Question

A particle executing SHM. The phase difference between velocity and displacement is

• A. 0
• B. $\frac{\pi}{2}$
• C. $\pi$
• D. 2$\pi$

Answer 1

Answer: (B)

$\frac{\pi}{2}$

Answer 2

Le the displacement of a particle executing SHM at any instant t is

$x = A\cos\omega t$

Velocity, $v = - A\omega\sin\omega t = A\omega\cos\left( \omega t + \frac{\pi}{2} \right)$

Phase of displacement, $\varphi_{1} = \omega t$

Phase of velocity $\varphi_{2} = \left( \omega t + \frac{\pi}{2} \right)$

$\therefore$ Phase difference $= \varphi_{2} - \varphi_{1} = \omega t + \frac{\pi}{2} - \omega t = \frac{\pi}{2}$