JEE Physics Oscillations and Waves MCQs Set 01

Answer: (c) 0.5 m s⁻¹

Explanation:
1. Compare with \( y=A\sin(kx-\omega t) \): \( k=8\ \text{m}^{-1},\ \omega=4\ \text{rad/s} \).
2. \( v=\frac\omega k=\frac48=0.5\ \text{m/s} \).

Teacher's Note:
1. Wave speed is ω/k.
2. Do not confuse k with the frequency.

Answer: (a) 400 Hz

Explanation:
1. Source approaching: \( f'=f\frac{v}{v-v_s}=320\times\frac{330}{330-66}=320\times\frac{330}{264}=400\ \text{Hz} \).

Teacher's Note:
1. The apparent frequency rises when the source moves towards the observer.
2. Use v - v_s in the denominator.

Answer: (b) 55 Hz

Explanation:
1. Train A (approaching): \( f_A=300\times\frac{330}{300}=330\ \text{Hz} \).
2. Train B (receding): \( f_B=300\times\frac{330}{360}=275\ \text{Hz} \).
3. Difference \( =55\ \text{Hz} \).

Teacher's Note:
1. Use v - vs for the approaching train and v + vs for the receding one.
2. Compute each frequency before subtracting.

Answer: (a) 20 cm

Explanation:
1. \( y_2=5\cdot2\left[\frac12\sin\omega t+\frac{\sqrt3}2\cos\omega t\right]=10\sin\left(\omega t+\frac\pi3\right) \).
2. Both waves have amplitude 10 cm and the same phase, so the resultant amplitude is \( 10+10=20 \) cm.

Teacher's Note:
1. Convert a sin + b cos into a single sine term.
2. Waves in phase add their amplitudes.

Answer: (b) 100 m/s

Explanation:
1. \( v=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{70}{7\times10^{-3}}}=\sqrt{10^4}=100\ \text{m/s} \).

Teacher's Note:
1. Speed of a transverse wave on a string is √(tension/linear density).
2. Check that μ is in kg per metre.

Answer: (a) 5

Explanation:
1. Springs on the two sides act in parallel: \( k_{eff}=20+20=40\ \text{N/m} \).
2. \( T=2\pi\sqrt{\frac{m}{k}}=2\pi\sqrt{\frac{2}{40}}=\frac{2\pi}{\sqrt{20}}=\frac{\pi}{\sqrt5} \), so x = 5.

Teacher's Note:
1. When the block is displaced, one spring stretches and the other compresses, and both push it back.
2. The effective stiffness is the sum of the two constants.

Answer: (c) 2 : 3

Explanation:
1. The spring force supplies the centripetal force: \( k\Delta=m\omega^2(L+\Delta) \).
2. \( 12.5\Delta=0.2\times25\,(L+\Delta)=5(L+\Delta) \).
3. \( 7.5\Delta=5L\Rightarrow\frac\Delta L=\frac23 \).

Teacher's Note:
1. The radius of the circle is the natural length plus the extension.
2. The spring is the only horizontal force on the body.

Answer: (a) 1 kg

Explanation:
1. \( T=2\pi\sqrt{\frac mk}\Rightarrow\frac{T'}{T}=\sqrt{\frac{m+3}{m}}=\frac21 \).
2. \( \frac{m+3}{m}=4\Rightarrow m=1\ \text{kg} \).

Teacher's Note:
1. T is proportional to √m, so the ratio of periods fixes the ratio of masses.
2. New period = 1 + 1 = 2 s.

Answer: (d) 4 s

Explanation:
1. With \( x=A\sin\omega t \): x = A/2 at \( \omega t=\frac\pi6 \), i.e. at \( t=\frac T{12} \) = 2 s, so T = 24 s.
2. x = A at \( t=\frac T4=6\ \text{s} \).
3. Time from A/2 to A \( =6-2=4\ \text{s} \).

Teacher's Note:
1. The particle slows down as it approaches the extreme position, so the second part takes longer.
2. Use the reference circle: π/6 corresponds to T/12.

Answer: (a) 120°

Explanation:
1. \( A^2=A_1^2+A_2^2+2A_1A_2\cos\phi\Rightarrow64=64+64+128\cos\phi \).
2. \( \cos\phi=-\frac12\Rightarrow\phi=120^\circ \).

Teacher's Note:
1. Use the formula for the resultant amplitude of two superposed waves.
2. When the amplitudes are equal, A = 2A1 cos(φ/2); here cos(φ/2) = 1/2.

Answer: (a) 40 cm

Explanation:
1. Force constant k = 25 N/m, so \( \omega=\sqrt{\frac km}=\sqrt{\frac{25}{0.25}}=10\ \text{rad/s} \).
2. \( v_{max}=A\omega\Rightarrow A=\frac4{10}=0.4\ \text{m}=40\ \text{cm} \).

Teacher's Note:
1. Convert 250 g to 0.25 kg.
2. Maximum speed occurs at the mean position and equals Aω.

Answer: (a) 8

Explanation:
1. \( U=\frac12kA^2\sin^2\omega t\Rightarrow\frac{dU}{dt}=\frac12kA^2\omega\sin2\omega t \).
2. The slope is maximum when \( 2\omega t=\frac\pi2 \), i.e. \( t=\frac{\pi}{4\omega}=\frac T8 \).
3. So \( \beta=8 \).

Teacher's Note:
1. Differentiate U with respect to time and use sin 2θ = 2 sinθ cosθ.
2. Potential energy oscillates at twice the frequency of the displacement.

Answer: (b) \( \frac1{\sqrt2} \)

Explanation:
1. \( \omega=\sqrt{\frac km}\propto\frac1{\sqrt m} \).
2. \( \frac{\omega_2}{\omega_1}=\sqrt{\frac{m_1}{m_2}}=\sqrt{\frac12}=\frac1{\sqrt2} \).

Teacher's Note:
1. The spring constant is the same in both cases.
2. A heavier block oscillates more slowly.

Answer: (a) 88

Explanation:
1. \( v^2=\frac{50-x^2}{4}=\omega^2(A^2-x^2)\Rightarrow\omega^2=\frac14 \), so \( \omega=\frac12\ \text{rad/s} \).
2. \( T=\frac{2\pi}{\omega}=4\pi=4\times\frac{22}7=\frac{88}7\ \text{s} \), so x = 88.

Teacher's Note:
1. Compare with v² = ω²(A² - x²).
2. The coefficient of x² in v² is -ω².

Answer: (c) 18.75 J

Explanation:
1. Total energy \( E=\frac12kA^2=25\ \text{J} \).
2. At x = A/2 the potential energy is \( \frac14\times25=6.25\ \text{J} \).
3. \( KE=25-6.25=18.75\ \text{J} \).

Teacher's Note:
1. Potential energy is proportional to x².
2. Total energy equals the maximum potential energy.

Answer: (a) 20

Explanation:
1. Parallel springs: \( k_{eff}=2K=4\ \text{N/m} \).
2. \( T=2\pi\sqrt{\frac{0.49}{4}}=2\pi\times0.35=0.7\pi\ \text{s} \).
3. Number of oscillations \( =\frac{14\pi}{0.7\pi}=20 \).

Teacher's Note:
1. The effective stiffness of two springs on opposite sides is the sum of the two constants.
2. Convert 490 g to 0.49 kg.

Answer: (a) 2 cm

Explanation:
1. \( KE=1.25\,PE\Rightarrow E=KE+PE=2.25\,PE \).
2. \( \frac12kA^2=2.25\times\frac12kx^2\Rightarrow x=\frac{A}{1.5}=2\ \text{cm} \).

Teacher's Note:
1. Total energy is the sum of KE and PE at every position.
2. PE is proportional to x².

Answer: (c) A straight line through the origin with positive slope

Explanation:
1. \( T=2\pi\sqrt{\frac Lg}\Rightarrow T^2=\frac{4\pi^2}{g}L \).
2. T² is directly proportional to L, so the graph of T² against L is a straight line through the origin with positive slope \( \frac{4\pi^2}g \).

Teacher's Note:
1. This graph is used in the laboratory to find g from the slope.
2. T² = (4π²/g) L has the form y = mx.

Answer: (a) 50 N/m

Explanation:
1. The amplitude is 10 cm and the total mechanical energy is constant: \( E=\frac12kA^2=0.25\ \text{J} \).
2. \( k=\frac{2\times0.25}{(0.1)^2}=50\ \text{N/m} \).

Teacher's Note:
1. The block starts from rest at 10 cm, so 10 cm is the amplitude.
2. Energy in SHM is the same at every position.