Q 11-13-101JEE MainJEE Main 2021 (27 Jul, Shift 2)Easy
An object of mass $0.5$ kg is executing simple harmonic motion. Its amplitude is $5$ cm and time period $(T)$ is $0.2$ s. What will be the potential energy of the object at an instant $t = \dfrac T4$ s starting from mean position? Assume that the initial phase of the oscillation is zero.
Answer: (A) $0.62$ J
$x = A\sin\omega t$; at $t = \dfrac T4$, $x = A\sin\dfrac\pi2 = A$ (extreme position).
$\omega = \dfrac{2\pi}{0.2} = 10\pi\ \text{rad/s}$.
$$U = \frac12m\omega^2A^2 = \frac12(0.5)(100\pi^2)(0.05)^2 \approx 0.62\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics