Question

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

• A. 0
• B. $\frac{\pi}{2}$
• C. $\pi$
• D. $\frac{3}{2}\pi$

Answer 1

Answer: (C)

$\pi$

Answer 2

Let the displacement of a particle executing simple harmonic motion at any instant t is

$x = A\cos\omega t$

Velocity $v = \frac{dx}{dt} = \frac{d}{dt}(A\cos\omega t) = - A\omega\sin\omega t$

Acceleration, $a = \frac{dv}{dt} = - A\omega^{2}\cos\omega t = A\omega^{2}\cos(\omega t + \pi)$

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

Phase of acceleration $\varphi_{2} = \omega t + \pi$

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