Physics - SHM Question with Solution | TestHub

The displacement of a particle performing SHM starting from the mean position is given by $x=A\sin (\omega t)$.

The kinetic energy is $K=\frac{1}{2}m{\omega }^2(A^2-x^2)$ and the total energy is $E=\frac{1}{2}m{\omega }^2{A}^2$.

Given that $K=75\%\text{ of }E=\frac{3}{4}E$.

$\frac{1}{2}m{\omega }^2(A^2-x^2)=\frac{3}{4}\left(\frac{1}{2}m{\omega }^2{A}^2\right)$

$A^2-x^2=\frac{3}{4}{A}^2$

$x^2=\frac{1}{4}{A}^2\Rightarrow x=\frac{A}{2}$

Substituting $x=\frac{A}{2}$ in the displacement equation:

$\frac{A}{2}=A\sin (\omega t)$

$\sin (\omega t)=\frac{1}{2}=\sin \left(\frac{\pi }{6}\right)$

$\omega t=\frac{\pi }{6}$

$\frac{2\pi }{T}t=\frac{\pi }{6}$

$t=\frac{T}{12}$

Answer: $\frac{\mathrm{T}}{12}$