JEE Physics Oscillations MCQs Set 04

Answer: (c) \( 2\pi \)

Explanation:
1. For S.H.M. \( |a| = \omega^2|x| \).
2. \( 1 = \omega^2\times1 \), so \( \omega = 1 \) rad/s.
3. \( T = \dfrac{2\pi}{\omega} = 2\pi \) s.

Teacher's Note:
1. Only the acceleration and displacement are needed.
2. The velocity value is consistent with an amplitude of √2 cm.

Answer: (d) \( \dfrac{3}{\pi} \)

Explanation:
1. The path length is 2A, so \( A = 2 \) cm.
2. \( v_{max} = A\omega \Rightarrow \omega = \dfrac{12}{2} = 6 \) rad/s.
3. \( f = \dfrac{\omega}{2\pi} = \dfrac{3}{\pi} \) Hz.

Teacher's Note:
1. Amplitude is half the length of the path.

Answer: (d) the velocity at the mean position is zero

Explanation:
1. At the mean position the speed is maximum, \( v = A\omega \), so statement 4 is wrong.
2. The other three statements are true: acceleration is maximum at the extremes, the net work over a full oscillation is zero, and energy changes between kinetic and potential.

Teacher's Note:
1. Mean position: maximum speed, zero acceleration.

Answer: (b) \( 16\pi\ \text{m s}^{-1} \)

Explanation:
1. \( \omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{0.05} = 40\pi \) rad/s.
2. \( v_{max} = A\omega = 0.4\times40\pi = 16\pi \) m/s.

Teacher's Note:
1. Convert 40 cm to 0.4 m.

Answer: (c) \( 5\sqrt{3} \) cm

Explanation:
1. \( \omega = \dfrac{v_{max}}{A} = \dfrac{100}{10} = 10 \) s⁻¹.
2. \( v = \omega\sqrt{A^2 - x^2} \Rightarrow 50 = 10\sqrt{100 - x^2} \).
3. \( 100 - x^2 = 25 \), so \( x = \sqrt{75} = 5\sqrt{3} \) cm.

Teacher's Note:
1. Half the maximum speed occurs at √3/2 times the amplitude.

Answer: (a) 25 Hz

Explanation:
1. The velocity completes one full cycle between t = 0 and t = 0.04 s.
2. So \( T = 0.04 \) s and \( f = \dfrac{1}{T} = 25 \) Hz.

Teacher's Note:
1. The velocity has the same period as the displacement.

Answer: (d) \( 60\pi \) cm/s

Explanation:
1. \( v_{max} = A\omega = A(2\pi f) = 0.1\times2\pi\times300 = 60\pi \) cm/s.

Teacher's Note:
1. Keep the amplitude in cm to get cm/s.

Answer: (d) maximum velocity

Explanation:
1. Frequency and period do not depend on amplitude.
2. \( v_{max} = A\omega \) doubles.
3. Energy \( \propto A^2 \) becomes four times.

Teacher's Note:
1. Energy goes as the square of amplitude.

Answer: (c) when v is maximum, a is zero

Explanation:
1. At the mean position \( x = 0 \): \( v = v_{max} \) and \( a = -\omega^2x = 0 \).
2. At the extremes \( v = 0 \) and a is maximum.

Teacher's Note:
1. Speed and acceleration are 90° out of phase.

Answer: (d) the acceleration of the particle is maximum at the equilibrium position

Explanation:
1. \( a = -\omega^2x \) is zero at the equilibrium position and maximum at the extremes.
2. So statement 4 is wrong; the others are correct.

Teacher's Note:
1. Force and acceleration are proportional to displacement.

Answer: (c) at the central position

Explanation:
1. \( v = \omega\sqrt{A^2 - x^2} \) is largest when \( x = 0 \), the central (mean) position.

Teacher's Note:
1. Kinetic energy is maximum there.

Answer: (b) 5.0, 0, \( -20\pi^2 \)

Explanation:
1. Phase at t = 1.5 s: \( 2\pi\times1.5 + \pi = 4\pi \).
2. \( x = 5\cos4\pi = 5.0 \) m.
3. \( v = -10\pi\sin4\pi = 0 \).
4. \( a = -\omega^2x = -(2\pi)^2\times5 = -20\pi^2 \).

Teacher's Note:
1. A phase that is a multiple of 2π means the particle is at the positive extreme.

Answer: (a) 1 Hz

Explanation:
1. \( v_{max} = A\omega \Rightarrow \omega = \dfrac{31.4}{5} = 6.28 \) rad/s.
2. \( f = \dfrac{\omega}{2\pi} = 1 \) Hz.

Teacher's Note:
1. 6.28 is 2π.

Answer: (c) 300

Explanation:
1. Amplitude \( A = 3 \), \( \omega = 100 \).
2. \( v_{max} = A\omega = 300 \) (in the units of y per second).

Teacher's Note:
1. The phase constant π/6 does not affect the maximum velocity.

Answer: (b) 0.01 s

Explanation:
1. \( \omega = \dfrac{v_{max}}{A} = \dfrac{4.4}{7\times10^{-3}} = 628.6 \) rad/s.
2. \( T = \dfrac{2\pi}{\omega} = \dfrac{6.283}{628.6} = 0.01 \) s.

Teacher's Note:
1. Convert mm to m first.

Answer: (c) zero

Explanation:
1. Average velocity is displacement divided by time.
2. In one period the net displacement is zero, so the average velocity is zero (the average speed is 4A/T).

Teacher's Note:
1. Do not confuse average velocity with average speed.

Answer: (a) \( \dfrac{T}{6} \)

Explanation:
1. Take \( x = A\cos\omega t \) from the extreme.
2. \( \dfrac{A}{2} = A\cos\omega t \Rightarrow \omega t = \dfrac{\pi}{3} \).
3. \( t = \dfrac{\pi}{3}\times\dfrac{T}{2\pi} = \dfrac{T}{6} \).

Teacher's Note:
1. From the mean position the same distance takes T/12.

Answer: (b) displacement

Explanation:
1. \( a = -\omega^2x \), so the ratio \( \dfrac{a}{x} = -\omega^2 \) is constant.

Teacher's Note:
1. This is the defining property of S.H.M.

Answer: (a) 2 s

Explanation:
1. Comparing with \( A\cos(\omega t + \phi) \): \( \omega = \pi \) rad/s.
2. \( T = \dfrac{2\pi}{\omega} = 2 \) s.

Teacher's Note:
1. The coefficient of t is the angular frequency.

Answer: (c) totally kinetic

Explanation:
1. At the mean position the displacement is zero, so the potential energy (measured from the mean position) is zero.
2. All the energy is kinetic.

Teacher's Note:
1. At the extremes all the energy is potential.