Starting from mean position, a body oscillates simple harmonically with a period 'T'. After what time will its kinetic energy be $75 \%$ of the total energy? $\left(\sin 30^{\circ}=0.5\right)$

Question

Aicharya · Accepted Answer

$\begin{array}{ll} & \mathrm{KE}=75 \% 	ext { of TE } \ 	herefore & \frac{1}{2} \mathrm{ma}^2 \omega^2 \cos ^2 \omega \mathrm{t}=\frac{75}{100} 	imes \frac{1}{2} \mathrm{ma}^2 \omega^2 \ 	herefore & \cos ^2 \omega \mathrm{t}=\frac{3}{4} \ 	herefore \quad & \cos \omega \mathrm{t}=\frac{\sqrt{3}}{2} \ 	herefore \quad & \omega \mathrm{t}=\frac{\pi}{6} \ 	herefore \quad & \frac{2 \pi}{\mathrm{~T}} \mathrm{t}=\frac{\pi}{6} \quad 	herefore \mathrm{t}=\frac{\mathrm{T}}{12}\end{array}$

Starting from mean position, a body oscillates simple harmonically with a period 'T'. After what time will…

Solution