Physics - SHM Question with Solution | TestHub

Total energy is SHM $E=\frac{1}{2}m{\omega }^2{a}^2$ (where $a=$ amplitude) The potential energy at displacement $y$ is $U=\frac{1}{2}m{\omega }^2{y}^2$ The kinetic energy is $K=E-U=\frac{1}{2}m{\omega }^2\left(a^2-y^2\right)=E-\frac{1}{2}m{\omega }^2{y}^2$ Now at $y=\frac{a}{2}$ $K=E-\frac{1}{2}m{\omega }^2\left(\frac{a^2}{4}\right)=E-\frac{E}{4}=\frac{3E}{4}$