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)

Let the displacement of a particle executing simple harmonic motion at any instant $t$ is, $x=Acos\omega t$
Velocity, $v=\frac{dx}{dt}=\frac{d}{dt}\left(Acos\omega t\right)=-A\omega sin\omega t$
Acceleration, $a=\frac{dv}{dt}=-A{\omega }^{2}cos\omega t$
$=A{\omega }^{2}cos\left(\omega t+\pi \right)$
Phase of displacement, ${\varphi }_1=\omega t$
Phase of acceleration, ${\varphi }_2=\omega t+\pi$
$\therefore$ Phase difference $={\varphi }_2-{\varphi }_1$
$=\left(\omega t+\pi \right)-\omega t=\pi$