Q 11-13-008NEETNEET 2021Top questionEasy
A body is executing simple harmonic motion with frequency 'n', the frequency of its potential energy is :
Answer: (C) $2$n
$x = A\sin(\omega t)$, so
$$U = \frac{1}{2}kx^2 = \frac{1}{2}kA^2\sin^2\omega t = \frac{1}{4}kA^2(1 - \cos 2\omega t)$$
The potential energy oscillates at angular frequency $2\omega$, i.e. with frequency $2n$.
Solution by Sreeraj P, M.Sc Physics