The $ P.E.$ of a particle executing $SHM$ at a distance $x$ from its equilibrium position is
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We know that potential energy is given by,
$PE =\frac{1}{2} kx ^2$
and we know, $\omega^2=\frac{k}{m}$
or, $k =\omega^2 m$
Using value of $k$ in equation $(1)$,
$PE =\frac{1}{2} m \omega^2 x ^2$ is our required answer.
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