Q 11-13-096JEE MainJEE Main 2022 (27 Jun, Shift 1)Easy
The displacement of simple harmonic oscillator after $3$ seconds starting from its mean position is equal to half of its amplitude. The time period of harmonic motion is
Answer: (C) $36$ s
Starting from the mean position, $x = A\sin\omega t$.
$$\frac{A}{2} = A\sin(3\omega) \Rightarrow 3\omega = \frac{\pi}{6} \Rightarrow \frac{2\pi}{T}(3) = \frac{\pi}{6}$$
$$T = 36\ \text{s}$$
Solution by Sreeraj P, M.Sc Physics