Question

The velocity and acceleration of a particle performing simple harmonic motion have a steady phase relationship. The acceleration shows a phase lead over the velocity in radians of

• A. $+\pi$
• B. 0
• C. $+\pi /2$
• D. $-\pi /2$
• E. $-\pi$

Answer 1

Answer: (C)

Let displacement equation of a particle executing $SHM$ is given as
$x=a\sin \omega t$
then velocity $(v)=\frac{dx}{dt}=a\omega \cos \omega t..(i)$
and acceleration $(a)=\frac{d^{2}x}{d{t}^2}=\frac{dv}{dt}=-a{\omega }^2\sin \omega t$
$a=a{\omega }^2\cos \left(\omega t+\frac{\pi }{2}\right)...(ii)$
$\left(\because \cos \left(\theta +\frac{\pi }{2}\right)=-\sin \theta \right)$
Now, comparing Eqs (i) and (ii), we can conclude that the acceleration lead the velocity by a phase of $+\frac{\pi }{2}$ radians.