$\mathrm{The} \mathrm{equation} \mathrm{of} \mathrm{displacement} \mathrm{of} \mathrm{a} \mathrm{particle} \mathrm{executing} \mathrm{SHM}:$
$\mathrm{x}=\mathrm{Asin\omega t}$
$\mathrm{where},$
$\mathrm{A}=\mathrm{Amplitude} \mathrm{of} \mathrm{oscillation}$
$\mathrm{\omega }=\mathrm{Angular} \mathrm{frequency}=\sqrt{\frac{\mathrm{k}}{\mathrm{M}}}$
$\mathrm{The} \mathrm{velocity} \mathrm{of} \mathrm{the} \mathrm{particle} \mathrm{is}:$
$\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{A\omega cos\omega t}$
$\mathrm{The} \mathrm{kinetic} \mathrm{energy} \mathrm{of} \mathrm{the} \mathrm{particle} \mathrm{is}:$
${\mathrm{E}}_{\mathrm{k}}=\frac{1}{2}{\mathrm{Mv}}^2=\frac{1}{2}{\mathrm{MA}}^2{\mathrm{\omega }}^2{\cos }^2\mathrm{\omega t}$
$\mathrm{The} \mathrm{potential} \mathrm{energy} \mathrm{of} \mathrm{the} \mathrm{particle} \mathrm{is}:$
${\mathrm{E}}_{\mathrm{P}}=\frac{1}{2}{\mathrm{kx}}^2=\frac{1}{2}{\mathrm{M\omega }}^2{\mathrm{A}}^2{\sin }^2\mathrm{\omega t}$
$\mathrm{For} \mathrm{time} \mathrm{period} \mathrm{T}, \mathrm{the} \mathrm{average} \mathrm{kinetic} \mathrm{energy} \mathrm{over} \mathrm{a} \mathrm{single}$
$\mathrm{cycle}:$
$\begin{array}{l}\begin{array}{rl}{\left({\mathrm{E}}_{\mathrm{k}}\right)}_{\mathrm{Avg}}& =\frac{1}{\mathrm{T}}{\int }_0^{\mathrm{T}}{\mathrm{E}}_{\mathrm{k}}\mathrm{dt}\\ & =\frac{1}{\mathrm{T}}{\int }_0^{\mathrm{T}}\frac{1}{2}{\mathrm{MA}}^2{\mathrm{\omega }}^2{\cos }^2\mathrm{\omega tdt}\\ =& \frac{1}{2\mathrm{T}}{\mathrm{MA}}^2{\mathrm{\omega }}^2{\int }_0^{\mathrm{T}}\frac{(1+\cos 2\mathrm{\omega t})}{2}\mathrm{dt}\end{array}\end{array}$
$\begin{array}{rl}& =\frac{1}{4\mathrm{T}}{\mathrm{MA}}^2{\mathrm{\omega }}^2{[\mathrm{t}+\frac{\sin 2\mathrm{\omega t}}{2\mathrm{\omega }}]}_0^{\mathrm{T}}\\ & =\frac{1}{4\mathrm{T}}{\mathrm{MA}}^2{\mathrm{\omega }}^2\left(\mathrm{T}\right)\\ & =\frac{1}{4}{\mathrm{MA}}^2{\mathrm{\omega }}^2 ...\left(\mathrm{i}\right)\end{array}$
$\mathrm{And}, \mathrm{average} \mathrm{potential} \mathrm{energy} \mathrm{over} \mathrm{one} \mathrm{cycle}:$
$\begin{array}{rl}{\left({\mathrm{E}}_{\mathrm{P}}\right)}_{\mathrm{Avg}}& =\frac{1}{\mathrm{T}}{\int }_0^{\mathrm{T}}{\mathrm{E}}_{\mathrm{p}}\mathrm{dt}\\ & =\frac{1}{\mathrm{T}}{\int }_0^{\mathrm{T}}\frac{1}{2}{\mathrm{M\omega }}^2{\mathrm{A}}^2{\sin }^2\mathrm{\omega tdt}\\ & =\frac{1}{2\mathrm{T}}{\mathrm{M\omega }}^2{\mathrm{A}}^2{\int }_0^{\mathrm{T}}\frac{(1-\cos 2\mathrm{\omega t})}{2}\mathrm{dt}\end{array}$
$\begin{array}{l}=\frac{1}{4\mathrm{T}}{\mathrm{M\omega }}^2{\mathrm{A}}^2{[\mathrm{t}-\frac{\sin 2\mathrm{\omega t}}{2\mathrm{\omega }}]}_0^{\mathrm{T}}\\ =\frac{1}{4\mathrm{T}}{\mathrm{M\omega }}^2{\mathrm{A}}^2\left(\mathrm{T}\right)\\ =\frac{{\mathrm{M\omega }}^2{\mathrm{A}}^2}{4} ...\left(\mathrm{ii}\right)\end{array}$

So the average kinetic energy for a given time period is equal to the average potential energy for the same time period.