Question

A particle executes S.H.M. with a period $8s.$ Find the time in which half the total energy is potential.

Answer 1

K.E = P.E

$\frac{1}{2}k{x}^2=\frac{1}{2}m{v}^2$

$v=\omega \sqrt{A^2-x^2}$

$\frac{1}{2}k{x}^2=\frac{1}{2}m\left(\omega \sqrt{A^2-x^2}\right)$

$\Rightarrow x=\frac{A}{\sqrt{2}}$

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

$x=\frac{A}{\sqrt{2}}$

$\frac{A}{\sqrt{2}}=A\sin \omega t$

$\omega t=\left(\frac{\pi }{4}\right)$

$r=\left(\frac{\pi }{4\omega }\right)=\frac{\pi \times T}{42\pi }=\frac{8}{8}=1\sec$