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

The correct option is C $\pi$

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, $a=\frac{dv}{dt}.=-A{\omega }^{2}cos\omega t=A{\omega }^{2}cos(\omega t+\pi )$
Phase of displacement, ${\varphi }_1=\omega t$
Phase of acceleration, ${\varphi }_2=\omega t+\pi$
$\therefore$ Phase difference = ${\varphi }_2-{\varphi }_1=(\omega t+\pi )-\omega t=\pi$