Energy in SHM: Practice Problem & Solution

Solution Explained:

To solve this problem, we apply the core principles of Energy in SHM. Understanding the underlying formula is key to arriving at the correct answer below:

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 $2n$.