(C) The kinetic energy $K$ of a particle performing $S.H.M.$ is given by the formula: $K = \frac{1}{2} m \omega^2 (a^2 - y^2)$.Given: mass $m = 10 \,

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(C) The kinetic energy $K$ of a particle performing $S.H.M.$ is given by the formula: $K = \frac{1}{2} m \omega^2 (a^2 - y^2)$.Given: mass $m = 10 \, g$,period $T = 2 \, s$,amplitude $a = 10 \, cm$,and displacement $y = 5 \, cm$.The angular frequency $\omega = \frac{2\pi}{T} = \frac{2\pi}{2} = \pi \, rad/s$.Substituting the values into the formula:$K = \frac{1}{2} \times 10 \times (\pi)^2 \times (10^2 - 5^2)$$K = 5 \times \pi^2 \times (100 - 25)$$K = 5 \times \pi^2 \times 75$$K = 375 \pi^2 \, ergs$.

Class 11 Physics Oscillations Energy of Simple Harmonic Motion

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$A$ particle of mass $10 \, g$ is describing $S.H.M.$ along a straight line with a period of $2 \, s$ and an amplitude of $10 \, cm$. Its kinetic energy when it is at $5 \, cm$ from its equilibrium position is

A
$37.5 \pi^2 \, ergs$
B
$3.75 \pi^2 \, ergs$
C
$375 \pi^2 \, ergs$
D
$0.375 \pi^2 \, ergs$

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Energy of Simple Harmonic Motion

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$A$ particle of mass $10 \, g$ is describing $S.H.M.$ along a straight line with a period of $2 \, s$ and an amplitude of $10 \, cm$. Its kinetic energy when it is at $5 \, cm$ from its equilibrium position is

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$A$ block is in simple harmonic motion $(S.H.M)$ at the end of a spring with position given by $x = 5 \cos \left(\omega t + \frac{\pi}{4}\right)$. If the total mechanical energy is $100 \ J$,then the potential energy at time $t = 0$ is: (in $J$)

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