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(D) In an undamped simple harmonic motion,the total energy $E = PE + KE$ remains constant throughout the motion.At the mean position $(x = 0)$,the pot

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$(A)$ and $(C)$ both.

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(D) In an undamped simple harmonic motion,the total energy $E = PE + KE$ remains constant throughout the motion.At the mean position $(x = 0)$,the potential energy $PE = 0$,so the total energy $E$ is equal to the maximum kinetic energy $(KE_{\max})$.Since the total energy is constant,the average total energy per cycle is equal to the total energy at any point,which is $KE_{\max}$. Thus,statement $(A)$ is correct.For velocity,if $x = A \sin(\omega t)$,then $v = \frac{dx}{dt} = A\omega \cos(\omega t)$.The maximum velocity is $v_{\max} = A\omega$.The root mean square $(RMS)$ velocity is given by $v_{	ext{rms}} = \sqrt{\langle v^2 angle} = \sqrt{\langle A^2\omega^2 \cos^2(\omega t) angle} = A\omega \sqrt{\langle \cos^2(\omega t) angle}$.Since the average value of $\cos^2(	heta)$ over a cycle is $\frac{1}{2}$,we get $v_{	ext{rms}} = A\omega \sqrt{\frac{1}{2}} = \frac{v_{\max}}{\sqrt{2}}$. Thus,statement $(C)$ is also correct.Therefore,both $(A)$ and $(C)$ are correct.

List-$I$	List-$II$
$A$. Velocity is maximum	$I$. Acceleration is maximum
$B$. $K.E.$ is $\left(\frac{3}{4}\right)^{\text{th}}$ of total energy	$II$. At mean position
$C$. $P.E.$ is $\left(\frac{3}{4}\right)^{\text{th}}$ of total energy	$III$. At half of the amplitude
$D$. Acceleration is maximum	$IV$. At $\frac{\sqrt{3}}{2}$ times the amplitude

$A$ system is oscillating with undamped simple harmonic motion. Then the

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