JEE Physics Sound Waves MCQs Set 03

Answer: (d) both the elastic property and the inertia property of the medium

Explanation:
1. \( v=\sqrt{\frac{\text{elastic modulus}}{\text{density}}} \).
2. The modulus is the elastic property and the density is the inertia property.

Teacher's Note:
1. Stiffer media carry sound faster and denser media carry it slower.
2. Newton-Laplace type expressions always contain both factors.

Answer: (c) 1 cm

Explanation:
1. Displacing the sliding tube by d changes the path by \( 2d=10\ \text{cm} \).
2. Ten maxima and a final minimum correspond to a path change of ten wavelengths: \( 10\lambda=10\ \text{cm}\Rightarrow\lambda=1\ \text{cm} \).

Teacher's Note:
1. Moving the slider by d changes the path difference by 2d because the sound goes in and out.
2. From minimum to the next minimum the path difference changes by one wavelength.

Answer: (c) 1.0 m

Explanation:
1. Consecutive frequencies differ by 170 Hz; since 425 = 2.5 x 170, the pipe is closed with \( 2f_1=170\Rightarrow f_1=85\ \text{Hz} \) (425 = 5 x 85).
2. \( L=\frac{v}{4f_1}=\frac{340}{340}=1.0\ \text{m} \).

Teacher's Note:
1. For a closed pipe only odd harmonics occur, and successive frequencies differ by twice the fundamental.
2. If all harmonics were present the differences would equal the fundamental, and 425 would have to be an integer multiple of 170.

Answer: (b) all of these: the displacement wave reflects with a phase change of π at the closed end, the pressure wave reflects without phase change, and the wave reflected at the closed end is again reflected at the open end

Explanation:
1. At the closed end displacement is zero, so the displacement wave reflects with a phase change of π.
2. Pressure is maximum there: the pressure wave reflects without phase change.
3. The wave then travels back and is reflected again at the open end, giving a standing wave.

Teacher's Note:
1. A closed end is like a fixed end for displacement; an open end behaves like a free end for displacement.
2. Pressure phase changes are opposite to displacement phase changes.

Answer: (a) 200 m/s

Explanation:
1. \( v_2=\frac{\omega}{k}=\frac{92\pi}{0.46\pi}=200\ \text{m/s} \) (the first wave has speed 100 m/s).

Teacher's Note:
1. Wave speed is omega over k.
2. Compare cos(kx - omega t) with the given equation.

Answer: (d) the length of B to that of A is 5 : 6, and the frequencies of the first overtones of A and B are in the ratio 10 : 9

Explanation:
1. Open pipe second overtone: \( \frac{3v}{2L_A} \). Closed pipe second overtone: \( \frac{5v}{4L_B} \). Equal gives \( \frac{L_B}{L_A}=\frac56 \).
2. Fundamentals: \( \frac{f_A}{f_B}=\frac{2L_B}{L_A}=\frac53 \). First overtones: \( \frac{2f_A}{3f_B}=\frac{10}{9} \).

Teacher's Note:
1. Second overtone is the 3rd harmonic for an open pipe and the 5th harmonic for a closed pipe.
2. Both of the statements about 5 : 6 (lengths of B to A) and 10 : 9 hold.

Answer: (d) 10 g

Explanation:
1. Pipe fundamental \( =\frac{320}{4\times0.8}=100\ \text{Hz} \).
2. String second harmonic: \( 100=\frac{v_s}{0.5}\Rightarrow v_s=50\ \text{m/s} \).
3. \( \mu=\frac{T}{v_s^2}=\frac{50}{2500}=0.02\ \text{kg/m} \), mass \( =0.02\times0.5=10\ \text{g} \).

Teacher's Note:
1. The second harmonic of a string fixed at both ends has wavelength equal to its length.
2. Mass = mu times length.

Answer: (c) 0.3 m

Explanation:
1. \( \lambda=\frac{v-v_s}{f}=\frac{332-32}{1000}=0.3\ \text{m} \).

Teacher's Note:
1. A source moving towards the wall squeezes the wavelength between source and wall.
2. The motion of the wall does not change the wavelength.

Answer: (b) 1%

Explanation:
1. \( T=\sqrt{\frac{2L}{g}}+\frac Lv\Rightarrow\frac{dT}{dL}=\frac{1}{\sqrt{2gL}}+\frac1v=\frac1{20}+\frac1{300}=0.0533 \).
2. \( \delta L=\frac{0.01}{0.0533}=0.1875\ \text{m}\Rightarrow\frac{\delta L}{L}=0.94\%\approx1\% \).

Teacher's Note:
1. Total time is fall time plus sound travel time; differentiate with respect to L.
2. Because the fall time dominates, the sound term gives only a small correction.

Answer: (a) 2286 m

Explanation:
1. Slant distance \( =330\times8=2640\ \text{m} \).
2. Vertical height \( =2640\sin60^\circ=2640\times0.866=2286\ \text{m} \).

Teacher's Note:
1. Light travels almost instantly, so the delay is the travel time of sound alone.
2. Height = slant distance times sine of the elevation angle.

Answer: (a) 2.5 m/s

Explanation:
1. Moving away from A: \( f_A=680\frac{340-u}{340} \); towards B: \( f_B=680\frac{340+u}{340} \).
2. \( f_B-f_A=\frac{2\times680\,u}{340}=4u=10\Rightarrow u=2.5\ \text{m/s} \).

Teacher's Note:
1. The sound from the source behind the observer is Doppler lowered, from the source ahead it is raised.
2. Beat frequency is the difference of the two apparent frequencies.

Answer: (d) zero

Explanation:
1. The observer and both sources have the same velocity, so there is no relative motion between any of them.
2. Both frequencies stay at 100 Hz; the beat frequency is zero.

Teacher's Note:
1. Doppler shifts depend on relative motion, not on the common motion relative to the ground.
2. A uniform wind or uniform common velocity changes nothing in the frequencies heard.

Answer: (c) 313 m/s

Explanation:
1. \( \rho=\frac{0.032}{22.4\times10^{-3}}=1.43\ \text{kg/m}^3,\ \gamma=\frac{3.5}{2.5}=1.4 \).
2. \( v=\sqrt{\frac{\gamma P}{\rho}}=\sqrt{\frac{1.4\times10^5}{1.43}}\approx313\ \text{m/s} \).

Teacher's Note:
1. Use Laplace's formula with the adiabatic constant gamma = Cp/Cv.
2. One mole at STP occupies 22.4 litres.

Answer: (b) A to P, T; B to P, S; C to Q, S; D to Q, R

Explanation:
1. Pipes carry longitudinal waves, wires transverse waves.
2. Closed pipe: \( L=\lambda/4\Rightarrow\lambda=4L \). Open pipe: \( \lambda=2L \). Wire clamped at both ends: \( \lambda=2L \). Wire also clamped at its mid-point (node there): loop length L/2, so \( \lambda=L \).

Teacher's Note:
1. Count the node and antinode positions to relate L to lambda.
2. Fixing the mid-point forces a node there, doubling the lowest frequency.

Answer: (b) \( \frac{2\pi}{35}\ \text{rad} \)

Explanation:
1. \( \lambda=\frac{350}{100}=3.5\ \text{m} \).
2. \( \Delta\phi=\frac{2\pi}{\lambda}\Delta x=\frac{2\pi(0.1)}{3.5}=\frac{2\pi}{35}\ \text{rad} \).

Teacher's Note:
1. Phase difference between two points at a fixed time is 2 pi times the separation over the wavelength.
2. Convert 10 cm to 0.1 m.

Answer: (a) \( \frac{\pi^2\times10^{-9}}{4}\ \text{W/m}^2 \)

Explanation:
1. \( v=\sqrt{B/\rho}=20\ \text{m/s} \), \( \omega=500\pi \).
2. \( I=\frac12\rho v\omega^2A^2=\frac12(1)(20)(500\pi)^2(10^{-8})^2=\frac{\pi^2\times10^{-9}}{4}\approx2.47\times10^{-9}\ \text{W/m}^2 \).

Teacher's Note:
1. Intensity of a sound wave is (1/2) rho v omega^2 A^2.
2. Find v from the bulk modulus.

Answer: (d) above 10 km

Explanation:
1. Level difference 40 dB means intensity falls by \( 10^4 \).
2. Intensity \( \propto1/r^2 \), so distance increases by \( 10^2=100 \): \( 100\ \text{m}\times100=10\ \text{km} \).

Teacher's Note:
1. A point source has intensity inversely proportional to the square of the distance.
2. 40 dB corresponds to a factor 10^4 in intensity and 10^2 in distance.

Answer: (c) \( 1.08\times10^{-4} \)

Explanation:
1. Velocity amplitude \( =A\omega \); wave speed \( =\omega/k \).
2. \( \frac{A\omega}{\omega/k}=Ak=6\times10^{-5}\times1.8=1.08\times10^{-4} \).

Teacher's Note:
1. The ratio of particle speed amplitude to wave speed equals the product of amplitude and wave number.
2. It is the dimensionless strain amplitude of the wave.

Answer: (c) 330 Hz

Explanation:
1. Wall receives \( f_1=334\,\frac{330-2}{330} \) and re-emits as a moving source receding at 2 m/s.
2. \( f_2=f_1\frac{330}{330+2}=334\times\frac{328}{332}=330\ \text{Hz} \).

Teacher's Note:
1. A moving reflector acts first as a moving observer, then as a moving source.
2. Both steps lower the frequency here.

Answer: (b) 1.06

Explanation:
1. \( v=\sqrt{\gamma RT/M} \), so \( \frac{v_{Ne}}{v_{H_2O}}=\sqrt{\frac{5/3}{4/3}\times\frac{18}{20.2}} \).
2. \( =\sqrt{1.25\times0.891}=1.06 \).

Teacher's Note:
1. Monatomic gases have gamma = 5/3 and non-linear triatomic gases such as water vapour have gamma = 4/3.
2. Speed of sound in a gas is sqrt(gamma R T / M).

Answer: (a) \( \nu_1\gt \nu_3\gt \nu_2 \)

Explanation:
1. Clockwise motion: at A (top) the source moves to the right, towards O, so \( \nu_1\gt f \).
2. At B (bottom) it moves to the left, away from O: \( \nu_2\lt f \). At C the velocity is perpendicular to the line of sight: \( \nu_3=f \).

Teacher's Note:
1. Only the component of the source velocity along the line of sight matters.
2. At C the velocity is perpendicular to the line to O.

Answer: (d) at the middle of the pipe

Explanation:
1. Open ends are displacement antinodes, hence pressure nodes.
2. The fundamental has a displacement node (pressure antinode) at the middle.

Teacher's Note:
1. Pressure and displacement nodes are interchanged.
2. Open ends are always at atmospheric pressure.

Answer: (d) closed at both ends, of length 2 m

Explanation:
1. \( \cos\frac{3\pi x}{2} \) is maximum (a pressure antinode) at x = 0, so that end is closed. At the other end \( x=L \): a closed end needs \( \cos\frac{3\pi L}{2}=\pm1 \), i.e. \( L=\frac{2n}{3} \) (n = 3 gives L = 2 m).
2. A closed-open pipe would need \( \cos\frac{3\pi L}{2}=0 \) i.e. \( L=\frac13,1,\frac53 \) m, which excludes 0.5 m and 2/3 m; an open-open pipe would need a pressure node at x = 0, which is not so.

Teacher's Note:
1. A pressure antinode at x = 0 means a closed end there.
2. Test the other end: closed means antinode, open means node.

Answer: (c) 1.27

Explanation:
1. \( \lambda=0.40\ \text{m},\ v=f\lambda=440\ \text{m/s} \).
2. \( \gamma=\frac{v^2M}{RT}=\frac{440^2\times0.016}{8.314\times293}=1.27 \).

Teacher's Note:
1. Nodes are half a wavelength apart.
2. Use v = sqrt(gamma R T / M) and solve for gamma.

Answer: (b) 20 cm, 80 cm and 200 Hz

Explanation:
1. Closed piece \( L_1 \): \( \frac{v}{4L_1} \). Open piece \( L_2 \): first overtone \( \frac{v}{L_2} \). Equal gives \( L_2=4L_1 \).
2. \( L_1+L_2=100\Rightarrow L_1=20\ \text{cm},\ L_2=80\ \text{cm} \); fundamental of the open piece \( \frac{320}{2(0.8)}=200\ \text{Hz} \).

Teacher's Note:
1. The first overtone of an open pipe is its second harmonic, equal to v/L.
2. Convert all lengths to metres.

Answer: (a) 150 Hz

Explanation:
1. Closed column: successive resonances differ by \( 2f_1=300\ \text{Hz} \).
2. So \( f_1=150\ \text{Hz} \) (750 = 5 x 150, 1050 = 7 x 150).

Teacher's Note:
1. Only odd harmonics are present, so consecutive resonances differ by twice the fundamental.
2. Check that both frequencies are odd multiples of the answer.

Answer: (a) 25/4

Explanation:
1. \( \frac{I_{max}}{I_{min}}=\left(\frac{a_1+a_2}{a_1-a_2}\right)^2=\frac{49}{9}\Rightarrow\frac{a_1+a_2}{a_1-a_2}=\frac73 \).
2. \( \frac{a_1}{a_2}=\frac52\Rightarrow\frac{I_1}{I_2}=\frac{25}{4} \).

Teacher's Note:
1. Take the square root of the intensity ratio to get an amplitude ratio, then use componendo and dividendo.
2. Intensity of each component is proportional to the square of its amplitude.

Answer: (d) \( 1.7\times10^{-5} \)

Explanation:
1. \( \lambda=\frac{2\pi}{1.8}=3.49\ \text{m} \) and \( A=6\times10^{-5}\ \text{m} \).
2. \( \frac A\lambda=\frac{6\times10^{-5}\times1.8}{2\pi}=1.7\times10^{-5} \).

Teacher's Note:
1. Amplitude is given in units of 10^-5 m, so do not forget the factor.
2. Wave number k = 1.8 per metre.