The total energy of the body executing $S.H.M.$ is $E$. Then the kinetic energy when the displacement is half of the amplitude, is
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(c) Total energy in $SHM $ $E = \frac{1}{2}m{\omega ^2}{a^2}$; (where $a =$ amplitude)
Potential energy $U = \frac{1}{2}m{\omega ^2}({a^2} - {y^2}) $
$= E - \frac{1}{2}m{\omega ^2}{y^2}$
When $y = \frac{a}{2}$
$⇒$ $U = E - \frac{1}{2}m{\omega ^2}\left( {\frac{{{a^2}}}{4}} \right)$
$= E - \frac{E}{4} = \frac{{3E}}{4}$
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