During a simple harmonic motion,

In a simple harmonic motion, the displacement and velocity as functions of time are given by $x = A \sin(\omega t)$ and $v = A \omega \cos(\omega t)$.

The instantaneous potential energy is $U = \dfrac{1}{2} m \omega^2 x^2 = \dfrac{1}{2} m \omega^2 A^2 \sin^2(\omega t)$.

The instantaneous kinetic energy is $K = \dfrac{1}{2} m v^2 = \dfrac{1}{2} m \omega^2 A^2 \cos^2(\omega t)$.

The average potential energy over one complete time period $T$ is:

$\langle U \rangle = \dfrac{1}{T} \int_{0}^{T} \dfrac{1}{2} m \omega^2 A^2 \sin^2(\omega t) dt = \dfrac{1}{4} m \omega^2 A^2$

The average kinetic energy over one complete time period $T$ is:

$\langle K \rangle = \dfrac{1}{T} \int_{0}^{T} \dfrac{1}{2} m \omega^2 A^2 \cos^2(\omega t) dt = \dfrac{1}{4} m \omega^2 A^2$

Thus, the average potential energy over a single time period is exactly equal to the average kinetic energy for this period.