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. $- \pi/2$

Answer 1

Answer: (C)

$+ \pi /2$

Answer 2

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