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The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is
A
$\pi$
B
$0.707\pi$
C
zero
D
$0.5\pi$
Detailed Solution
Let the displacement be $y = A\sin\omega t$
Instantaneous velocity: $v = \dfrac{dy}{dt} = A\omega\cos\omega t = A\omega\sin\left(\omega t + \dfrac{\pi}{2}\right)$
Acceleration: $a = \dfrac{dv}{dt} = -A\omega^2\sin\omega t = A\omega^2\sin(\omega t + \pi)$
Phase of the velocity $= \omega t + \dfrac{\pi}{2}$ and phase of the acceleration $= \omega t + \pi$
Phase difference: $\phi = (\omega t + \pi) - \left(\omega t + \dfrac{\pi}{2}\right)$
$\phi = \dfrac{\pi}{2} = 0.5\pi$
Instantaneous velocity: $v = \dfrac{dy}{dt} = A\omega\cos\omega t = A\omega\sin\left(\omega t + \dfrac{\pi}{2}\right)$
Acceleration: $a = \dfrac{dv}{dt} = -A\omega^2\sin\omega t = A\omega^2\sin(\omega t + \pi)$
Phase of the velocity $= \omega t + \dfrac{\pi}{2}$ and phase of the acceleration $= \omega t + \pi$
Phase difference: $\phi = (\omega t + \pi) - \left(\omega t + \dfrac{\pi}{2}\right)$
$\phi = \dfrac{\pi}{2} = 0.5\pi$