JEE Physics Sound Waves MCQs Set 02

Answer: (a) 4

Explanation:
1. Frequencies: \( f_1=50\ \text{Hz},\ f_2=46\ \text{Hz} \).
2. Beat frequency \( =50-46=4 \): maximum intensity 4 times per second.

Teacher's Note:
1. Beats result from two close frequencies; the beat frequency is their difference.
2. Angular frequencies 100 pi and 92 pi correspond to 50 Hz and 46 Hz.

Answer: (d) \( f_C/f_D=\sqrt{11/28} \)

Explanation:
1. \( f=\frac{1}{2\ell}\sqrt{\frac{\gamma RT}{M}}\propto\frac{1}{\ell\sqrt M} \).
2. \( \frac{f_C}{f_D}=\frac{L/3}{2L/3}\sqrt{\frac{44}{28}}=\frac12\sqrt{\frac{44}{28}}=\sqrt{\frac{11}{28}} \).
3. Checking the others: \( \frac{f_A}{f_B}=\frac{L/2}{L}\sqrt{\frac{32}{2}}=2 \); \( \frac{f_B}{f_C}=\frac{2L/3}{L/2}\sqrt{\frac{28}{32}}=\frac43\sqrt{\frac78} \); \( \frac{f_D}{f_A}=3\sqrt{\frac{2}{44}} \); none matches the other options.

Teacher's Note:
1. For an open pipe the fundamental is proportional to 1/(length x root of molar mass) when gamma and T are the same.
2. Compare each option by direct substitution.

Answer: (d) 6 and 6

Explanation:
1. Closed: \( f=(2n+1)\frac{v}{4l}=(2n+1)100\ \text{Hz} \); below 1250: 100, 300, 500, 700, 900, 1100, which is 6 frequencies.
2. Open: \( f=(n+1)\frac{v}{2l}=(n+1)200\ \text{Hz} \); below 1250: 200, 400, ..., 1200, which is 6 frequencies.

Teacher's Note:
1. Closed pipe: odd multiples of v/4l. Open pipe: all multiples of v/2l.
2. Here v/4l = 100 Hz and v/2l = 200 Hz.

Answer: (c) 45 cm

Explanation:
1. Third overtone is the 7th harmonic: \( \lambda=\frac{4\times105}{7}=60\ \text{cm} \).
2. Pressure nodes (displacement antinodes) are at \( \frac\lambda4,\ \frac{3\lambda}{4},\dots=15,45,75,105\ \text{cm} \) from the closed end. 45 cm is listed.

Teacher's Note:
1. Pressure nodes are at displacement antinodes, which lie a quarter wavelength from the closed end and then every half wavelength.
2. The open end (105 cm) is also a pressure node.

Answer: (b) x > 54

Explanation:
1. In winter \( \frac{\lambda_w}{4}\approx18 \) so the second resonance would be at \( \frac{3\lambda_w}{4}=54\ \text{cm} \).
2. In summer the speed (and wavelength) is larger, so \( x=\frac{3\lambda_s}{4}\gt 54 \).

Teacher's Note:
1. Speed of sound increases with temperature, so the wavelength at the same frequency is longer.
2. Second resonance of a closed column is at three quarters of a wavelength.

Answer: (a) 1375 Hz

Explanation:
1. Silence at 12 cm, maximum at 24 cm, silence at 36 cm: \( \frac\lambda2=12\ \text{cm}\Rightarrow\lambda=24\ \text{cm} \).
2. \( f=\frac{330}{0.24}=1375\ \text{Hz} \).

Teacher's Note:
1. Successive minima are separated by one wavelength in path difference, and a minimum and the next maximum by half a wavelength.
2. Convert 24 cm to 0.24 m before dividing.

Answer: (a) 83 Hz

Explanation:
1. Path difference ranges from 2 m (at B) down towards 0 far away. A minimum needs \( \Delta=\frac\lambda2 \) for some position, so need \( \frac\lambda2\le2\ \text{m} \).
2. No destructive interference if \( \lambda\gt 4\ \text{m} \), i.e. \( f\lt \frac{332}{4}=83\ \text{Hz} \).

Teacher's Note:
1. The largest possible path difference is the speaker separation.
2. A destructive point needs the path difference to reach half a wavelength somewhere on the line.

Answer: (d) at a distance L/4 from either end inside the tube

Explanation:
1. Second harmonic of an open pipe: \( \lambda=L \); displacement antinodes at both ends and the middle, so displacement nodes (pressure antinodes) at \( L/4 \) and \( 3L/4 \).

Teacher's Note:
1. Pressure antinodes sit where displacement nodes are.
2. Adjacent node and antinode are lambda/4 apart.

Answer: (c) \( c_2=c_1\,(v_2/v_1) \)

Explanation:
1. Both \( v=\sqrt{\gamma RT/M} \) and \( c=\sqrt{3RT/M} \) are proportional to \( \sqrt T \).
2. Hence \( \frac{c_2}{c_1}=\frac{v_2}{v_1} \).

Teacher's Note:
1. For one gas, the speed of sound and the rms speed differ by a constant factor sqrt(gamma/3).
2. Both scale as the square root of absolute temperature.

Answer: (b) 8.50 kHz

Explanation:
1. Car speed \( =10\ \text{m/s} \). Building receives \( f_1=8\times\frac{320}{320-10} \) and re-emits it.
2. \( f_2=f_1\frac{320+10}{320}=8\times\frac{330}{310}=8.5\ \text{kHz} \).

Teacher's Note:
1. The car is first a moving source and then a moving observer.
2. Convert the speed to m/s: 36 km/h is 10 m/s.

Answer: (b) 9 and 6

Explanation:
1. Closed: first overtone \( =3f_c \). Open: first overtone \( =2f_o \). So \( f_o=1.5f_c \).
2. \( nf_c=mf_o=1.5mf_c\Rightarrow n=1.5m \) with n odd: m = 6, n = 9.

Teacher's Note:
1. Closed pipes have only odd harmonics, which restricts n.
2. Each of the options can be tested using n = 1.5 m.

Answer: (a) \( \lambda/2 \)

Explanation:
1. Destructive: path difference is an odd multiple of \( \lambda/2 \).
2. Adding \( \lambda/2 \) changes it to a whole number of wavelengths: constructive.

Teacher's Note:
1. Half a wavelength converts a minimum into a maximum.
2. Adding a full wavelength would leave the interference unchanged.

Answer: (d) 1 and 2 only

Explanation:
1. In the air frame the source speed is \( u\mp w \) and the observer speed \( u\pm w \), both towards each other.
2. \( f_2=f_1\frac{V+u+w}{V-u+w}\gt f_1 \) for wind from source to observer, and \( f_2=f_1\frac{V+u-w}{V-u-w}\gt f_1 \) for the opposite wind. In both cases \( f_2\gt f_1 \).

Teacher's Note:
1. Vehicles approaching each other always give a higher frequency, whatever the wind direction.
2. The wind changes the numbers but not the direction of the shift.

Answer: (c) 0.125 m

Explanation:
1. \( \lambda=\frac{330}{660}=0.5\ \text{m} \).
2. The wall is a displacement node; the first displacement antinode is \( \frac\lambda4=0.125\ \text{m} \) from it.

Teacher's Note:
1. A rigid wall prevents air particle motion, so it is a displacement node (pressure antinode).
2. Node to antinode is a quarter wavelength.

Answer: (c) 196 Hz

Explanation:
1. Loading fork 2 lowers its frequency. Beats increased from 4 to 6, so fork 2 was below fork 1 and moved further away.
2. \( f_2=200-4=196\ \text{Hz} \) (original); afterwards 194 Hz.

Teacher's Note:
1. Adding load (tape or wax) lowers a fork's frequency.
2. If fork 2 had been at 204 Hz, lowering it would have reduced the beats.

Answer: (b) \( \pi R \)

Explanation:
1. Path difference \( =\frac{3\pi R}{2}-\frac{\pi R}{2}=\pi R \).
2. Maximum: \( \pi R=n\lambda \); the largest wavelength is for n = 1: \( \lambda=\pi R \).

Teacher's Note:
1. Constructive interference needs path difference equal to a whole number of wavelengths.
2. The longest wavelength corresponds to the smallest n.

Answer: (a) 100

Explanation:
1. \( 10\log\frac{I_1}{I_2}=20\Rightarrow\frac{I_1}{I_2}=100 \).

Teacher's Note:
1. 20 dB is two factors of ten.
2. Every 10 dB corresponds to a factor 10 in intensity.

Answer: (d) 300 m/s

Explanation:
1. \( \frac{v}{1.0}-\frac{v}{1.02}=6\Rightarrow v\times0.0196=6 \).
2. \( v\approx306\approx300\ \text{m/s} \).

Teacher's Note:
1. Beat frequency is the difference of the two frequencies, each equal to v over its wavelength.
2. The answer is closest to 300 m/s.

Answer: (d) 17.3 GHz

Explanation:
1. For light the relativistic Doppler formula applies: \( f'=f\sqrt{\frac{1+\beta}{1-\beta}} \) with \( \beta=\frac12 \).
2. \( f'=10\sqrt{\frac{1.5}{0.5}}=10\sqrt3=17.3\ \text{GHz} \).

Teacher's Note:
1. The classical sound formula (1.5 f = 15 GHz) does not apply to light at relativistic speeds.
2. Relativistic Doppler shifts depend only on relative speed.

Answer: (c) 735 N

Explanation:
1. \( f=\frac{3v}{4L}=\frac{3\times330}{4}=247.5\ \text{Hz} \).
2. Wire: \( f=\frac{v_w}{2l}\Rightarrow v_w=2(0.3)(247.5)=148.5\ \text{m/s} \).
3. \( T=\mu v_w^2=\frac{0.01}{0.3}(148.5)^2=735\ \text{N} \).

Teacher's Note:
1. The first overtone of a closed pipe is the third harmonic.
2. Mass per unit length of the wire is 0.01 kg over 0.3 m.

Answer: (b) \( \frac{uvt}{u-v} \)

Explanation:
1. Time difference \( t=\frac dv-\frac du=d\,\frac{u-v}{uv} \).
2. So \( d=\frac{uvt}{u-v} \).

Teacher's Note:
1. The faster wave (in water) arrives first, so subtract its travel time from the air travel time.
2. Check units: speed squared times time over speed gives length.

Answer: (a) All three results occur

Explanation:
1. Path difference \( =6-4=2\ \text{m} \).
2. \( \lambda=1\ \text{m}:\ 2=2\lambda \) (constructive); \( \lambda=\tfrac23\ \text{m}:\ 2=3\lambda \) (constructive); \( \lambda=4\ \text{m}:\ 2=\tfrac\lambda2 \) (destructive).

Teacher's Note:
1. Constructive interference needs a path difference that is a whole number of wavelengths.
2. Destructive interference needs an odd number of half wavelengths.

Answer: (a) \( \frac12n_0\left(\frac{1-f^2}{f}\right) \)

Explanation:
1. \( f_1=\frac{n_0v}{v-u},\ f_2=\frac{n_0v}{v+u},\ f=\frac{v-u}{v+u}\Rightarrow\frac uv=\frac{1-f}{1+f} \).
2. \( f_1-f_2=n_0\frac{2uv}{v^2-u^2}=n_0\frac{2(1-f)(1+f)}{(1+f)^2-(1-f)^2}=\frac{n_0}{2}\frac{1-f^2}{f} \).

Teacher's Note:
1. Write both Doppler frequencies and express u/v through the ratio f.
2. The denominators combine through (1+f)^2 - (1-f)^2 = 4f.

Answer: (d) 6

Explanation:
1. \( f=(2n+1)\frac{v}{4L}=(2n+1)\times100\ \text{Hz} \).
2. Below 1250 Hz: 100, 300, 500, 700, 900, 1100: six oscillations.

Teacher's Note:
1. A closed pipe supports only odd harmonics of v/4L.
2. v/4L = 340/3.4 = 100 Hz.

Answer: (c) 420 Hz

Explanation:
1. Moving the board by d changes the path of the reflected wave by \( 2d=40\ \text{cm} \).
2. Maximum to minimum is \( \frac\lambda2=0.4\ \text{m}\Rightarrow\lambda=0.8\ \text{m} \), so \( f=\frac{336}{0.8}=420\ \text{Hz} \).

Teacher's Note:
1. The reflected path changes by twice the distance moved.
2. A maximum to the next minimum is a half wavelength.

Answer: (b) the energy is redistributed and the distribution remains constant in time

Explanation:
1. Total energy is conserved: it is merely redistributed, with more at the maxima and less at the minima.
2. For coherent sources the pattern is stationary in time.

Teacher's Note:
1. Interference never creates or destroys energy.
2. Constant phase difference (coherence) gives a fixed pattern.

Answer: (b) 3/8

Explanation:
1. First overtone of a closed pipe: \( f=\frac{3v}{4L_1} \). Third overtone of an open pipe: \( f=\frac{4v}{2L_2} \).
2. \( \frac{3}{4L_1}=\frac{2}{L_2}\Rightarrow\frac{L_1}{L_2}=\frac38 \).

Teacher's Note:
1. Closed pipe: only odd harmonics; first overtone is the third harmonic.
2. Open pipe: third overtone is the fourth harmonic.

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

Explanation:
1. Towards the observer (leftwards) the sound moves at \( V-V_w \) relative to the ground and the source recedes from the observer at \( V_s \).
2. \( \lambda=\frac{(V-V_w)+V_s}{f} \).

Teacher's Note:
1. The wavelength is the sound speed relative to the ground, plus or minus the source speed, divided by f.
2. A source moving away from the observer lengthens the wavelength in that direction.