Where is maximum acceleration and zero veloci | vedclass

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(B) For a particle executing $SHM$,the displacement is given by $x = A \sin(\omega t + \phi)$.The velocity is $v = \frac{dx}{dt} = A\omega \cos(\omega

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At the extreme positions

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(B) For a particle executing $SHM$,the displacement is given by $x = A \sin(\omega t + \phi)$.The velocity is $v = \frac{dx}{dt} = A\omega \cos(\omega t + \phi)$.The acceleration is $a = \frac{dv}{dt} = -A\omega^2 \sin(\omega t + \phi) = -\omega^2 x$.At the extreme positions,the displacement $x = \pm A$.Therefore,the velocity $v = \pm A\omega \cos(\pm \pi/2) = 0$.The acceleration $a = -\omega^2(\pm A) = \mp A\omega^2$,which is the maximum magnitude of acceleration.Thus,both conditions are satisfied at the extreme positions.

Where is maximum acceleration and zero velocity of a particle executing $SHM$?

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