A body is executing simple harmonic motion with frequency $n,$ the frequency of its potential energy is

The correct option is C $2n$
Equation of displacement of particle executing SHM is given by

$x=A\sin (\omega t+\varphi )......\left(1\right)$

Potential energy of a particle executing SHM is given by

$U=\frac{1}{2}k{x}^2=\frac{1}{2}k{A}^2{\sin }^2(\omega t+\varphi )$

$U=\frac{1}{4}k{A}^2-\frac{1}{4}k{A}^2\cos (2\omega t+2\varphi )....\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$, it is clear that the time period of $x=A\sin (\omega t+\varphi )\text{is}$

$T_1=\frac{2\pi }{\omega }\Rightarrow \text{frequency},n=\frac{\omega }{2\pi }$

While time period of $x^2=A^2{\sin }^2(\omega t+\varphi )\text{is}$

$T_2=\frac{2\pi }{2\omega }=\frac{\pi }{\omega }\Rightarrow \text{frequency},n^{\prime }=\frac{\omega }{\pi }$

$\therefore n^{\prime }=2n$

Hence, option $\left(\text{C}\right)$ is correct.

Alternate solution: For every oscillation, the particle reaches the point of maximum energy twice. Hence, frequency becomes $2n$.