Physics - SHM Question with Solution | TestHub

The total mechanical energy of a particle in SHM is given by $E=\frac{1}{2}m{\omega }^2{A}^2$, where $A$ is the amplitude. The potential energy at a displacement $x$ is $U=\frac{1}{2}m{\omega }^2{x}^2$. When $x=\frac{A}{2}$, the potential energy is $U=\frac{1}{2}m{\omega }^2{\left(\frac{A}{2}\right)}^2=\frac{1}{4}\left(\frac{1}{2}m{\omega }^2{A}^2\right)=\frac{E}{4}$. The kinetic energy $K$ is $E-U=E-\frac{E}{4}=\frac{3E}{4}$. Therefore, the ratio $\frac{E}{K}=\frac{E}{3E/4}=\frac{4}{3}$.