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(C) For a particle executing $SHM$,the displacement $y$ is given by $y = A \sin(\omega t + \phi)$.The acceleration $a$ is the second derivative of dis

Vedclass · Accepted Answer

$180^{\circ}$

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(C) For a particle executing $SHM$,the displacement $y$ is given by $y = A \sin(\omega t + \phi)$.The acceleration $a$ is the second derivative of displacement with respect to time: $a = \frac{d^2y}{dt^2} = -\omega^2 A \sin(\omega t + \phi)$.We can rewrite this as $a = \omega^2 A \sin(\omega t + \phi + \pi)$.Comparing the arguments of the sine functions,the phase of displacement is $(\omega t + \phi)$ and the phase of acceleration is $(\omega t + \phi + \pi)$.The phase difference is $(\omega t + \phi + \pi) - (\omega t + \phi) = \pi 	ext{ radians}$ or $180^{\circ}$.

What is the phase difference between the displacement and acceleration of a particle executing $SHM$?

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