The amplitude of a particle executing $SHM$ about $O$ is $10\, cm.$ Then :
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$K.E.$ $=0.64 \mathrm{KE}_{\max }$
$v=0.8 v_{\max }$
$\therefore x=0.6 A=6 \mathrm{cm}$
$x=\frac{A}{2} P . E .=\frac{P E_{\max }}{4} K E=\frac{3}{4} P E_{\max }$
$\mathrm{KE}_{\max }=\mathrm{TE}$ at mean position
$x=\frac{A}{2} v=\frac{\sqrt{3} v_{\max }}{2}$
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