JEE Physics Sound Waves MCQs Set 04

Answer: (d) \( 64\pi \) rad

Explanation:
1. At a fixed position \( \phi = \omega t \), so \( \Delta\phi = 2\pi f\,\Delta t \).
2. \( \Delta\phi = 2\pi \times 80 \times 0.4 = 64\pi \) rad.

Teacher's Note:
1. At a fixed point the phase advances by 2π in every period.
2. In 0.4 s there are 32 complete periods, giving 64π.

Answer: (c) \( \dfrac{1}{20\pi} \)

Explanation:
1. Amplitude \( A = 2 \) cm = 0.02 m; \( k = 5 \) rad/m, so \( \lambda = 2\pi/5 \) m.
2. \( A/\lambda = 0.02\times 5/(2\pi) = 1/(20\pi) \).

Teacher's Note:
1. Convert the amplitude to metres before dividing by the wavelength.
2. The wave number k is the coefficient of x.

Answer: (a) \( \dfrac{8\times10^{5}}{\pi} \) N/m\(^2\)

Explanation:
1. Pressure amplitude \( \Delta p_0 = (100020-99980)/2 = 20 \) Pa.
2. \( k = 2\pi/\lambda = 2\pi/0.4 = 5\pi \) rad/m.
3. \( \Delta p_0 = B k s_0 \Rightarrow B = \dfrac{20}{5\pi\times5\times10^{-6}} = \dfrac{8\times10^{5}}{\pi} \) N/m\(^2\).

Teacher's Note:
1. Pressure amplitude, displacement amplitude and bulk modulus are linked by \( \Delta p_0 = B k s_0 \).
2. Half of the total pressure swing is the amplitude.

Answer: (c) 333.3 m/s

Explanation:
1. Interval between claps = 3/5 = 0.6 s.
2. Echo returns after travelling 2 x 100 = 200 m in this interval.
3. \( v = 200/0.6 = 333.3 \) m/s \( = 1000/3 \) m/s.

Teacher's Note:
1. An echo merging with the next clap means the round-trip time equals the clapping interval.
2. The round trip is twice the distance to the wall.

Answer: (b) 16

Explanation:
1. Let distance be d km: \( d/4 - d/8 = 240 \) s.
2. \( d/8 = 240 \Rightarrow d = 1920 \) km.
3. \( 120\eta = 1920 \Rightarrow \eta = 16 \).

Teacher's Note:
1. The time lag between the two waves comes only from their different speeds over the same distance.
2. Convert 4 minutes to 240 seconds.

Answer: (a) about 401 m/s

Explanation:
1. \( C_V \) of the mixture per mole \( = \dfrac{1(1.5R)+2(2.5R)}{3} = \dfrac{6.5R}{3} \), \( C_P = \dfrac{9.5R}{3} \), so \( \gamma = 19/13 = 1.462 \).
2. Mean molar mass \( M = (4+64)/3 = 22.67 \) g/mol.
3. \( v = \sqrt{\gamma RT/M} = \sqrt{1.462\times8.314\times300/0.02267} \approx 401 \) m/s.

Teacher's Note:
1. For a gas mixture, average the heat capacities and the molar masses by mole fraction.
2. Then use the usual formula for the speed of sound.

Answer: (d) \( \tfrac16\% \) increase

Explanation:
1. \( v\propto\sqrt{T} \), so \( \dfrac{\Delta v}{v} = \dfrac12\dfrac{\Delta T}{T} = \dfrac12\times\dfrac1{300} \).
2. In percent this is 1/6 per cent, i.e. an increase.

Teacher's Note:
1. Small changes follow \( \Delta v/v = \tfrac12\,\Delta T/T \).
2. The speed rises with temperature.

Answer: (c) about 303.5 m/s

Explanation:
1. Take 100 g: moles are Ar 0.6, O\(_2\) 1.0, CO\(_2\) 1.0, total 2.6 mol.
2. \( C_V = \dfrac{R}{\gamma-1} \): Ar 1.5R, O\(_2\) 2.5R, CO\(_2\) 3R. Mixture \( C_V = 0.6(1.5R)+2.5R+3R = 6.4R \), \( C_P = 6.4R+2.6R = 9R \), so \( \gamma = 9/6.4 = 1.406 \).
3. \( M = 100/2.6 = 38.46 \) g/mol, so \( v = \sqrt{1.406\times8.4\times300/0.03846} \approx 303.5 \) m/s.

Teacher's Note:
1. Convert percentages by mass to moles before averaging.
2. The mixture is treated as one ideal gas with effective gamma and molar mass.

Answer: (b) 60

Explanation:
1. Intensity \( I = \dfrac{p_0^2}{2\rho v} \), so for equal I: \( \dfrac{p_w}{p_a} = \sqrt{\dfrac{\rho_w v_w}{\rho_a v_a}} \).
2. \( = \sqrt{\dfrac{1000\times1530}{1.25\times340}} = \sqrt{3600} = 60 \).

Teacher's Note:
1. Water has a much larger acoustic impedance (density times speed) than air.
2. Equal intensity therefore needs a much larger pressure amplitude in water.

Answer: (a) \( 1/3600 \)

Explanation:
1. \( I = \dfrac{p_0^2}{2\rho v} \), so for equal \( p_0 \): \( \dfrac{I_w}{I_a} = \dfrac{\rho_a v_a}{\rho_w v_w} \).
2. \( = \dfrac{1.25\times340}{1000\times1530} = \dfrac{1}{3600} \).

Teacher's Note:
1. Here the intensity varies inversely with the impedance rho times v.
2. This is why sound passes poorly from air into water.

Answer: (d) 5 N/m\(^2\)

Explanation:
1. Intensity at 1.5 m: \( I = \dfrac{P}{4\pi r^2} = \dfrac{0.8}{4\pi(2.25)} = \dfrac{0.8}{9\pi} \) W/m\(^2\).
2. \( I = \dfrac{(\Delta p)_m^2}{2\rho v} \Rightarrow (\Delta p)_m^2 = 2\times\dfrac{225\pi}{544}\times340\times\dfrac{0.8}{9\pi} = 25 \).
3. So \( (\Delta p)_m = 5 \) N/m\(^2\).

Teacher's Note:
1. Intensity falls as the inverse square of distance from a point source.
2. The pressure amplitude then follows from the intensity formula.

Answer: (c) \( \pi \)

Explanation:
1. \( BD = \sqrt{2.4^2+1^2} = 2.6 \) m, so path difference \( = 2.6-2.4 = 0.2 \) m.
2. \( \lambda = 340/850 = 0.4 \) m, so \( \Delta\phi = 2\pi\times0.2/0.4 = \pi \).

Teacher's Note:
1. Compute the path difference with the right triangle, then compare it with the wavelength.
2. A path difference of half a wavelength gives a phase difference of pi.

Answer: (b) \( (25/312)^2 \approx 6.4\times10^{-3} \) W/m\(^2\)

Explanation:
1. Intensities at D: \( I_A = \dfrac{25\pi}{4\pi(2.4)^2} \), \( I_B = \dfrac{25\pi}{4\pi(2.6)^2} \), so \( \sqrt{I_A} = \dfrac{5}{4.8} \), \( \sqrt{I_B} = \dfrac{5}{5.2} \).
2. The phase difference is pi, so amplitudes subtract.
3. \( I = (\sqrt{I_A}-\sqrt{I_B})^2 = (1.0417-0.9615)^2 = (25/312)^2 \approx 6.4\times10^{-3} \) W/m\(^2\).

Teacher's Note:
1. Resultant intensity for coherent waves is (sqrt I1 plus or minus sqrt I2) squared.
2. Opposite phase means subtraction of amplitudes.

Answer: (a) 83 Hz

Explanation:
1. Along BC the path difference runs from 2 m (at B) down towards 0 (far away).
2. Destructive interference needs a path difference of \( \lambda/2 \) (or an odd multiple), which exists only if \( \lambda/2 \le 2 \) m.
3. So no destructive point exists if \( \lambda \gt 4 \) m, i.e. \( f \lt 332/4 = 83 \) Hz.

Teacher's Note:
1. The largest possible path difference equals the separation of the sources.
2. Long wavelengths cannot produce a half-wavelength path difference.

Answer: (d) 12.5 cm

Explanation:
1. \( \lambda = 330/660 = 0.5 \) m.
2. Moving the wall by d changes the reflected path by 2d.
3. Maximum to minimum needs a change of \( \lambda/2 \): \( 2d = \lambda/2 \Rightarrow d = \lambda/4 = 12.5 \) cm.

Teacher's Note:
1. The reflected wave travels the extra distance twice (to the wall and back).
2. A change of half a wavelength in path takes maximum to minimum.

Answer: (c) 8.5 cm

Explanation:
1. A minimum needs a path difference of \( \lambda/2 = 1 \) cm.
2. For a shift y: \( \sqrt{24^2+(y+1.5)^2}-\sqrt{24^2+(y-1.5)^2} = 1 \).
3. Solving, \( y = 8.5 \) cm (check: \( \sqrt{576+100}-\sqrt{576+49} = 26-25 = 1 \)).

Teacher's Note:
1. Write the path difference exactly instead of using a small-angle approximation.
2. The numbers form neat Pythagorean triples (7.5, 24, 25 and 10, 24, 26).

Answer: (a) about 16.4 kHz

Explanation:
1. Speed in the rod: \( v = \sqrt{Y/\rho} = \sqrt{7.5\times10^{10}/2500} = 5477 \) m/s.
2. Both ends are nodes, and six antinodes means \( L = 6\lambda/2 \), so \( \lambda = 1/3 \) m.
3. \( f = v/\lambda = 3\times5477 \approx 1.64\times10^{4} \) Hz.

Teacher's Note:
1. Clamped ends are displacement nodes, so the rod holds a whole number of half-wavelengths.
2. The speed of a longitudinal wave in a solid is the square root of Y over rho.

Answer: (d) 10 m/s

Explanation:
1. \( k = 10\pi \), so \( \lambda = 2\pi/k = 0.2 \) m; \( \omega = 100\pi \), so \( f = 50 \) Hz.
2. \( v = f\lambda = 50\times0.2 = 10 \) m/s.

Teacher's Note:
1. Read k and omega directly from the standing-wave equation.
2. Each of the two progressive waves has the same wavelength and frequency.

Answer: (c) 100 N/m\(^2\) and \( 20\pi \) N/m\(^2\)

Explanation:
1. Density \( \rho = 10^{-3}\times10^{3} = 1 \) kg/m\(^3\); wave speed 10 m/s, so \( B = \rho v^2 = 100 \) N/m\(^2\).
2. Displacement amplitude of the standing wave = 20 mm, \( k = 10\pi \).
3. \( \Delta p_{max} = B k s_0 = 100\times10\pi\times0.02 = 20\pi \) N/m\(^2\).

Teacher's Note:
1. Bulk modulus follows from \( v=\sqrt{B/\rho} \).
2. Pressure amplitude is B k s0, with s0 in metres.

Answer: (b) 5 cm

Explanation:
1. \( \lambda = 2\pi/10\pi = 0.2 \) m.
2. Pressure antinodes coincide with displacement nodes, and the nearest displacement antinode is a quarter wavelength away.
3. Distance \( = \lambda/4 = 0.05 \) m = 5 cm.

Teacher's Note:
1. Displacement nodes are pressure antinodes and vice versa.
2. A node and the next antinode are lambda/4 apart.

Answer: (a) a

Explanation:
1. Third overtone of a closed pipe: \( \ell = 7\lambda/4 \), so \( \lambda = 4\ell/7 \).
2. Displacement amplitude at distance x from the closed end: \( a\sin(2\pi x/\lambda) \).
3. At \( x = \ell/7 \): \( 2\pi x/\lambda = 2\pi(\ell/7)(7/4\ell) = \pi/2 \), so the amplitude is a (an antinode).

Teacher's Note:
1. The closed end is a node, and antinodes are lambda/4 from it.
2. Here lambda/4 equals l/7 exactly.

Answer: (d) 10

Explanation:
1. Natural frequencies of a closed pipe: \( f = (2n+1)\dfrac{v}{4L} = (2n+1)\dfrac{320}{3.2} = 100(2n+1) \) Hz.
2. Values: 100, 300, ..., 1900 Hz (n = 0 to 9); the next, 2100 Hz, exceeds 2000 Hz.
3. So there are 10 natural frequencies in the range.

Teacher's Note:
1. Closed pipes support only odd harmonics.
2. Count the odd multiples of the fundamental that fit in the range.

Answer: (c) 4.2 kHz

Explanation:
1. Adjacent nodes are \( \lambda/2 \) apart, so \( \lambda = 8 \) cm.
2. \( f = v/\lambda = 336/0.08 = 4200 \) Hz.

Teacher's Note:
1. The node-to-node distance is half a wavelength.
2. Frequency is speed divided by wavelength.

Answer: (b) 0.2 m

Explanation:
1. Third harmonic of the closed pipe: \( 3\,\dfrac{v}{4(0.3)} = 2.5v \).
2. First harmonic of the open pipe: \( \dfrac{v}{2L_2} \).
3. Equating, \( L_2 = 1/5 = 0.2 \) m.

Teacher's Note:
1. For a closed pipe the third harmonic is 3 times the fundamental.
2. Equate the two frequencies and solve for the length.

Answer: (d) Closed 20 cm, open 80 cm, 200 Hz

Explanation:
1. Closed piece \( \ell_1 \): fundamental \( v/4\ell_1 \). Open piece \( \ell_2 \): first overtone \( v/\ell_2 \).
2. Equating, \( \ell_2 = 4\ell_1 \), and \( \ell_1+\ell_2 = 1 \) m, so \( \ell_1 = 0.2 \) m, \( \ell_2 = 0.8 \) m.
3. Fundamental of the open piece \( = 320/(2\times0.8) = 200 \) Hz.

Teacher's Note:
1. The first overtone of an open pipe is its second harmonic, equal to v/L.
2. Combine the frequency condition with the total length.

Answer: (c) 56

Explanation:
1. End correction \( e = 0.3d = 1.5 \) cm, so \( \lambda/4 = 16+1.5 = 17.5 \) cm.
2. \( v = 4f(\ell+e) = 4\times480\times0.175 = 336 \) m/s.
3. \( 6\eta = 336 \Rightarrow \eta = 56 \).

Teacher's Note:
1. Include the end correction 0.3 d when finding the quarter wavelength.
2. The first resonance of a closed column is at lambda/4.

Answer: (d) 0.0201

Explanation:
1. \( f\propto\sqrt{T} \). New frequency = 606 Hz.
2. \( T'/T = (606/600)^2 = 1.0201 \).
3. Fractional increase = 0.0201 (about 2%).

Teacher's Note:
1. Frequency of a string varies as the square root of tension.
2. The exact result is slightly more than twice the fractional frequency change.

Answer: (c) 95 Hz

Explanation:
1. Fundamental \( f = \dfrac{1}{2L}\sqrt{T/\mu} \), so \( f\propto\sqrt{T} \): the wire gives \( f_1 \) at 100 N and \( f_2 = 0.9f_1 \) at 81 N.
2. Same beat rate on reducing tension means the fork lies between them: \( f_1 - f = f - f_2 = 5 \), so \( f_1 - f_2 = 10 \) Hz.
3. \( 0.1f_1 = 10 \Rightarrow f_1 = 100 \) Hz, \( f_2 = 90 \) Hz, and the fork is at 95 Hz.

Teacher's Note:
1. The fork frequency lies exactly halfway between the two wire frequencies.
2. Use the ratio of frequencies as the square root of the ratio of tensions (10 : 9).