JEE Physics Sound Waves MCQs Set 05

Answer: (b) \( \dfrac{4}{\pi}\times10^{4} \) kg/m\(^3\) (about \( 1.27\times10^{4} \))

Explanation:
1. \( f_1 = \dfrac{1}{2L}\sqrt{T/\mu} = \sqrt{100/\mu} = 100 \) (L = 0.5 m), so \( \mu = 0.01 \) kg/m.
2. \( \mu = \rho\,\pi d^2/4 \Rightarrow \rho = \dfrac{0.01}{\pi(10^{-3})^2/4} = \dfrac{4}{\pi}\times10^{4} \) kg/m\(^3\).

Teacher's Note:
1. Linear density is mass per unit length = density times cross-sectional area.
2. Wire diameter must be converted to metres.

Answer: (d) \( \dfrac{2vf}{c-v} \)

Explanation:
1. Frequency of the wave striking the wall: \( f_1 = \dfrac{fc}{c-v} \), which is reflected unchanged in frequency.
2. Observer moving towards the wall hears \( f'' = f_1\dfrac{c+v}{c} = f\dfrac{c+v}{c-v} \).
3. Beat frequency \( = f''-f = \dfrac{2vf}{c-v} \).

Teacher's Note:
1. Treat the wall as a stationary observer first, then as a stationary source.
2. The beats arise between the direct and the reflected sound.

Answer: (c) It decreases by about 0.2%

Explanation:
1. Frequency reaching the wall \( f_1 = f(v+u)/v \); the wall then acts as a source approaching at u: \( f_2 = f\dfrac{v+u}{v-u} \).
2. \( \lambda_2/\lambda_0 = \dfrac{v-u}{v+u} \), so the fractional decrease is \( \dfrac{2u}{v+u} = \dfrac{0.66}{330.33}\approx0.2\% \).

Teacher's Note:
1. A wall moving towards the source raises the frequency of the echo.
2. A higher frequency at the same speed means a shorter wavelength.

Answer: (b) R\(_1\), about 1.0 Hz

Explanation:
1. R\(_2\) hears the direct wave at \( f\dfrac{v}{v-u} \), and the echo (reflected from a stationary wall, reaching R\(_2\) on the way back) also has \( f\dfrac{v}{v-u} \): no beats.
2. R\(_1\) hears the direct wave at \( f\dfrac{v}{v+u} \) and the echo at \( f\dfrac{v}{v-u} \).
3. Beats at R\(_1\): \( f\,v\left[\dfrac1{v-u}-\dfrac1{v+u}\right] = \dfrac{2fvu}{v^2-u^2}\approx\dfrac{2fu}{v} = 1.0 \) Hz.

Teacher's Note:
1. Compare which receivers receive two different frequencies.
2. For a stationary receiver the direct wave from a receding source is lower than the echo from the approaching direction.

Answer: (c) 160 Hz

Explanation:
1. Beats between Q and its echo: \( f_Q\left(\dfrac{c+u}{c-u}-1\right) = f_Q\dfrac{2u}{c-u} = 5 \), so \( f_Q = 5\times327/10 = 163.5 \) Hz.
2. P differs from Q by 7/2 = 3.5 Hz, so P is 160 Hz or 167 Hz.
3. Loading wax lowers the frequency of P and the beats rise from 3.5 to 5, so P must lie below Q: it is 160 Hz (it becomes 158.5 Hz).

Teacher's Note:
1. Waxing a fork lowers its frequency, which tells you which side of Q it started on.
2. Use the Doppler beat first to find the frequency of Q.

Answer: (b) Frequency

Explanation:
1. The frequency is fixed by the source and does not change at a boundary.
2. Wave velocity changes with the medium, and \( \lambda = v/f \) and \( k = 2\pi/\lambda \) change with it.

Teacher's Note:
1. The source decides the frequency; the medium decides the speed.
2. Wavelength and wave number follow from speed and frequency.

Answer: (a) 0.44 m/s

Explanation:
1. \( \omega = 2\pi f = 2\pi v/\lambda = 2\times\dfrac{22}{7}\times\dfrac{0.21}{0.015} = 88 \) rad/s.
2. \( v_{max} = A\omega = 0.005\times88 = 0.44 \) m/s.

Teacher's Note:
1. The cork moves in SHM with the frequency of the wave.
2. Maximum speed is amplitude times angular frequency.

Answer: (d) 0.1 s

Explanation:
1. Speed of the bullet relative to the ground = 340 - 20 = 320 m/s, so \( t_b = 544/320 = 1.7 \) s.
2. Sound travels at 340 m/s relative to air (independent of the car): \( t_s = 544/340 = 1.6 \) s.
3. The sound arrives first, and the difference is 0.1 s.

Teacher's Note:
1. The speed of sound does not depend on the motion of the source, but the speed of the bullet does.
2. Subtract the car's speed because the gun fires backwards.

Answer: (c) 0.75

Explanation:
1. \( v = \sqrt{\gamma RT/M} \), so \( \dfrac{v_1}{v_2} = \sqrt{\dfrac{\gamma_1 M_2}{\gamma_2 M_1}} \).
2. \( = \sqrt{\dfrac{(5/3)(18)}{(4/3)(40)}} = \sqrt{0.5625} = 0.75 \).

Teacher's Note:
1. Monatomic gases have gamma = 5/3 and non-linear polyatomic gases 4/3.
2. The speed depends on both gamma and molar mass.

Answer: (b) H\(_2\)

Explanation:
1. \( v = \sqrt{\gamma RT/M} \). The value of \( \gamma/M \) per mole: H\(_2\) \( 1.4/2 = 0.70 \), He \( 1.67/4 = 0.42 \), N\(_2\) \( 1.4/28 = 0.05 \), CO\(_2\) \( 1.3/44 = 0.03 \).
2. Hydrogen has the largest value, hence the highest speed of sound.

Teacher's Note:
1. Light molecules carry sound faster.
2. Helium is lighter-than-air but hydrogen is lighter still and has a larger gamma than the heavier gases.

Answer: (a) \( v_{rms} > v_{av} > v_{mp} > v_s \)

Explanation:
1. In units of \( \sqrt{RT/M} \): \( v_{rms} = \sqrt3 = 1.73 \), \( v_{av} = \sqrt{8/\pi} = 1.60 \), \( v_{mp} = \sqrt2 = 1.41 \).
2. \( v_s = \sqrt{\gamma} \) in the same units, and \( \gamma \le 5/3 \) gives \( v_s \le 1.29 \), which is below \( v_{mp} \).

Teacher's Note:
1. Compare all four speeds in the same unit.
2. Gamma never exceeds 5/3 for ordinary gases.

Answer: (b) 10 W

Explanation:
1. \( P = I\times4\pi r^2 = 0.008\times4\pi\times100 \).
2. This is about 10.05 W, i.e. approximately 10 W.

Teacher's Note:
1. A point source spreads its power over a sphere of area 4 pi r squared.
2. Multiply intensity by the area of the sphere.

Answer: (a) 1375 Hz

Explanation:
1. Silence at 12 cm, maximum at 24 cm, silence at 36 cm: consecutive silences are 24 cm apart, hence half-wavelength steps of 12 cm.
2. \( \lambda/2 = 12 \) cm, so \( \lambda = 24 \) cm.
3. \( f = v/\lambda = 330/0.24 = 1375 \) Hz.

Teacher's Note:
1. A move from minimum to maximum is half a wavelength in path difference.
2. Use v = f lambda with the wavelength in metres.

Answer: (d) \( \dfrac{nv}{r(\pi-2)} \)

Explanation:
1. Path difference between the semicircle and the straight path \( = \pi r-2r = r(\pi-2) \).
2. Maximum intensity needs path difference \( n\lambda \): \( r(\pi-2) = nv/f \).
3. \( f = \dfrac{nv}{r(\pi-2)} \).

Teacher's Note:
1. Both paths start and end together, so only the path difference matters.
2. Constructive interference needs a whole number of wavelengths.

Answer: (b) 69 dB

Explanation:
1. Intensities add for incoherent sources: \( I' = 8I \).
2. \( \Delta B = 10\log_{10}8 = 9 \) dB, so the level is 69 dB.

Teacher's Note:
1. Independent sources add intensities, not amplitudes.
2. Eight times the intensity adds about 9 dB.

Answer: (a) 0

Explanation:
1. At a rigid wall the air cannot move, so the displacement has a node there and the displacement wave is inverted (phase change pi).
2. The pressure has an antinode at the wall, so the pressure wave is reflected without a phase change.

Teacher's Note:
1. Displacement and pressure waves are 90 degrees out of step in space.
2. A node of displacement is an antinode of pressure.

Answer: (d) \( 1 : \tfrac12 : \tfrac13 \)

Explanation:
1. Open pipe wavelengths: \( \lambda_n = 2L/n \) with n = 1, 2, 3 for the fundamental and the first two overtones.
2. Hence \( \lambda_1:\lambda_2:\lambda_3 = 1:\tfrac12:\tfrac13 \).

Teacher's Note:
1. An open pipe supports all harmonics.
2. Wavelength is inversely proportional to the harmonic number.

Answer: (c) the middle of the pipe

Explanation:
1. In the fundamental mode of an open pipe there are displacement antinodes at both ends and a displacement node in the middle.
2. Pressure variation is maximum at a displacement node, i.e. at the middle.

Teacher's Note:
1. Pressure antinodes are displacement nodes.
2. The open ends always stay near atmospheric pressure.

Answer: (b) 12.5 cm

Explanation:
1. First overtone of the open pipe: \( v/L_o = v/0.5 \).
2. Fundamental of the closed pipe: \( v/4L_c \). Equating, \( L_c = L_o/4 = 12.5 \) cm.

Teacher's Note:
1. The first overtone of an open pipe equals its second harmonic.
2. The closed pipe fundamental is v over four times its length.

Answer: (a) \( f_0 \)

Explanation:
1. Tube A is open at both ends: \( f_0 = v/2\ell \).
2. Tube B is closed at the water surface, with an air column of length \( \ell/2 \): \( f_B = \dfrac{v}{4(\ell/2)} = \dfrac{v}{2\ell} = f_0 \).

Teacher's Note:
1. Dipping a tube in water turns it into a closed pipe of shorter length.
2. Halving the length of a closed pipe compares with the open pipe at full length.

Answer: (d) \( \dfrac{\ell+0.6d}{2(\ell+0.3d)} \)

Explanation:
1. Open at both ends: \( \nu_1 = \dfrac{v}{2(\ell+0.6d)} \).
2. One end closed: \( \nu_2 = \dfrac{v}{4(\ell+0.3d)} \).
3. \( \dfrac{\nu_2}{\nu_1} = \dfrac{\ell+0.6d}{2(\ell+0.3d)} \).

Teacher's Note:
1. An end correction of 0.3 d applies at each open end.
2. Take the ratio so that the speed cancels.

Answer: (c) 5 : 3

Explanation:
1. Second overtone of the open pipe = third harmonic = \( 3f_A \).
2. Second overtone of the closed pipe = fifth harmonic = \( 5f_B \).
3. \( 3f_A = 5f_B \Rightarrow f_A : f_B = 5 : 3 \).

Teacher's Note:
1. Count overtones carefully: open pipes go 2nd, 3rd harmonic; closed pipes go 3rd, 5th.
2. Equate the actual frequencies, not the harmonic numbers.

Answer: (a) 506 Hz

Explanation:
1. The string is at 506 Hz or 518 Hz.
2. Stretching raises the tension and so raises the string frequency. For the beats to decrease, the string must move closer to 512 Hz, so it started below 512 Hz.
3. The original frequency was 506 Hz.

Teacher's Note:
1. Increasing tension raises frequency.
2. Check which side of the source frequency makes the beat rate fall.

Answer: (c) 2

Explanation:
1. Fundamentals: closed \( v/4L \), open \( v/2L \). Beat frequency \( = v/4L = 4 \) Hz.
2. With lengths doubled the beat frequency is \( v/8L = 2 \) Hz.

Teacher's Note:
1. Beat frequency is proportional to 1 over the length.
2. Doubling the length halves every frequency.

Answer: (b) Distance between source and observer

Explanation:
1. The apparent frequency depends on the source frequency and on the velocities of the source and the observer relative to the medium.
2. The distance between them does not enter the Doppler formula.

Teacher's Note:
1. Only relative motion matters, not separation.
2. Distance affects loudness, not pitch.

Answer: (a) 1.6 kHz

Explanation:
1. Frequency received by the wall: \( 1200\times\dfrac{350}{300} = 1400 \) Hz.
2. The driver moves towards the wall: \( 1400\times\dfrac{350+50}{350} = 1600 \) Hz = 1.6 kHz.

Teacher's Note:
1. Treat the reflection in two steps: wall as observer, then wall as source.
2. Both steps raise the frequency.

Answer: (c) \( p = \sum p_{0n}\sin(k_nx-\omega_nt) \)

Explanation:
1. A clap is a sudden pulse, not a pure single-frequency tone.
2. Such a pulse is a superposition of many harmonic components of different frequencies and wave numbers.

Teacher's Note:
1. A pure sine wave has one frequency; a clap contains many.
2. A pulse can be built from a sum (Fourier series) of sinusoids.