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(D) The kinetic energy $(K.E.)$ of a particle performing $SHM$ is given by: $K.E. = \frac{1}{2} m \omega^2 (A^2 - x^2)$ Given $x = \frac{A}{2}$,substi

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$\frac{3 m \pi^2 A^2}{2 T^2}$

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(D) The kinetic energy $(K.E.)$ of a particle performing $SHM$ is given by: $K.E. = \frac{1}{2} m \omega^2 (A^2 - x^2)$ Given $x = \frac{A}{2}$,substitute this into the equation: $K.E. = \frac{1}{2} m \omega^2 (A^2 - (\frac{A}{2})^2)$ $K.E. = \frac{1}{2} m \omega^2 (A^2 - \frac{A^2}{4}) = \frac{1}{2} m \omega^2 (\frac{3 A^2}{4}) = \frac{3}{8} m \omega^2 A^2$ Since the angular frequency $\omega = \frac{2 \pi}{T}$,substitute this value: $K.E. = \frac{3}{8} m (\frac{2 \pi}{T})^2 A^2$ $K.E. = \frac{3}{8} m (\frac{4 \pi^2}{T^2}) A^2$ $K.E. = \frac{3 m \pi^2 A^2}{2 T^2}$

$A$ bob of a simple pendulum of mass $m$ performs $SHM$ with amplitude $A$ and period $T$. The kinetic energy of the pendulum at displacement $x = \frac{A}{2}$ will be:

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