JEE Physics Sound Waves MCQs Set 07

Answer: (d) 80

Explanation:
1. Reflected sound heard by the driver: \( f' = f\dfrac{c+v}{c-v} \).
2. Beats \( = f'-f = f\dfrac{2v}{c-v} = 6 \Rightarrow f = \dfrac{6\times320}{24} = 80 \) Hz.

Teacher's Note:
1. The beat is between the driver's own whistle and the echo.
2. Substituting c - v = 320 makes the arithmetic simple.

Answer: (c) 2

Explanation:
1. Wall receding: reflected frequency \( f' = f\dfrac{v-u}{v+u} \).
2. Beat frequency \( = f-f' = f\dfrac{2u}{v+u} \), so k = 2.

Teacher's Note:
1. A receding reflector lowers the frequency of the echo.
2. The direct sound is at f, the echo at f times (v-u)/(v+u).

Answer: (d) 12

Explanation:
1. Angular speed \( = 2\pi/(\pi/2) = 4 \) rad/s, so the speed of the source \( v = 5\times4 = 20 \) m/s.
2. At the tangent points the velocity lies along the line of sight: \( f_{max} = 320\times\dfrac{300}{280} \), \( f_{min} = 320\times\dfrac{300}{320} \) (receding).
3. Difference \( = 320\times300\left(\dfrac1{280}-\dfrac1{320}\right) = \dfrac{300}{7} = \dfrac{25}{7}\times12 \), so P = 12.

Teacher's Note:
1. The maximum Doppler shift occurs when the source's velocity points straight at or away from the observer.
2. The observer is outside the circle, so tangent lines to the circle exist.

Answer: (c) 59

Explanation:
1. Echo from a receding wall: \( f' = 1700\times\dfrac{340-6}{340+6} = 1641.04 \) Hz.
2. Beats \( = 1700-1641.04 \approx 59 \) Hz.

Teacher's Note:
1. The direct sound has frequency 1700 Hz at the stationary receiver.
2. The wall receding lowers the frequency of the echo.

Answer: (b) Upper end open, lower end closed

Explanation:
1. Wavelength \( \lambda = 80 \) cm. At the top (effective position \( y+1 = 0 \)) the displacement amplitude is a maximum (cosine = 1): an antinode, so the end is open.
2. At the bottom (\( y = 99 \)) \( y+1 = 100 \) cm and \( \cos(2\pi\times100/80) = \cos(5\pi/2) = 0 \): a node, so the end is closed.

Teacher's Note:
1. An open end is a displacement antinode and a closed end is a displacement node.
2. Check the amplitude factor at both ends.

Answer: (a) Second overtone

Explanation:
1. The effective length is \( 99+1 = 100 \) cm and \( \lambda = 80 \) cm, so \( L = 5\lambda/4 \).
2. A closed pipe has \( L = (2n+1)\lambda/4 \); here \( 2n+1 = 5 \), n = 2: the fifth harmonic, i.e. the second overtone.

Teacher's Note:
1. A closed pipe holds an odd number of quarter wavelengths.
2. The fifth harmonic is the second overtone.

Answer: (d) \( P_{ex} = (125\pi\ \text{N/m}^2)\sin\left[\dfrac{2\pi}{80}(y+1\ \text{cm})\right]\cos2\pi(400)t \)

Explanation:
1. \( P_{ex} = -B\,\dfrac{\partial\xi}{\partial y} \), with \( k = \dfrac{2\pi}{0.8} \) rad/m.
2. Amplitude \( = Bks_0 = 5\times10^5\times\dfrac{2\pi}{0.8}\times10^{-4} = 125\pi \) N/m\(^2\).
3. Differentiating the cosine gives a sine in position while the time factor stays cos 2π(400)t.

Teacher's Note:
1. Pressure variation is the space derivative of the displacement.
2. The position factor changes from cosine to sine; the time factor and frequency are unchanged.

Answer: (c) \( \dfrac{10\pi}{36\times22.4} \)

Explanation:
1. End correction \( 0.3d = 1 \) cm, so \( d = 10/3 \) cm and the radius is \( 5/3 \) cm.
2. Volume \( \approx \pi(5/3)^2\times100 \) cm\(^3\) \( = \dfrac{2500\pi}{9} \) cm\(^3\) \( = \dfrac{2.5\pi}{9} \) litre \( = \dfrac{10\pi}{36} \) litre.
3. Moles \( = \dfrac{10\pi}{36\times22.4} \).

Teacher's Note:
1. Find the diameter from the given end correction before the volume.
2. Take the effective length as about 100 cm for an approximate answer.

Answer: (b) 50 m/s

Explanation:
1. Reflection from a fixed island: \( f' = f\dfrac{c+u}{c-u} = 1.1f \).
2. \( c+u = 1.1(c-u) \Rightarrow 2.1u = 0.1\times1050 \Rightarrow u = 50 \) m/s.

Teacher's Note:
1. The island acts first as a stationary observer and then as a stationary source.
2. The submarine is both the moving source and the moving detector.

Answer: (a) 50 m/s towards the Indian submarine

Explanation:
1. The submarine moves at 50 m/s towards the ship; for a ship moving at w towards the submarine: \( f'' = f\dfrac{c+u}{c-u}\cdot\dfrac{c+w}{c-w} \).
2. \( 1.21 = 1.1\times\dfrac{c+w}{c-w} \Rightarrow \dfrac{c+w}{c-w} = 1.1 \Rightarrow w = 50 \) m/s.
3. The ratio is greater than 1, so the ship moves towards the submarine.

Teacher's Note:
1. The factor from the submarine's own speed is already 1.1.
2. A factor above 1 for the reflector means it approaches the submarine.

Answer: (d) 1.1

Explanation:
1. The ship receives the wavelength emitted by the moving submarine: \( \lambda' = (c-u)/f \).
2. The ship re-emits as a source moving at w towards the submarine, with frequency \( f_1 = f\dfrac{c+w}{c-u} \), so \( \lambda'' = \dfrac{c-w}{f_1} = \dfrac{(c-w)(c-u)}{f(c+w)} \).
3. \( \lambda'/\lambda'' = \dfrac{c+w}{c-w} = 1.1 \).

Teacher's Note:
1. A moving observer does not change the wavelength; a moving source does.
2. The ratio is the same Doppler factor as found earlier.

Answer: (c) \( 10^{9} \) N/m\(^2\)

Explanation:
1. \( v = \sqrt{B/\rho} \Rightarrow B = \rho v^2 = 1000\times1050^2 = 1.1\times10^9 \) N/m\(^2\).

Teacher's Note:
1. Bulk modulus equals density times the square of the sound speed.
2. The result is of order 10 to the 9 for liquids such as water.

Answer: (b) (A), (B), (C) and (D)

Explanation:
1. The expression is just a travelling sinusoid, and any periodic physical quantity in a wave can vary this way.
2. An electromagnetic wave has sinusoidal electric and magnetic fields; a sound wave has sinusoidal displacement and pressure change.
3. So all four can be represented.

Teacher's Note:
1. The form of the equation does not depend on the type of wave.
2. y is a generic wave variable.

Answer: (d) (A), (B) and (D)

Explanation:
1. Path difference is 2 m.
2. 1 m: 2 = 2 lambda, constructive (A true). 2/3 m: 2 = 3 lambda, constructive (B true). 2 m: 2 = 1 lambda, constructive, not destructive (C false). 4 m: 2 = lambda/2, destructive (D true).

Teacher's Note:
1. Compare the path difference with whole and half wavelengths.
2. A path difference equal to one wavelength is constructive.

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

Explanation:
1. The path difference of 5 m is \( \lambda/4 \), which corresponds to a phase of \( \pi/2 \).
2. At B (nearer to P) the Q wave travels 5 m extra, so it lags by a further pi/2: phase difference = pi/2 + pi/2 = pi, giving zero intensity.
3. At A (nearer to Q) the P wave lags by pi/2, cancelling its lead: phase difference 0, intensity 4I. At C the phase difference is pi/2, giving 2I.
4. The ratio A : B : C is 4I : 0 : 2I = 2 : 0 : 1.

Teacher's Note:
1. Add the initial phase difference to the phase from the path difference, with the correct sign at each point.
2. Resultant intensity for phase difference phi is 4 I cos squared (phi over 2).

Answer: (a) (A), (C) and (D)

Explanation:
1. A plane wave spreads over a constant area so its intensity is constant (A true).
2. A spherical wave's power is spread over \( 4\pi r^2 \), so the intensity falls as \( 1/r^2 \) (C true, B false).
3. The total power through any sphere around the source is constant in the absence of absorption (D true).

Teacher's Note:
1. Intensity is power per unit area.
2. Amplitude of a spherical wave falls as 1 over r, but intensity as 1 over r squared.

Answer: (b) (B) and (D)

Explanation:
1. The hole forces an antinode at \( L/3 \) from the closed end, so the pipe now behaves like a closed pipe of length \( L/3 \): the fundamental becomes \( v/(4L/3) = 3v/4L \), three times the old \( v/4L \) (B true, C false).
2. The open pipe of length L has v/2L, so the new fundamental is 1.5 times that, not 2 times (A false).
3. The allowed modes are the 3rd, 9th, 15th, ... harmonics of the original, whose ratios 1 : 3 : 5 are still odd (D true).

Teacher's Note:
1. A hole acts as an open end, giving an antinode at the hole.
2. Compare with the pipe of length L/3 closed at one end.

Answer: (d) (A), (B) and (C)

Explanation:
1. \( \lambda = 340/340 = 1 \) m. Resonant air column lengths: 25 cm, 75 cm (125 cm exceeds 120 cm).
2. Water columns are 120 - 75 = 45 cm (minimum) and 120 - 25 = 95 cm (maximum): A and C true.
3. Successive nodes are lambda/2 = 50 cm apart: B true, D false.

Teacher's Note:
1. Water column length is the tube length minus the air column length.
2. Resonance in a closed tube occurs at odd multiples of lambda over 4.

Answer: (c) (B) and (C)

Explanation:
1. Let \( a = v/2\ell_1 \) and \( b = v/4\ell_2 \). Fundamentals: \( |a-b| = 5 \). First overtones: \( 2a \) and \( 3b \): \( |2a-3b| = 5 \).
2. Positive solutions: \( a = b+5 \) gives \( b = 5, a = 10 \) or \( b = 15, a = 20 \), so \( a = 2b \) or \( a = \tfrac43b \).
3. \( \ell_1/\ell_2 = 2b/a \cdot 1 = 1 \) or \( 3/2 \).

Teacher's Note:
1. Write both beat conditions with an absolute value and test the cases.
2. Reject negative frequencies.

Answer: (b) (A) and (D)

Explanation:
1. Frequency and the time period are set by the source and are not altered by the wind (A and D).
2. The speed of sound relative to the ground becomes V + w, and the wavelength becomes (V + w)/f, so B and C change.

Teacher's Note:
1. Wind adds a drift to the medium, changing the wave speed over the ground.
2. Source frequency is unaffected.

Answer: (a) (A), (B) and (C)

Explanation:
1. The moving surface receives \( \nu(c+u)/c \) waves per second (A true).
2. It re-emits as a source moving towards the observer: \( \nu_r = \nu\dfrac{c+u}{c-u} \) (C true), so the wavelength is \( c/\nu_r = \dfrac{c(c-u)}{\nu(c+u)} \) (B true).
3. Beats \( = \nu_r-\nu = \dfrac{2u\nu}{c-u} \), not \( u\nu/(c-u) \), so D is false.

Teacher's Note:
1. The surface acts as a moving observer then a moving source.
2. The beat frequency here is the difference between the reflected and the original frequency.

Answer: (d) (A), (C) and (D)

Explanation:
1. O\(_1\) in air: the source approaches at V/5, so \( \lambda_1 = (V-V/5)/f = 4V/5f \) (A true, B false) and the frequency is \( f_1 = 5f/4 \).
2. The frequency does not change on entering the water. O\(_2\) moves towards the source at V/5 in water (speed 4V): \( f_2 = \dfrac54f\cdot\dfrac{4V+V/5}{4V} = \dfrac{21f}{16} \) (C true).
3. The wavelength in water is \( 4V/(5f/4) = 16V/5f \) (D true).

Teacher's Note:
1. The boundary preserves frequency, not wavelength.
2. A moving observer changes the frequency but not the wavelength.

Answer: (b) (A) and (D)

Explanation:
1. Power \( \propto \Delta p_0^2/(\rho v) \): same pressure amplitude, same medium, so \( P_1 = P_2 \) (A).
2. \( \Delta p_0 = Bks_0 \), so \( s_0\propto1/k\propto\lambda \): \( s_1/s_2 = 2 \) (D).

Teacher's Note:
1. Pressure amplitude fixes the intensity in a given medium.
2. Displacement amplitude is proportional to wavelength at fixed pressure amplitude.

Answer: (a) (B) and (D)

Explanation:
1. Let the unknown be f. Condition: \( |f-262| = 2|f-256| \).
2. For \( f \lt 256 \): \( 262-f = 2(256-f) \Rightarrow f = 250 \) Hz. For \( 256 \lt f \lt 262 \): \( 262-f = 2(f-256) \Rightarrow f = 258 \) Hz. For \( f \gt 262 \): no solution.
3. So 250 Hz and 258 Hz.

Teacher's Note:
1. Treat the three ranges of f separately because of the absolute values.
2. Check each candidate: 250 gives beats 6 and 12; 258 gives 2 and 4.

Answer: (d) (A) and (D)

Explanation:
1. The sound is weaker at the second resonance because of greater damping in the longer column (A true).
2. The prongs are held vertically so that their plane of vibration lets sound enter the tube properly, and their amplitude is only about a millimetre or less (B and C false).
3. The end correction means the measured column is shorter than a quarter wavelength (D true).

Teacher's Note:
1. The end correction is added to the measured length to reach lambda/4.
2. Typical prong amplitudes are tiny.

Answer: (c) (B) and (D)

Explanation:
1. At an open end the pressure must stay near atmospheric, so the reflected pressure wave is inverted: a high-pressure pulse comes back as a low-pressure pulse (B true, A false).
2. At a closed end the pressure wave is reflected without inversion: a high-pressure pulse returns as a high-pressure pulse (D true, C false).

Teacher's Note:
1. Think of the pressure antinode at a closed end and the pressure node at an open end.
2. Phase change of pi for pressure at an open end.

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

Explanation:
1. Frequencies: \( 50 \) Hz and \( 46 \) Hz, so the beat frequency (maximum intensity) is 4 per second (A true).
2. Speed \( = \omega/k = 100\pi/0.5\pi = 200 \) m/s (and \( 92/0.46 = 200 \)), so B and D are false.
3. At x = 0: \( y_1+y_2 = 2A\cos(96\pi t)\cos(4\pi t) \); the first factor vanishes 96 times and the second 4 times in a second: 100 times (C true).

Teacher's Note:
1. Wave speed is omega over k.
2. Count the zeros of both factors of the product form.