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(C) Let the displacement of a particle executing simple harmonic motion be $y = A \sin(\omega t)$.The instantaneous velocity $v$ is given by the deriv

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$0.5\pi$

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(C) Let the displacement of a particle executing simple harmonic motion be $y = A \sin(\omega t)$.The instantaneous velocity $v$ is given by the derivative of displacement with respect to time:$v = \frac{dy}{dt} = A\omega \cos(\omega t) = A\omega \sin(\omega t + \frac{\pi}{2})$.The instantaneous acceleration $a$ is given by the derivative of velocity with respect to time:$a = \frac{dv}{dt} = -A\omega^2 \sin(\omega t) = A\omega^2 \sin(\omega t + \pi)$.The phase of velocity is $(\omega t + \frac{\pi}{2})$ and the phase of acceleration is $(\omega t + \pi)$.The phase difference between acceleration and velocity is $(\omega t + \pi) - (\omega t + \frac{\pi}{2}) = \frac{\pi}{2}$ or $0.5\pi$.

The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is

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