The mass of the particle executing simple harmonic motion is $m$. The particle’s displacement at a given time $t$ is given by

$x=A \sin \omega t$

Velocity of the particle is given as $\mathrm{v}=\mathrm{d} x / \mathrm{dt}=\mathrm{A} \omega \mathrm{cos} \omega \mathrm{t}$

Instantaneous Kinetic Energy can be calculated as,

$K=(1 / 2) m v^{2}$

$=(1 / 2) \mathrm{m}(\mathrm{A} \omega \cos \omega \mathrm{t})^{2}$

$=(1 / 2) m\left(A^{2} \omega^{2} \cos ^{2} \omega t\right)$

Average value of kinetic energy over one complete cycle can be calculated as,

$\begin{aligned}
&K_{a v}=\frac{1}{T} \int_{0}^{T} \frac{1}{2} m A^{2} \omega^{2} \cos ^{2} \omega t d t=\frac{m A^{2} \omega^{2}}{2 T} \int_{0}^{T} \cos ^{2} \omega t d t=\frac{m A^{2} \omega^{2}}{2 T} \int_{0}^{T} \frac{(1+\cos 2 \omega t)}{2} d t \\
&=\frac{m A^{2} \omega^{2}}{4 T}\left[t+\frac{\sin 2 \omega t}{2 \omega}\right]_{0}^{T} \\
&=\frac{m A^{2} \omega^{2}}{4 T}\left[(T-0)+\left(\frac{\sin 2 \omega t-\sin 0}{2 \omega}\right)\right] \\
&=\frac{1}{4} m A^{2} w^{2} \\
&\text { Average instantaneous potential energy, } U=(1 / 2) k x^{2}=(1 / 2) m \omega^{2} x^{2}=(1 / 2) m \omega^{2} A^{2} \sin ^{2} \omega t
\end{aligned}$

Average value of potential energy over one complete cycle can be calculated as

$\begin{aligned}
U_{a v} &=\frac{1}{T} \int_{0}^{T} \frac{1}{2} m \omega^{2} A^{2} \sin ^{2} \omega t=\frac{m \omega^{2} A^{2}}{2 T} \int_{0}^{T} \sin ^{2} \omega t d t \\
&=\frac{m \omega^{2} A^{2}}{2 T} \int_{0}^{T} \frac{(1-\cos 2 \omega t)}{2} d t
\end{aligned}$

$=\frac{m \omega^{2} A^{2}}{4 T}\left[t-\frac{\sin 2 \omega t}{2 \omega}\right]_{0}^{T}$

$\begin{aligned}
&=\frac{m \omega^{2} A^{2}}{4 T}\left[(T-0)-\frac{(\sin 2 \omega t-\sin 0)}{2 \omega}\right] \\
&=\frac{1}{4} m \omega^{2} A^{2}
\end{aligned}$

Kinetic energy $=$ Potential energy