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

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(B) Let the displacement of a particle executing $SHM$ be $x = A \sin(\omega t + \phi)$.The velocity $v$ is given by $v = \frac{dx}{dt} = A \omega \cos(\omega t + \phi) = A \omega \sin(\omega t + \phi + \frac{\pi}{2})$.The acceleration $a$ is given by $a = \frac{dv}{dt} = -A \omega^2 \sin(\omega t + \phi) = A \omega^2 \sin(\omega t + \phi + \pi)$.The phase of velocity is $(\omega t + \phi + \frac{\pi}{2})$ and the phase of acceleration is $(\omega t + \phi + \pi)$.The phase difference is $(\omega t + \phi + \pi) - (\omega t + \phi + \frac{\pi}{2}) = \frac{\pi}{2}$ radians or $90^{\circ}$.

What is the phase difference between the velocity and acceleration of a particle executing $SHM$ (in $^{\circ}$)?

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