A particle of mass m performs SHM along a str | vedclass

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(D) For a particle in $SHM$,the displacement is $x = A \sin(\omega t)$,where $\omega = 2\pi f$.The velocity is $v = \frac{dx}{dt} = A\omega \cos(\omeg

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$(B)$ and $(C)$ both

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(D) For a particle in $SHM$,the displacement is $x = A \sin(\omega t)$,where $\omega = 2\pi f$.The velocity is $v = \frac{dx}{dt} = A\omega \cos(\omega t)$.The kinetic energy is $K = \frac{1}{2}mv^2 = \frac{1}{2}m A^2 \omega^2 \cos^2(\omega t) = \frac{1}{2}m A^2 (4\pi^2 f^2) \cos^2(\omega t) = 2m\pi^2 f^2 A^2 \cos^2(\omega t)$.Using the identity $\cos^2(	heta) = \frac{1 + \cos(2	heta)}{2}$,we get $K = m\pi^2 f^2 A^2 (1 + \cos(2\omega t))$.The average value of $\cos(2\omega t)$ over a cycle is $0$,so the average kinetic energy is $\langle K angle = m\pi^2 f^2 A^2$.Similarly,the average potential energy $\langle U angle$ is also $m\pi^2 f^2 A^2$.The kinetic energy expression contains the term $\cos(2\omega t)$,which indicates that the frequency of oscillation of kinetic energy is $2\omega / 2\pi = 2f$.Thus,both statements $(B)$ and $(C)$ are correct.

Column-$I$	Column-$II$
$(a)$ $K = 10\,J$	$(i)$ $U = 40\,J$
$(b)$ $K = 60\,J$	$(ii)$ $U = 90\,J$
	$(iii)$ $U = 50\,J$

$A$ particle of mass $m$ performs $SHM$ along a straight line with frequency $f$ and amplitude $A.$

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