Let the particle executing SHM starts oscillating from its mean position. Then displacement equation is $x=\mathrm{A} \sin \omega t$
$\therefore \quad$ Particle velocity, $v=\mathrm{A} \omega \cos \omega t$
$\therefore \quad$ Instantaneous K.E.,
$$
K=\frac{1}{2} m v^2=\frac{1}{2} m A^2 \omega^2 \cos ^2 \omega t
$$
$\therefore \quad$ Average value of K.E. over one complete cycle
$$
\begin{aligned}
K_{a v} &=\frac{1}{T} \int_0^T \frac{1}{2} m A^2 \omega^2 \cos ^2 \omega t d t \\
&=\frac{m A^2 \omega^2}{2 T} \int_0^T \cos ^2 \omega t d t \\
&=\frac{m A^2 \omega^2}{2 T} \int_0^T \frac{(1+\cos 2 \omega t)}{2} d t \\
&=\frac{m A^2 \omega^2}{4 T}\left[t+\frac{\sin 2 \omega t}{2 \omega}\right]_0^T \\
=\frac{m A^2 \omega^2}{4 T}\left[(T-0)+\left(\frac{\sin 2 \omega T-\sin 0}{2 \omega}\right)\right] \\
&=\frac{1}{4} m A^2 \omega^2
\end{aligned}
$$
Again instantaneous P.E.,
$$
U=\frac{1}{2} k x^2=\frac{1}{2} m \omega^2 x^2=\frac{1}{2} m \omega^2 A^2 \sin ^2 \omega t
$$
$\therefore \quad$ Average value of P.E. over one complete cycle
$$
U_{a v}=\frac{1}{T} \int_0^T \frac{1}{2} m \omega^2 A^2 \sin ^2 \omega t
$$
$$
\begin{aligned}
=& \frac{m \omega^2 A^2}{2 T} \int_0^T \sin ^2 \omega t d t \\
=& \frac{m \omega^2 A^2}{2 T} \int_0^T \frac{(1-\cos 2 \omega t)}{2} d t \\
=& \frac{m \omega^2 A^2}{4 T}\left[t-\frac{\sin 2 \omega t}{2 \omega}\right]_0^T \\
=\frac{m \omega^2 A^2}{4 T}\left[(T-0)-\frac{(\sin 2 \omega T-\sin 0)}{2 \Theta}\right] \\
=& \frac{1}{4} m \omega^2 A^2
\end{aligned}
$$
Simple comparison of (i) and (ii), shows that
$$
K_{a v}=U_{a v}=\frac{1}{4} m \omega^2 A^2
$$