Equation of displacement of particle executing SHM is $x = A\sin(\omega t + \phi)$. Potential energy $U = \frac{1}{2}kx^2 = \frac{1}{2}kA^2\sin^2(\omega t + \phi)$. Using $\sin^2\theta = \frac{1-\cos(2\theta)}{2}$, $U = \frac{1}{4}kA^2(1-\cos(2\omega t + 2\phi))$. The angular frequency of potential energy is $2\omega$. Since $\omega = 2\pi n$, the new frequency $n_2 = \frac{2\omega}{2\pi} = 2n$.