Physics - SHM Question with Solution | TestHub

Kinetic energy $K=\frac{1}{2}m{\omega }^2\left(A^2-y^2\right)$ At mean position $y=0$ $$K=\frac{1}{2}m{\omega }^2\left(A^2\right)$$ Potential energy $U=\frac{1}{2}m{\omega }^2{y}^2$ $U$ at $y=\frac{A}{2}$ $$U=\frac{1}{2}\frac{m{A}^2}{4}{\omega }^2$$ Dividing Eq (i) by Eq (ii), we get $$\frac{\mathrm{KE}}{U}=\frac{4}{1}$$