Question

A particle is executing S.H.M. After 2 seconds of crossing the equilibrium position it is at a distance of $\frac{\sqrt{3}}{2}$ of its amplitude. Find its time period?

• A. 2 s
• B. 3 s
• C. 4 s
• D. 12 s

Answer 1

The correct option is D 12 s

$y=A\sin \omega t$ ; $\frac{\sqrt{3}}{2}.A=A\sin \omega t$
$\sin \omega t=\frac{\sqrt{3}}{2}or\omega t=\frac{\pi }{3};$ $\omega =\frac{\pi }{3t}=\frac{\pi }{6}$
$T=\frac{2\pi }{\omega }=\frac{2\pi }{\left(\frac{\pi }{6}\right)}=12;$ T = 12 seconds.