JEE Physics Waves MCQs Set 13

Answer: (a) \( v_u,\ v_i,\ v_a \) are equal

Explanation:
1. The speed of a sound wave depends only on the elastic and inertial properties of the medium (and its temperature), not on the frequency.
2. Ultrasonic, audible and infrasonic waves are all sound waves, so in the same medium they travel with the same speed.

Teacher's Note:
1. Speed of sound in a given medium is independent of frequency.
2. The three types differ only in their frequency range.

Answer: (b) 480 Hz

Explanation:
1. The fork forces the wire to vibrate: these are forced vibrations.
2. A body under forced vibration oscillates at the frequency of the driving force, not at its natural frequency.
3. So the wire vibrates at 480 Hz (with a small amplitude since it is not in resonance).

Teacher's Note:
1. In forced oscillations the system follows the driving frequency.
2. Large amplitude arises only when the driving frequency equals the natural frequency (resonance).

Answer: (a) (1)

Explanation:
1. For \( y=A\sin(kx-\omega t) \) the wave speed is \( v=\omega/k \).
2. (1): 3/1 = 3 m/s; (2): 1/4 = 0.25 m/s; (3): 1/2 = 0.5 m/s; (4): 2/1 = 2 m/s.
3. The fastest is wave (1).

Teacher's Note:
1. Speed depends only on ω/k, not on the amplitude.
2. Read k from the coefficient of x and ω from the coefficient of t.

Answer: (d) 6 s slow

Explanation:
1. The sound takes time \( t=\frac{2000}{330}\approx6\ \text{s} \) to reach him.
2. He hears the whistle 6 s late and sets his watch at that moment, so his watch is behind (slow) by about 6 s.

Teacher's Note:
1. Sound reaches the man after a delay, so he starts his watch late.
2. Starting late means the watch shows an earlier time: it runs slow.

Answer: (a) 10 cm/s

Explanation:
1. Distance between consecutive crests is the wavelength: \( \lambda=5\ \text{cm} \).
2. Two waves pass per second: \( f=2\ \text{Hz} \).
3. \( v=f\lambda=2\times5=10\ \text{cm/s} \).

Teacher's Note:
1. v = fλ for any wave.
2. Wavelength is the crest-to-crest distance.

Answer: (b) Frequency

Explanation:
1. The frequency is fixed by the source and does not change when the wave crosses from one medium to another.
2. The speed changes (about 340 m/s in air, about 1500 m/s in water), so by \( \lambda=v/f \) the wavelength changes as well.

Teacher's Note:
1. At a boundary the number of waves arriving per second equals the number leaving per second.
2. Speed and wavelength are properties of the medium; frequency belongs to the source.

Answer: (c) 9 m/s

Explanation:
1. Frequency: \( f=\frac{54}{60}=0.9\ \text{Hz} \).
2. \( v=f\lambda=0.9\times10=9\ \text{m/s} \).

Teacher's Note:
1. Convert waves per minute to waves per second before using v = fλ.
2. Check units: Hz × m = m/s.

Answer: (d) Only I and IV are true

Explanation:
1. \( v=\sqrt{\frac{\gamma P}{\rho}}=\sqrt{\frac{\gamma RT}{M}} \).
2. It increases with temperature, as \( v\propto\sqrt T \) (I true).
3. At constant temperature, a change of pressure changes the density in the same proportion, so P/ρ and hence v do not change (IV true).

Teacher's Note:
1. For an ideal gas P/ρ = RT/M, which depends on temperature alone.
2. Hence pressure by itself has no effect on the speed of sound.

Answer: (c) 2 : 1

Explanation:
1. For gases of the same \( \gamma \) at the same pressure, \( v=\sqrt{\gamma P/\rho}\propto\frac{1}{\sqrt\rho} \).
2. \( \frac{v_1}{v_2}=\sqrt{\frac{\rho_2}{\rho_1}}=\sqrt4=2 \).
3. So the ratio is 2 : 1.

Teacher's Note:
1. Both gases are monoatomic, so γ = 5/3 for both.
2. The lighter gas carries sound faster.

Answer: (c) 1092 K

Explanation:
1. Speed of sound is proportional to \( \sqrt T \) (T in kelvin).
2. Initial temperature \( T_0=273\ \text{K} \). For double speed, \( \sqrt{T/T_0}=2\Rightarrow T=4T_0=1092\ \text{K} \).

Teacher's Note:
1. Always use the absolute (kelvin) temperature.
2. Doubling the speed needs four times the absolute temperature.

Answer: (a) directly on the square of the wave amplitude and the square of the wave frequency

Explanation:
1. For a mechanical wave the average power is \( P=\frac12\mu v\omega^2A^2 \) for a string (similar for sound).
2. Since \( \omega=2\pi f \), the power is proportional to \( A^2f^2 \).

Teacher's Note:
1. Energy per unit length depends on ω²A²; multiplying by the wave speed gives the power.
2. Doubling either the amplitude or the frequency quadruples the power.

Answer: (a) \( \sqrt{\frac{32}{17}} \)

Explanation:
1. Take molar masses 2 for hydrogen and 32 for oxygen. Equal volumes means equal moles, so the mixture has \( M_{mix}=\frac{2+32}{2}=17 \).
2. Both gases are diatomic (same \( \gamma \)), so \( v\propto\frac1{\sqrt M} \).
3. \( \frac{v_{mix}}{v_{O_2}}=\sqrt{\frac{32}{17}} \).

Teacher's Note:
1. Equal volumes at the same temperature and pressure contain equal numbers of molecules.
2. Both gases are diatomic, so γ is the same and cancels.

Answer: (a) \( A\omega \)

Explanation:
1. The particle velocity is \( \frac{\partial y}{\partial t}=-A\omega\cos(kx-\omega t) \).
2. Its maximum magnitude is \( A\omega \).

Teacher's Note:
1. Do not confuse the particle velocity (Aω) with the wave velocity (ω/k).
2. Particle velocity changes with time; wave velocity is constant.

Answer: (b) 300

Explanation:
1. Compare with \( y=A\sin(\omega t-kx+\phi) \): \( \omega=600\ \text{rad/s} \) and \( k=2\ \text{rad/m} \).
2. \( v=\frac\omega k=\frac{600}{2}=300\ \text{m/s} \).

Teacher's Note:
1. The wave speed is ω/k, independent of the phase constant.
2. The sign of ωt and kx tells that the wave travels along +x.

Answer: (b) 12.5 cm

Explanation:
1. Wavelength: \( \lambda=\frac vf=\frac{300}{400}=0.75\ \text{m} \).
2. Path difference \( \Delta x=\frac{\lambda}{2\pi}\Delta\phi=\frac{0.75}{360^\circ}\times60^\circ=0.125\ \text{m} \).
3. That is 12.5 cm.

Teacher's Note:
1. A phase difference of 60° is 1/6 of a full cycle, which corresponds to λ/6.
2. Convert the final answer from metres to centimetres.

Answer: (d) 500

Explanation:
1. Source moving towards a stationary observer: \( f'=f\frac{v}{v-v_s} \).
2. \( f'=450\times\frac{330}{330-33}=450\times\frac{330}{297}=500\ \text{Hz} \).

Teacher's Note:
1. When the source approaches, the observer hears a higher frequency.
2. Use v_s in the denominator with a minus sign for approach.

Answer: (a) 436 to 586

Explanation:
1. Angular speed \( \omega=\frac{2\pi\times400}{60}=41.9\ \text{rad/s} \), so the speed of the whistle is \( v_s=\omega r=41.9\times1.2\approx50\ \text{m/s} \).
2. Maximum frequency (approaching): \( 500\times\frac{340}{340-50}\approx586\ \text{Hz} \).
3. Minimum frequency (receding): \( 500\times\frac{340}{340+50}\approx436\ \text{Hz} \).

Teacher's Note:
1. The extremes occur when the whistle moves directly towards or directly away from the listener.
2. The speed of the whistle is ωr, with ω converted to rad/s.

Answer: (d) \( \frac{19}{18} \)

Explanation:
1. \( f_1=f\frac{340}{340-34}=f\frac{340}{306} \) and \( f_2=f\frac{340}{340-17}=f\frac{340}{323} \).
2. \( \frac{f_1}{f_2}=\frac{323}{306}=\frac{19}{18} \).

Teacher's Note:
1. The faster the approaching source, the higher the observed frequency.
2. 323/306 simplifies by dividing both numbers by 17.

Answer: (b) 1000

Explanation:
1. \( \beta=10\log_{10}\frac{I}{I_0}=30 \).
2. \( \log_{10}\frac{I}{I_0}=3\Rightarrow\frac I{I_0}=10^3=1000 \).

Teacher's Note:
1. Each 10 dB corresponds to a factor 10 in intensity.
2. 30 dB means three such factors: 10 × 10 × 10.

Answer: (b) 1.90 Å

Explanation:
1. For speeds much smaller than the speed of light, \( \frac{\Delta\lambda}{\lambda}=\frac vc \).
2. \( \Delta\lambda=5700\times\frac{10^5}{3\times10^8}=1.9\ \text{Å} \) (red shift, since the star recedes).

Teacher's Note:
1. The shift in wavelength is a fraction v/c of the original wavelength.
2. A receding source produces an increase in wavelength (red shift).

Answer: (b) 12 cm

Explanation:
1. Wavelength \( \lambda=\frac{360}{500}=0.72\ \text{m} \).
2. Distance for 60°: \( \frac\lambda6=0.12\ \text{m}=12\ \text{cm} \).

Teacher's Note:
1. Nearest points with 60° phase difference are λ/6 apart.
2. Convert metres to centimetres.

Answer: (b) 2 N

Explanation:
1. For a given string \( f\propto\sqrt T \).
2. \( \frac{f_2}{f_1}=\frac32=\sqrt{\frac{T+2.5}{T}}\Rightarrow\frac{T+2.5}{T}=\frac94 \).
3. \( 4T+10=9T\Rightarrow T=2\ \text{N} \).

Teacher's Note:
1. Frequency of a stretched string is proportional to the square root of tension (length and mass per unit length fixed).
2. Square the frequency ratio to get the tension ratio.

Answer: (b) 0.12 N

Explanation:
1. Wave speed: \( v=\frac\omega k=\frac{30}{1}=30\ \text{m/s} \).
2. \( T=\mu v^2=1.3\times10^{-4}\times900=0.117\approx0.12\ \text{N} \).

Teacher's Note:
1. Speed of a transverse wave on a string is √(T/μ).
2. Here k = 1 rad/m and ω = 30 rad/s.

Answer: (b) 1.07 rad

Explanation:
1. Write the cosine as a sine: \( y_2=10^{-6}\sin\left[100t+\frac x{50}+\frac\pi2\right] \).
2. Phase difference \( =\frac\pi2-0.5=1.57-0.5=1.07\ \text{rad} \).

Teacher's Note:
1. Convert both waves to the same trigonometric function before comparing phases.
2. cos θ = sin(θ + π/2).

Answer: (b) 98 m

Explanation:
1. The motorcycle moves away from the stationary siren: \( f'=f\frac{v-v_o}{v}=0.94f \).
2. \( v_o=0.06\times330=19.8\ \text{m/s} \).
3. From rest with a = 2 m/s²: \( s=\frac{v_o^2}{2a}=\frac{19.8^2}{4}\approx98\ \text{m} \).

Teacher's Note:
1. For an observer receding from a stationary source, the heard frequency is lower by the factor (v − v_o)/v.
2. Use v² = 2as for the distance.