The acceleration of a particle in S.H.M. is | vedclass

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(C) The acceleration $a$ of a particle executing $S.H.M.$ is given by the formula $a = -{\omega ^2}x$,where $\omega$ is the angular frequency and $x$

Vedclass · Accepted Answer

Maximum at the extreme position

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(C) The acceleration $a$ of a particle executing $S.H.M.$ is given by the formula $a = -{\omega ^2}x$,where $\omega$ is the angular frequency and $x$ is the displacement from the equilibrium position.At the equilibrium position,$x = 0$,so the acceleration is zero.At the extreme positions,the displacement $x$ is equal to the amplitude $A$ (i.e.,$x = \pm A$).Therefore,the magnitude of acceleration is $|a| = {\omega ^2}A$,which is the maximum value.Thus,the acceleration is maximum at the extreme position.

The acceleration of a particle in $S.H.M.$ is

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