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(D) For a particle executing $SHM$,the displacement is given by $y = A \sin(\omega t)$.The velocity is $v = \frac{dy}{dt} = A \omega \cos(\omega t)$.T

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(D) For a particle executing $SHM$,the displacement is given by $y = A \sin(\omega t)$.The velocity is $v = \frac{dy}{dt} = A \omega \cos(\omega t)$.The kinetic energy $(KE)$ is given by:$KE = \frac{1}{2} m v^2 = \frac{1}{2} m (A \omega \cos(\omega t))^2 = \frac{1}{2} m \omega^2 A^2 \cos^2(\omega t)$.Using the identity $\cos^2(	heta) = \frac{1 + \cos(2	heta)}{2}$,we get:$KE = \frac{1}{4} m \omega^2 A^2 (1 + \cos(2\omega t))$.At $t = 0$,$KE = \frac{1}{2} m \omega^2 A^2$,which is the maximum value.At $t = \frac{T}{4}$,$\omega t = \frac{\pi}{2}$,so $KE = 0$.At $t = \frac{T}{2}$,$\omega t = \pi$,so $KE = \frac{1}{2} m \omega^2 A^2$ (maximum).The kinetic energy oscillates with a frequency double that of the displacement,and it starts from a maximum value at $t = 0$ and becomes zero at $t = \frac{T}{4}$.

$A$ particle is executing simple harmonic motion with a time period $T$. At time $t = 0$,it is at its position of equilibrium. The kinetic energy-time graph of the particle will look like:

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