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(C) The given equation for position in $SHM$ is $x(t) = 2 \cos \left(\frac{\pi}{15} t - \frac{\pi}{2}\right)$.Comparing this with the standard equatio

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$15$

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(C) The given equation for position in $SHM$ is $x(t) = 2 \cos \left(\frac{\pi}{15} t - \frac{\pi}{2}ight)$.Comparing this with the standard equation $x(t) = A \cos(\omega t + \phi)$, we get the angular frequency $\omega = \frac{\pi}{15} 	ext{ rad/s}$.The time period of the $SHM$ is given by $T = \frac{2\pi}{\omega} = \frac{2\pi}{\pi/15} = 30 	ext{ s}$.The kinetic energy of a particle in $SHM$ is given by $KE = \frac{1}{2} m v^2 = \frac{1}{2} m \omega^2 A^2 \sin^2(\omega t + \phi)$.Using the identity $\sin^2 	heta = \frac{1 - \cos(2	heta)}{2}$, the kinetic energy varies with a frequency of $2\omega$.Therefore, the time period of the kinetic energy is $T_{KE} = \frac{T}{2} = \frac{30}{2} = 15 	ext{ s}$.

The position of a particle executing simple harmonic motion is given by $x(t) = 2 \cos \left(\frac{\pi}{15} t - \frac{\pi}{2}\right)$, where $x$ is in centimetre and $t$ is in seconds. The time period of the kinetic energy of the particle in seconds is

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