For a particle executing SHM. The KE ‘K’ is given by K = K₀ cos² wt. The maximum value of PE is

$\mathrm{K}.\mathrm{E}={\mathrm{K}}_0{\cos }^2\omega \mathrm{t}$
But $\mathrm{K}.\mathrm{E}=\frac{1}{2}{\mathrm{mv}}^2\mathrm{x}=\mathrm{Asin}\mathrm{\omega t}$
$\begin{array}{l}\mathrm{V}=\mathrm{A\omega cos}\mathrm{\omega t}; \mathrm{K}.\mathrm{E}=\frac{1}{2}{\mathrm{mA}}^2{\mathrm{\omega }}^2{\cos }^2\mathrm{\omega t}\\ \text{ compare with }\mathrm{K}.\mathrm{E}={\mathrm{K}}_0{\cos }^2\mathrm{\omega t}\\ {\mathrm{K}}_0=\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{A}}^2\to \text{ max value of P.E }\\ \text{ P.E }{\mathrm{E}}_{max}=\frac{1}{2}{\mathrm{m}}^2{\mathrm{A}}^2={\mathrm{k}}_0\end{array}$