pi}=5 m/s \): \( (c), (d) \) true.

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 Hz, 240 Hz and 400 Hz

Explanation:
1. Fundamental \( =\dfrac{320}{4(1)}=80 \) Hz; allowed frequencies are odd multiples: 80, 240, 400, 560 Hz...
2. So 80 Hz, 240 Hz and 400 Hz can resonate, but 320 Hz (an even multiple) cannot.

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) 31.25 cm and 93.75 cm

Explanation:
1. \( \lambda=\dfrac{330}{264}=1.25 m \), \( \dfrac\lambda4=31.25 \) cm.
2. Resonant lengths are odd multiples of \( \lambda/4 \): 31.25, 93.75, 156.25 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) f

Explanation:
1. Open: \( f=v/2L \). Half in water: closed pipe of length \( L/2 \): \( v/(4\cdot L/2)=v/2L=f \).
2. The frequency is unchanged.

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) 240 Hz

Explanation:
1. Tension falls from \( 50.7g \) to \( 50.7g-0.0075\times1000\,g=43.2g \).
2. \( f'=260\sqrt{\dfrac{43.2}{50.7}}\approx240 \) 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) 0.125 m

Explanation:
1. \( \lambda=\dfrac{330}{660}=0.5 m \). The wall is a displacement node.
2. The first displacement antinode is at \( \lambda/4=0.125 m \).

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) 5

Explanation:
1. The phase difference is \( \pi/2 \), so the phasors are perpendicular.
2. \( A=\sqrt{4^2+3^2}=5 \).

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) 5

Explanation:
1. \( \mu=\dfrac{10^{-3}}{0.2}=5\times10^{-3} \) kg/m, \( v=\sqrt{0.5/(5\times10^{-3})}=10 m/s \).
2. \( \lambda=0.1 m \), node separation \( \lambda/2=5 \) 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: (b) \( \dfrac{\pi^2a^2T}{4l} \)

Explanation:
1. \( y=a\sin\dfrac{\pi x}l\cos\omega t \), \( \omega=\dfrac\pi l\sqrt{T/\mu} \). Total energy \( =\tfrac12\mu\omega^2a^2\cdot\dfrac l2 \).
2. \( E=\dfrac{\mu a^2l}4\dfrac{\pi^2T}{l^2\mu}=\dfrac{\pi^2a^2T}{4l} \).

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) \( \dfrac{400}{189} \)

Explanation:
1. \( \dfrac{v_A}{L}=\dfrac{3v_B}{4L}\Rightarrow\dfrac{v_A}{v_B}=\dfrac34 \).
2. \( v=\sqrt{\gamma RT/M} \): \( \dfrac{\gamma_AM_B}{\gamma_BM_A}=\dfrac9{16} \) with \( \gamma_A=\dfrac53,\gamma_B=\dfrac75 \), so \( \dfrac{M_A}{M_B}=\dfrac{25}{21}\cdot\dfrac{16}9=\dfrac{400}{189} \).

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) \( \dfrac{3}{4} \)

Explanation:
1. A (open both ends): \( f_A=\dfrac{v_A}{2L} \). \( B \) (closed both ends): \( f_B=\dfrac{v_B}{2L} \).
2. \( \dfrac{f_A}{f_B}=\dfrac{v_A}{v_B}=\dfrac34 \).

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) 3.2 m, 2.4 m, 1.6 m, 0.8 m

Explanation:
1. \( \lambda=\dfrac{340}{212.5}=1.6 m \). Air-column lengths for resonance: \( 0.4, 1.2, 2.0, 2.8, 3.6 m \) (odd multiples of \( 0.4 m \)).
2. Water height \( =3.6-\) air column: \( 3.2, 2.4, 1.6, 0.8 m \) (the last, \( 0 m \), is the full pipe).

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) 43 s

Explanation:
1. \( A\dfrac{dH}{dt}=-a\sqrt{2gH}\Rightarrow\dfrac{dH}{dt}=-\dfrac{a}{A}\sqrt{2g}\sqrt H=-(1.11\times10^{-2})\sqrt H \) with \( \dfrac aA=\dfrac1{400} \).
2. \( t=\dfrac{2(\sqrt{3.2}-\sqrt{2.4})}{1.118\times10^{-2}}\approx43 s \).

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: (c) \( \dfrac{15}{16} \) m

Explanation:
1. Second overtone of a closed pipe is the 5th harmonic: \( L=\dfrac{5\lambda}4 \).
2. \( \lambda=\dfrac{330}{440}=0.75 m \), so \( L=\dfrac{15}{16} m \).

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) (a), (b) and (c) only

Explanation:
1. The open end is a pressure node (no variation, pressure \( p_0 \)); the closed end is a pressure antinode (\( p_0\pm\Delta p_0 \)): \( (b), (c) \) true, \( (d) \) false.
2. At the middle, distance \( \dfrac L2=\dfrac{15}{32} m \) from the open end, the amplitude is \( \Delta p_0|\sin(kL/2)|=\Delta p_0\sin\left(\dfrac{5\pi}{4}\right) \) in magnitude \( \Delta p_0/\sqrt2 \): \( (a) \) true.

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 \( f = n v/(4L) \).