JEE Physics Sound Waves MCQs Set 08

Answer: (d) 344 Hz

Explanation:
1. Closed pipe third harmonic: \( f = 3\times\dfrac{340}{4\times0.75} = 340 \) Hz, which is the string frequency.
2. The fork is at 336 Hz or 344 Hz. Raising the tension raises the string frequency above 340 Hz: with 336 Hz the beats would increase; with 344 Hz they would decrease.
3. So n = 344 Hz.

Teacher's Note:
1. Tension up means frequency up.
2. Beats reduce only if the string frequency moves towards the fork.

Answer: (c) 7

Explanation:
1. Reflected frequency \( \approx f_0\left(1+\dfrac{2v}{c}\right) \) for \( v\ll c \).
2. Difference \( \dfrac{2\Delta v}{c} = 0.012 \Rightarrow \Delta v = 0.006\times330 = 1.98 \) m/s \( = 7.13 \) km/h \( \approx 7 \) km/h.

Teacher's Note:
1. For small speeds the Doppler shift is linear in v.
2. Reflection from a moving car shifts the frequency twice.

Answer: (b) 10 grams

Explanation:
1. Pipe fundamental: \( f = 320/(4\times0.8) = 100 \) Hz.
2. String second harmonic: \( f = \dfrac{v_s}{L} = 100 \Rightarrow v_s = 50 \) m/s. \( v_s = \sqrt{T/\mu} \Rightarrow \mu = 50/2500 = 0.02 \) kg/m.
3. Mass \( = \mu L = 0.02\times0.5 = 0.01 \) kg = 10 g.

Teacher's Note:
1. The second harmonic of a string fixed at both ends has a frequency v over L.
2. Linear density times length gives the mass.

Answer: (a) 8.50 kHz

Explanation:
1. The car's speed is 10 m/s. Frequency at the building: \( 8000\times\dfrac{320}{310} \).
2. The driver moving towards the building hears \( 8000\times\dfrac{320}{310}\times\dfrac{330}{320} = 8000\times\dfrac{330}{310} = 8516 \) Hz, i.e. about 8.50 kHz.

Teacher's Note:
1. Use the two-step method for echoes from a fixed reflector.
2. Both the moving source and the moving observer effects raise the frequency.

Answer: (d) 15.2

Explanation:
1. \( \lambda/4 = \dfrac{336}{4\times512} = 0.164 \) m = 16.4 cm, and this equals the air column length plus the end correction.
2. End correction \( = 0.3d = 1.2 \) cm, so the water level reading is \( 16.4-1.2 = 15.2 \) cm.

Teacher's Note:
1. The scale reading is the length of the air column, which is shorter than lambda/4 by the end correction.
2. End correction for a tube of diameter d is 0.3 d.

Answer: (c) (A) and (B)

Explanation:
1. With the wind opposing the sound: \( f_2 = f_1\dfrac{V-w+u}{V-w-u} \), which exceeds \( f_1 \).
2. With the wind along the sound: \( f_2 = f_1\dfrac{V+w+u}{V+w-u} \), which also exceeds \( f_1 \).
3. Wind only changes the speed of sound relative to the ground, so the approach of the vehicles raises the frequency in both cases.

Teacher's Note:
1. Wind adds to or subtracts from V, but the vehicles still approach each other.
2. In both cases the ratio of numerator to denominator is greater than 1.

Answer: (b) Argon (M = 36, \( \sqrt{10/M} = 17/32 \))

Explanation:
1. Minimum resonance length is \( \lambda/4 = 0.350 \) m, so \( \lambda = 1.40 \) m and \( v = 244\times1.40 = 341.6 \) m/s.
2. Monatomic gases (\( \gamma = 5/3 \)): \( v = 640\sqrt{10/M} \); diatomic gases (\( \gamma = 7/5 \)): \( v = 590\sqrt{10/M} \).
3. Neon: \( 640\times0.7 = 448 \); nitrogen: \( 590\times0.6 = 354 \); oxygen: \( 590\times0.5625 = 331.9 \); argon: \( 640\times17/32 = 340 \) m/s, which matches 341.6 m/s within the uncertainty.

Teacher's Note:
1. Compute the speed of sound in each candidate gas using the right gamma.
2. Argon is monatomic, so use the 640 value.

Answer: (a) 3

Explanation:
1. Add the phasors: real part \( 1+\cos60^\circ+\cos120^\circ+\cos180^\circ = 1+0.5-0.5-1 = 0 \).
2. Imaginary part \( 0+\sin60^\circ+\sin120^\circ+0 = \sqrt3 \), so the resultant amplitude is \( \sqrt3A_0 \).
3. Intensity \( = 3I_0 \), so n = 3.

Teacher's Note:
1. Treat each wave as a phasor with equal length.
2. Intensity is proportional to the amplitude squared.

Answer: (d) 6

Explanation:
1. Frequency received by the car: \( f_1 = 492\times\dfrac{330+2}{330} \).
2. Reflected frequency: \( f_2 = f_1\times\dfrac{330}{330-2} = 492\times\dfrac{332}{328} = 498 \) Hz.
3. Beat frequency = 498 - 492 = 6 Hz.

Teacher's Note:
1. The car first acts as a moving observer and then as a moving source.
2. The beats are between the original and reflected signals.

Answer: (c) 5

Explanation:
1. The stationary man is 5 m horizontally from each, so each is 13 m away and the cosine of the angle with the line is 5/13.
2. The front man moves away: component 1 x 5/13 along the line of sight; the rear man moves towards: 2 x 5/13.
3. \( f_{rear} = 1430\times\dfrac{330}{330-10/13} \), \( f_{front} = 1430\times\dfrac{330}{330+5/13} \); the difference is \( \approx5.0 \) Hz.

Teacher's Note:
1. Use only the velocity component along the line from each source to the listener.
2. The listener is at the same distance from both.

Answer: (b) (A) and (C)

Explanation:
1. Successive resonances differ by \( \lambda/2 = 33.2 \) cm, so \( \lambda = 66.4 \) cm (C true) and \( v = 500\times0.664 = 332 \) m/s (A true).
2. The quarter wavelength is 16.6 cm, so a column of 50.7 cm cannot be the fundamental (D false); it is the third harmonic (3\(\lambda\)/4 = 49.8 cm), and the end correction would be \( 49.8-50.7 = -0.9 \) cm: negative, so B is not a valid end correction.

Teacher's Note:
1. The difference of two successive resonance lengths gives half a wavelength.
2. A fundamental needs a column shorter than lambda over 4 plus nothing, here only 16.6 cm.

Answer: (d) x > 54

Explanation:
1. In winter, \( \lambda/4 = 18 \) cm (neglecting end correction), so the second resonance is at \( 3\lambda/4 = 54 \) cm.
2. In summer the speed of sound, and hence the wavelength, is larger, so the second resonance occurs at a length greater than 54 cm.

Teacher's Note:
1. Warm air carries sound faster so lambda is larger at the same frequency.
2. The second resonance is at three times the first resonance length.

Answer: (c) 98 m

Explanation:
1. Receding observer: \( f' = f\dfrac{c-v}{c} = 0.94f \Rightarrow v = 0.06\times330 = 19.8 \) m/s.
2. \( v^2 = 2as \Rightarrow s = \dfrac{19.8^2}{2\times2} \approx 98 \) m.

Teacher's Note:
1. The motorcycle moves away from the siren so the frequency drops.
2. Use v squared equals 2as for the distance.

Answer: (b) 12%

Explanation:
1. Approaching: \( 1000\times\dfrac{320}{300} = 1066.7 \) Hz. Receding: \( 1000\times\dfrac{320}{340} = 941.2 \) Hz.
2. The change is 125.5 Hz, about 12% of the true frequency.

Teacher's Note:
1. Compute the frequency before and after the train passes.
2. Compare the change with the 1000 Hz natural frequency.

Answer: (a) 17.3 GHz

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

Teacher's Note:
1. At speeds comparable to c, the classical sound formulas do not apply.
2. Use the relativistic form for electromagnetic waves.

Answer: (d) 5 kHz

Explanation:
1. The middle is a node and both ends are free (antinodes): \( L = \lambda/2 \), so \( \lambda = 1.2 \) m.
2. \( v = \sqrt{Y/\rho} = \sqrt{9.27\times10^{10}/2.7\times10^3} = 5860 \) m/s.
3. \( f = v/\lambda = 5860/1.2 \approx 4.9 \) kHz, i.e. about 5 kHz.

Teacher's Note:
1. A rod clamped in the middle with free ends is like an open pipe.
2. Use the sound speed in a solid, root of Y over rho.

Answer: (c) 484 Hz

Explanation:
1. Speed of the whistle \( v = r\omega = 1.5\times20 = 30 \) m/s.
2. Maximum frequency when approaching: \( 440\times\dfrac{330}{330-30} = 484 \) Hz (the minimum is \( 440\times\dfrac{330}{360} = 403.3 \) Hz).

Teacher's Note:
1. The highest frequency is heard when the whistle moves directly towards the observer.
2. Rotation speed is radius times angular velocity.

Answer: (b) about 7.25

Explanation:
1. A hears B's source receding: \( f = 256\times\dfrac{343}{353} = 248.75 \) Hz.
2. Beat frequency \( = 256-248.75 = 7.25 \) Hz.

Teacher's Note:
1. A hears its own source at 256 Hz and B's source Doppler shifted down.
2. Check the arithmetic: 256 x 10/353.

Answer: (a) 4 mm

Explanation:
1. \( y_1+y_2 = 8\sin2\pi t\cos\pi x \) (a stationary wave).
2. Amplitude at \( x = 7/3 \): \( 8|\cos(7\pi/3)| = 8\times0.5 = 4 \) mm.

Teacher's Note:
1. Add the sines with the sum-to-product identity.
2. The resulting amplitude depends on position.

Answer: (d) 3.6%

Explanation:
1. \( f\propto1/\sqrt M \): \( \dfrac{f_{mix}}{f_{O_2}} = \sqrt{\dfrac{32}{M}} = 1+\dfrac1{440} \Rightarrow M = \dfrac{32}{(1+1/440)^2} = 31.855 \).
2. If the mole fraction of nitrogen is x, \( M = 32-4x \), so \( x = 0.0362 \), i.e. about 3.6%. For an ideal gas the partial pressure fraction equals the mole fraction.

Teacher's Note:
1. Beats of 1 Hz at 440 Hz means a fractional change of 1 in 440.
2. Partial pressure is proportional to the mole fraction.

Answer: (c) 735 N

Explanation:
1. First overtone of the closed tube: \( f = 3\times\dfrac{330}{4\times1} = 247.5 \) Hz.
2. The wire fundamental \( f = \dfrac{1}{2L}\sqrt{T/\mu} \) with \( \mu = 0.01/0.3 = 0.0333 \) kg/m: \( \sqrt{T/\mu} = 2\times0.3\times247.5 = 148.5 \).
3. \( T = 0.0333\times148.5^2 \approx 735 \) N.

Teacher's Note:
1. Plucking at the mid-point excites the fundamental of the wire.
2. Linear density is mass over length.

Answer: (b) \( \dfrac{2v_b(v+v_m)f}{v^2-v_b^2} \)

Explanation:
1. Direct sound: the band is ahead of the motorist and moves away from the sound's travel direction, while the motorist moves towards the band: \( f_1 = f\dfrac{v+v_m}{v+v_b} \).
2. Echo: the wall receives \( f\dfrac{v}{v-v_b} \) and reflects it unchanged; the motorist moves towards the wall, against the echo: \( f_2 = f\dfrac{v}{v-v_b}\cdot\dfrac{v+v_m}{v} = f\dfrac{v+v_m}{v-v_b} \).
3. Beats \( = f_2-f_1 = f(v+v_m)\left[\dfrac1{v-v_b}-\dfrac1{v+v_b}\right] = \dfrac{2v_b(v+v_m)f}{v^2-v_b^2} \).

Teacher's Note:
1. Treat the direct and the reflected sound separately, each with its own Doppler factor.
2. Both waves reach the motorist travelling away from the wall.

Answer: (d) \( \dfrac{1}{1089} \) kg/m\(^3\)

Explanation:
1. \( \Delta p = v^2\Delta\rho \) for a sound wave, so \( \Delta\rho_0 = \dfrac{100}{330^2} = \dfrac{1}{1089} \) kg/m\(^3\).

Teacher's Note:
1. Pressure and density variations are related through the square of the sound speed.
2. The amplitude does not depend on the harmonic number.

Answer: (c) 15/16 m

Explanation:
1. Second overtone of a closed pipe is the fifth harmonic: \( f = 5v/4L \).
2. \( L = \dfrac{5\times330}{4\times440} = \dfrac{15}{16} \) m.

Teacher's Note:
1. Closed pipe: first overtone is the 3rd harmonic, second overtone is the 5th.
2. Rearrange f = 5v/4L for L.

Answer: (b) \( \Delta P_0/\sqrt2 \)

Explanation:
1. Pressure amplitude is maximum at the closed end and varies as \( \Delta P_0|\cos kx| \), with x measured from the closed end.
2. \( \lambda = v/f = 0.75 \) m and \( x = L/2 = 15/32 \) m, so \( kx = \dfrac{2\pi}{0.75}\times\dfrac{15}{32} = \dfrac{5\pi}{4} \).
3. \( |\cos(5\pi/4)| = 1/\sqrt2 \), so the amplitude is \( \Delta P_0/\sqrt2 \).

Teacher's Note:
1. The closed end is a pressure antinode.
2. Compute the phase kx at the middle of the pipe.

Answer: (a) about 1.35 kHz

Explanation:
1. The signal received at \( T = 10 \) s was emitted earlier at \( t_e \): \( t_e+\dfrac{\tfrac12at_e^2}{v} = 10 \), i.e. \( 0.0147t_e^2+t_e-10 = 0 \Rightarrow t_e = 8.85 \) s.
2. The source speed at the emission time is \( at_e = 88.5 \) m/s, so \( f = \dfrac{1700}{1+88.5/340} = 1349 \) Hz.

Teacher's Note:
1. Use the retarded time: the signal heard now left the source earlier.
2. Doppler formula for a receding source with the speed at the emission instant.

Answer: (d) 46 kHz

Explanation:
1. Speed of the enemy submarine \( = 100 \) m/s.
2. Reflection from an approaching target: \( f' = f\dfrac{c+v}{c-v} = 40\times\dfrac{1550}{1350} = 45.9 \) kHz, about 46 kHz.

Teacher's Note:
1. The target first receives the sound as a moving observer and then re-emits as a moving source.
2. Both steps raise the frequency.