JEE Physics Oscillations MCQs Set 27

Answer: (b) decrease

Explanation:
1. Standing up raises the centre of mass, so the effective length of the pendulum decreases.
2. \( T ∝ \sqrt{l} \), so the period decreases.

Teacher's Note:
1. Shorter length, smaller \( T \).

Answer: (b) The girl sitting on the swing stands up

Explanation:
1. The time period \( T = 2\pi\sqrt{l/g} \) is independent of the mass.
2. Standing up raises the centre of mass and shortens the effective length, so \( T \) changes.

Teacher's Note:
1. \( T \) depends on the effective length, not on the mass.

Answer: (c) motion of a planet around the Sun

Explanation:
1. A planet moves in an elliptical orbit under an inverse-square force, which is not a linear restoring force.
2. The other three motions are (approximately) simple harmonic.

Teacher's Note:
1. Orbital motion is not SHM.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. \( v^{2}/\omega^{2}A^{2} + x^{2}/A^{2} = 1 \) is an ellipse, so the assertion is true.
2. The reason is also true, but the ellipse comes from the algebraic relation, not from non-uniform change.

Teacher's Note:
1. The v–x graph is an ellipse.

Answer: (b) 0.5 s

Explanation:
1. The speed is maximum at the mean position, \( x = 0 \).
2. \( \cos \pit = 0 \) first at \( \pit = \pi/2 \), so \( t = 0.5 s \).

Teacher's Note:
1. Mean position first reached at \( T/4 \).

Answer: (c) \( e^{-2\omega t} \)

Explanation:
1. Sine and cosine functions repeat after 2π/ω.
2. \( e^{-2\omegat} \) decreases continuously and never repeats its value.

Teacher's Note:
1. Exponential decay is not periodic.

Answer: (a) periodic

Explanation:
1. The particle returns to the same point after a fixed time, so the motion is periodic.
2. It does not move to and fro about a mean position, so it is not oscillatory.

Teacher's Note:
1. Periodic but not oscillatory.

Answer: (d) zero

Explanation:
1. \( a = -\omega^{2}x \) and \( x = 0 \) at the mean position.
2. So the acceleration is zero.

Teacher's Note:
1. Zero displacement, zero acceleration.

Answer: (a) \( k_1k_2/(k_1+k_2) \)

Explanation:
1. For springs in series, \( 1/k = 1/k_{1} + 1/k_{2} \).
2. So \( k = k_{1}k_{2}/(k_{1} + k_{2}) \).

Teacher's Note:
1. Series: reciprocals add.

Answer: (c) 3 : 1

Explanation:
1. PE = \( E/4 \) and KE = \( E - E/4 = 3E/4 \).
2. KE : PE = 3 : 1.

Teacher's Note:
1. PE = \( E(x/A)^{2} \).

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. At the mean position the speed is maximum and the acceleration is zero.
2. The π/2 phase difference between \( x \) and \( v \) is the reason.

Teacher's Note:
1. \( v \) leads \( x \) by π/2.

Answer: (b) an ellipse

Explanation:
1. \( v^{2}/(\omega^{2}A^{2}) + x^{2}/A^{2} = 1 \).
2. This is the equation of an ellipse.

Teacher's Note:
1. \( v^{2} + \omega^{2}x^{2} = \omega^{2}A^{2} \).

Answer: (d) zero

Explanation:
1. Average acceleration = change in velocity over a period divided by the period.
2. The velocity is the same at the start and the end, so the average is zero.

Teacher's Note:
1. Net change of velocity is zero.

Answer: (d) A is false, but R is true.

Explanation:
1. A damped oscillation is not exactly periodic, so the assertion is false.
2. SHM is oscillatory, so the reason is true.

Teacher's Note:
1. Oscillatory need not be periodic.

Answer: (a) \( A=x_0\omega^2,\ \delta=3\pi/4 \)

Explanation:
1. \( a = -\omega^{2}x_{0} \cos(\omegat - \pi/4) = \omega^{2}x_{0} \cos(\omegat - \pi/4 + \pi) \).
2. So \( A = x_{0}\omega^{2} \) and δ = 3π/4.

Teacher's Note:
1. Multiply by −1 means add π.

Answer: (c) A is true, but R is false.

Explanation:
1. Damping makes the amplitude decay, so the assertion is true.
2. The frequency stays practically constant, so the reason is false.

Teacher's Note:
1. Damping reduces amplitude only.

Answer: (a) \( T=2\pi\sqrt{\frac{R_e}{g}} \)

Explanation:
1. For a long pendulum \( T = 2\pi\sqrt [1/(g(1/l + 1/R_e))] \).
2. As \( l \to \infty \) this becomes \( 2\pi\sqrt{R_e/g} \), about 84 minutes.

Teacher's Note:
1. The \( 1/l \) term vanishes.

Answer: (a) \( k_B:k_A \)

Explanation:
1. \( W = F^{2}/2k \) for the same force \( F \).
2. \( W_A \) : \( W_B = k_B \) : \( k_A \).

Teacher's Note:
1. \( W = F^{2}/2k \).

Answer: (a) \( \sqrt L \)

Explanation:
1. \( T = 2\pi\sqrt{L/g} \), so \( T \) is proportional to \( \sqrt{L} \).

Teacher's Note:
1. \( T ∝ \sqrt{L} \).

Answer: (a) at the turning points of the oscillations

Explanation:
1. The bob is highest at the turning points.
2. So the gravitational potential energy is maximum there.

Teacher's Note:
1. Highest point means maximum PE.

Answer: (b) K.E. is maximum when x = 0

Explanation:
1. At the mean position \( (x = 0) \) the speed is greatest, so \( K.E \). is maximum.
2. \( P.E \). is zero there and \( T.E \). is constant.

Teacher's Note:
1. KE is maximum at the mean position.

Answer: (c) kinetic energy is maximum, potential energy is minimum

Explanation:
1. At the mean position the speed is maximum and the displacement is zero.
2. So \( K.E \). is maximum and \( P.E \). is minimum.

Teacher's Note:
1. KE \( \max \), PE \( \min \) at the mean.

Answer: (d) \( \text{simple harmonic with amplitude }\sqrt{A^2+B^2} \)

Explanation:
1. A \( \sin \omegat + B \cos \omegat = R \sin(\omegat + \phi) \) with \( R = \sqrt{A^{2} + B^{2}} \).
2. So it is SHM of amplitude \( \sqrt{A^{2} + B^{2}} \).

Teacher's Note:
1. Combine into one sinusoid.

Answer: (a) \( 2\pi\sqrt{\frac{x_2^2-x_1^2}{V_1^2-V_2^2}} \)

Explanation:
1. \( V_{1}^{2} - V_{2}^{2} = \omega^{2}(x_{2}^{2} - x_{1}^{2}) \).
2. So \( \omega^{2} = (V_{1}^{2} - V_{2}^{2})/(x_{2}^{2} - x_{1}^{2}) \) and \( T = 2\pi\sqrt [(x_{2}^{2} - x_{1}^{2})/(V_{1}^{2} - V_{2}^{2})] \).

Teacher's Note:
1. Subtract the two \( v^{2} \) equations.

Answer: (c) damped oscillator

Explanation:
1. The term −bv is a velocity-dependent drag force.
2. The system therefore dissipates energy and is a damped oscillator.

Teacher's Note:
1. Drag force means damping.