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(B) The instantaneous kinetic energy $K$ of a particle executing $S.H.M.$ is given by $K = \frac{1}{2} m v_{inst}^2 = \frac{1}{2} m \omega^2 a^2 \cos^

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$\pi^2 m a^2 v^2$

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(B) The instantaneous kinetic energy $K$ of a particle executing $S.H.M.$ is given by $K = \frac{1}{2} m v_{inst}^2 = \frac{1}{2} m \omega^2 a^2 \cos^2(\omega t + \phi)$.At the equilibrium position,the displacement is $x = 0$,so $x = a \sin(\omega t)$,which implies $\omega t = 0$. At the extreme position,$x = a$,so $\omega t = \pi/2$.The average kinetic energy over an interval is defined as $\langle K angle = \frac{1}{T'} \int_0^{T'} K dt$,where $T'$ is the time taken to move from equilibrium to the extreme position $(T' = T/4 = 1/(4v))$.$\langle K angle = \frac{1}{T/4} \int_0^{T/4} \frac{1}{2} m \omega^2 a^2 \cos^2(\omega t) dt$Using $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$:$\langle K angle = \frac{4}{T} \cdot \frac{1}{2} m \omega^2 a^2 \int_0^{T/4} \frac{1 + \cos(2\omega t)}{2} dt$$\langle K angle = \frac{m \omega^2 a^2}{T} [t + \frac{\sin(2\omega t)}{2\omega}]_0^{T/4}$Since $\omega = 2\pi v$ and $T = 1/v$,at $t = T/4$,$2\omega t = 2(2\pi v)(1/4v) = \pi$.$\langle K angle = \frac{m \omega^2 a^2}{T} [T/4 + 0 - 0] = \frac{m \omega^2 a^2}{4} = \frac{m (2\pi v)^2 a^2}{4} = \pi^2 m a^2 v^2$.

$A$ particle of mass $m$ executes simple harmonic motion with amplitude $a$ and frequency $v$. The average kinetic energy during its motion from the position of equilibrium to the end is

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