A particle performing S.H.M. starts from equilibrium position and its time period is 12 second. After 2 seconds its velocity is π m/s. Amplitude of the oscillation is [sin 30° = cos 60° = 0.5, sin 60° = cos 30° = √(3)/2] (a) 6 m (b) 12 m (c) 12√(3) m (d) 6√(3) m

Answer: (b). ω = (2π)/(12) = π/6 rad/s; starting from the mean position, v = Aωcos ω t. At t = 2 s: ω t = π/3, so π = A·π/6·cos 60° = (Aπ)/(12). A = 12 m. So the answer is (b) 12 m.

A particle performing S.H.M. starts from equilibrium position and its… | MHT CET 2024 (11 May, Shift 1)

Q27MHT CET 2024 · 11 MAY · SHIFT 1Medium

Velocity & Acceleration in SHM

A particle performing S.H.M. starts from equilibrium position and its time period is 12 second. After 2 seconds its velocity is $\pi\text{ m/s}$ . Amplitude of the oscillation is

$\left[\sin 30^\circ = \cos 60^\circ = 0.5,\ \sin 60^\circ = \cos 30^\circ = \sqrt{3}/2\right]$

Show answer and solution

Answer: (b)

$\omega = \frac{2\pi}{12} = \frac{\pi}{6}$ rad/s; starting from the mean position, $v = A\omega\cos\omega t$ .
At $t = 2$ s: $\omega t = \frac{\pi}{3}$ , so $\pi = A\cdot\frac{\pi}{6}\cdot\cos 60^\circ = \frac{A\pi}{12}$ .
$A = 12$ m.
So the answer is (b) 12 m.

Report a mistake

Velocity & Acceleration in SHM question from MHT CET 2024 (11 May, Shift 1)

More questions on Velocity & Acceleration in SHM