The total energy of a body executing S.H.M. i | vedclass

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(C) The total energy of a body executing $S.H.M.$ is given by $E = \frac{1}{2} k a^2$,where $a$ is the amplitude.The potential energy at a displacemen

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$\frac{3E}{4}$

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(C) The total energy of a body executing $S.H.M.$ is given by $E = \frac{1}{2} k a^2$,where $a$ is the amplitude.The potential energy at a displacement $y$ is given by $U = \frac{1}{2} k y^2$.The kinetic energy $K$ is given by $K = E - U = \frac{1}{2} k a^2 - \frac{1}{2} k y^2 = \frac{1}{2} k (a^2 - y^2)$.Given that the displacement $y = \frac{a}{2}$,we substitute this into the kinetic energy formula:$K = \frac{1}{2} k (a^2 - (\frac{a}{2})^2)$$K = \frac{1}{2} k (a^2 - \frac{a^2}{4})$$K = \frac{1}{2} k (\frac{3a^2}{4})$$K = \frac{3}{4} (\frac{1}{2} k a^2)$Since $E = \frac{1}{2} k a^2$,we get $K = \frac{3E}{4}$.

The total energy of a body executing $S.H.M.$ is $E$. Then the kinetic energy when the displacement is half of the amplitude,is

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