A block is executing SHM. Let the time period | vedclass

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(B) Let the displacement of the block in $SHM$ be $x = A \sin(\omega t).$The velocity of the block is $v = \frac{dx}{dt} = A\omega \cos(\omega t).$The

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$T_1=2T_2$

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(B) Let the displacement of the block in $SHM$ be $x = A \sin(\omega t).$The velocity of the block is $v = \frac{dx}{dt} = A\omega \cos(\omega t).$The time period of velocity is the time taken for the velocity to repeat its value, which is $T_1 = \frac{2\pi}{\omega} = T.$The kinetic energy of the block is $K = \frac{1}{2}mv^2 = \frac{1}{2}m(A\omega \cos(\omega t))^2 = \frac{1}{2}mA^2\omega^2 \cos^2(\omega t).$Using the identity $\cos^2(	heta) = \frac{1 + \cos(2	heta)}{2},$ we get $K = \frac{1}{4}mA^2\omega^2 (1 + \cos(2\omega t)).$The kinetic energy varies with a frequency of $2\omega.$ Therefore, the time period of variation of kinetic energy is $T_2 = \frac{2\pi}{2\omega} = \frac{\pi}{\omega} = \frac{T}{2}.$Thus, $T_1 = 2T_2.$

$A$ block is executing $SHM.$ Let the time period of variation of velocity be $T_1$ and time period of variation of kinetic energy be $T_2.$ The relation between $T_1$ and $T_2$ is

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