The $K.E.$ and $P.E.$ of a particle executing $SHM$ with amplitude $A$ will be equal when its displacement is-
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$\mathrm{KE}=\mathrm{PE}$
$\frac{1}{2} \mathrm{m} \omega^{2}\left(\mathrm{A}^{2}-\mathrm{y}^{2}\right)=\frac{1}{2} \mathrm{m} \omega^{2} \mathrm{y}^{2}$
$y=\frac{A}{\sqrt{2}}$
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