Question

The shortest distance travelled by a particle executing S.H.M. from mean position in $2s$ is equal to $\left(\frac{\sqrt{3}}{2}\right)$ times its amplitude. Determine its time period.

Answer 1

Let equation of SHM be:

$x=A\sin \left(\omega t\right)$

From the given data,

$\frac{\sqrt{3}A}{2}=A\sin \left(2\omega \right)$

$2\omega =\pi /3$

$2\times 2\pi /T=\pi /3$

$T=12s$