JEE Physics Waves MCQs Set 30

Answer: (a) (i) and (iv) only

Explanation:
1. Wavelength \( \lambda = 300/25 = 12 \) m, so 6 m is half a wavelength and \( \phi = \pi \) (i true).
2. With a phase difference of π, the displacements are equal and opposite, so the maximum difference is 2 × 2.5 × 10^-5 = 5 × 10^-5 m (iv true).

Teacher's Note:
1. Phase difference = 2π × (path difference)/λ.

Answer: (b) 3I

Explanation:
1. Resultant intensity \( I_R = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi = I + 4I + 4I\cos 120^\circ \).
2. Since cos 120° = -1/2, I_R = 5I - 2I = 3I.

Teacher's Note:
1. Use \( I_R = I_1+I_2+2\sqrt{I_1I_2}\cos\phi \).

Answer: (b) 440 Hz

Explanation:
1. Speed of sound is proportional to \( \sqrt{T} \), and f is proportional to v.
2. T goes from 300 K to 363 K, so \( f' = 400\sqrt{363/300} = 400\times 1.1 = 440 \) Hz.

Teacher's Note:
1. Use kelvin.
2. 363/300 = 1.21, whose root is 1.1.

Answer: (c) -a cos(kx + ωt)

Explanation:
1. A stationary wave is formed by two identical waves travelling in opposite directions, so the other wave has \( (kx+\omega t) \).
2. At x = 0 the sum must be zero for all t: \( a\cos(\omega t) + y_2(0,t) = 0 \), so \( y_2(0,t) = -a\cos\omega t \).
3. Hence \( y_2 = -a\cos(kx + \omega t) \).

Teacher's Note:
1. A node at x = 0 needs the two waves to cancel there at all times.
2. Reflection from a fixed end gives a phase change of π.

Answer: (a) 1.22v

Explanation:
1. Tension is proportional to extension, so T' = 1.5T.
2. Speed \( v \propto \sqrt{T} \), so v' = \( \sqrt{1.5} \) v = 1.22v.

Teacher's Note:
1. Hooke's law: T is proportional to x.

Answer: (d) 6

Explanation:
1. In phase, amplitudes add, so the resultant amplitude is 2a and the intensity is 4 times.
2. Difference in loudness = \( 10\log_{10}4 \approx 6 \) dB.

Teacher's Note:
1. dB = 10 log (I₂/I₁).
2. Intensity is proportional to amplitude squared.

Answer: (d) 20%

Explanation:
1. Frequency \( f = \frac{1}{2L}\sqrt{T/\mu} \) is constant, so L is proportional to \( \sqrt{T} \).
2. New T = 1.44 T, so new L = \( \sqrt{1.44} \) L = 1.2 L.
3. Length increases by 20%.

Teacher's Note:
1. Keep f fixed and compare \( \sqrt{T}/L \).

Answer: (c) 25 : 9

Explanation:
1. Intensity falls as 1/r² and intensity is proportional to the square of the amplitude, so amplitude is proportional to 1/r.
2. Ratio = (1/9) : (1/25) = 25 : 9.

Teacher's Note:
1. Spherical waves: amplitude is proportional to 1/r.

Answer: (b) (ii), (iii) and (iv) only

Explanation:
1. For simple harmonic motion, a = -ω²y, so y and a are always in opposite phase (ii true, i false).
2. Velocity is the derivative of displacement, so it leads y by π/2 (iii true), and similarly a leads v by π/2 (iv true).

Teacher's Note:
1. Each derivative advances the phase by π/2.

Answer: (a) 0.5

Explanation:
1. The pulse shape is the same, but its peak has moved from x = 0 to x = 1 m in 2 s.
2. Velocity = 1 m / 2 s = 0.5 m/s.

Teacher's Note:
1. Track the position of the peak.
2. Replacing x by (x - vt) shifts the shape to the right.

Answer: (d) amplitude a/2, frequency 2n and wavelength λ/2

Explanation:
1. Use \( \cos^2\theta = (1+\cos 2\theta)/2 \): \( y = \frac{a}{2} + \frac{a}{2}\cos(4\pi n t - 4\pi x/\lambda) \).
2. The oscillating part has amplitude a/2, angular frequency \( 4\pi n \) (frequency 2n) and wave number \( 4\pi/\lambda \) (wavelength λ/2).

Teacher's Note:
1. Rewrite a squared cosine using the double-angle formula.
2. The constant a/2 only shifts the mean position.

Answer: (b) V = v, if A = λ/2π

Explanation:
1. Maximum particle velocity \( v = A\omega = A\cdot 2\pi V/\lambda \).
2. Setting v = V gives \( A\cdot 2\pi/\lambda = 1 \), so \( A = \lambda/2\pi \).

Teacher's Note:
1. Particle velocity and wave velocity are different things.
2. Use \( \omega = 2\pi f = 2\pi V/\lambda \).

Answer: (c) √2 a

Explanation:
1. Since \( \cos(kx - \omega t) = \cos(\omega t - kx) = \sin(\omega t - kx + \pi/2) \), the second wave leads the first by a phase of π/2.
2. Resultant amplitude \( = \sqrt{a^2 + a^2 + 2a^2\cos 90^\circ} = \sqrt2 a \).

Teacher's Note:
1. Convert cos into sin to find the phase difference.

Answer: (b) (ii), (iii) and (iv) only

Explanation:
1. Angular frequency is \( \alpha \), so the frequency is \( \alpha/2\pi \) (ii is true, i is false).
2. Wave number \( \beta = 2\pi/\lambda \), so \( \lambda = 2\pi/\beta \) (iii true).
3. Velocity \( = \omega/k = \alpha/\beta \) (iv true).

Teacher's Note:
1. α is an angular frequency in rad/s, not a frequency in Hz.

Answer: (a) 200 Hz

Explanation:
1. Open pipe: \( f_0 = v/2L \). Closed pipe, third harmonic: \( 3v/4L = 1.5 f_0 \).
2. So 1.5f₀ - f₀ = 100 Hz, giving f₀ = 200 Hz.

Teacher's Note:
1. A closed pipe has only odd harmonics.
2. Third harmonic = 3 × fundamental of closed pipe.

Answer: (a) 1/4

Explanation:
1. Frequency \( n = \frac{1}{2L}\sqrt{T/\mu} \) with \( \mu = \rho\pi d^2/4 \), so \( n \propto \frac{1}{L}\sqrt{\frac{T}{\rho d^2}} \).
2. Ratio = \( \frac{1}{2}\sqrt{\frac{2}{2\times 4}} = \frac12\times\frac12 = \frac14 \).

Teacher's Note:
1. μ is proportional to ρd².
2. Substitute the doubled factors.

Answer: (c) n_A < n_B

Explanation:
1. The end correction is proportional to the radius, so the wider pipe A has a larger effective length \( L + 0.6r \).
2. Frequency \( f = v/4L_{eff} \) is therefore lower for A: n_A < n_B.

Teacher's Note:
1. End correction makes the effective length longer.
2. Longer effective length gives lower frequency.

Answer: (b) directions of propagation

Explanation:
1. Both waves have angular frequency \( \alpha \) and wave number \( \beta \), so their frequency, wavelength and speed \( \alpha/\beta \) are the same.
2. The first has the form \( (\alpha t - \beta x) \) and travels in the +x direction; the second has \( (\alpha t + \beta x) \) and travels in the -x direction.
3. So they differ in direction of propagation.

Teacher's Note:
1. A minus sign between \( \omega t \) and \( kx \) means motion towards +x.
2. A plus sign means motion towards -x.

Answer: (b) T_A = 4T_B

Explanation:
1. Frequency of string A: \( f = \frac{1}{2L}\sqrt{T_A/\mu} \); of B in its second harmonic: \( f = \frac{2}{2L}\sqrt{T_B/\mu} \).
2. Both equal the fork frequency, so \( \sqrt{T_A} = 2\sqrt{T_B} \), giving T_A = 4T_B.

Teacher's Note:
1. Same strings means same L and μ.
2. n-th harmonic: \( f_n = \frac{n}{2L}\sqrt{T/\mu} \).

Answer: (b) (ii), (iii) and (iv) only

Explanation:
1. The terms in x and t have the same sign, so the wave travels in the negative x-direction.
2. Velocity \( = \omega/k = 15\pi/10\pi = 1.5 \) m/s.
3. Wavelength \( = 2\pi/k = 2\pi/10\pi = 0.2 \) m.

Teacher's Note:
1. Same sign of kx and ωt means motion towards -x.

Answer: (a) (i) and (iv) only

Explanation:
1. Always having the same velocity means the same phase, which needs a separation of one wavelength: min PR = λ (iv).
2. Always having the same speed means equal or opposite phase, so particles λ/2 apart have equal speeds in opposite directions: min PQ = λ/2 (i).

Teacher's Note:
1. Same velocity: separation λ. Same speed: separation λ/2.

Answer: (a) Aω

Explanation:
1. Particle velocity \( \partial y/\partial t = -A\omega\cos(kx-\omega t) \).
2. Its maximum magnitude is Aω.

Teacher's Note:
1. ω/k is the wave speed, not the particle speed.