Rankers Physics
Topic: Oscillation

A body is executing simple harmonic motion with frequency '$n$', the frequency of its potential energy is: (2021)
$2n$
$3n$
$4n$
$n$

Solution:

In SHM, the displacement is $x = A\sin(\omega t)$. The potential energy is $U = \frac{1}{2}kx^2 = \frac{1}{2}kA^2\sin^2(\omega t)$.\nSince $\sin^2(\omega t) = \frac{1 - \cos(2\omega t)}{2}$, the frequency of PE is twice the frequency of displacement, so $2n$.

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