Physics - SHM Question with Solution | TestHub

When the displacement of a particle executing SHM is $y$, then its $K.E.=\frac{1}{2}m{\omega }^2\left(a^2-y^2\right)$ and $P.E.=\frac{1}{2}m{\omega }^2{y}^2$ For $K.E.=P.E$. or $2y^2=a^2$ or, $y=a/\sqrt{2}$. Since total energy remains constant through out the motion, which is $E=K.E.+P.E$. So, when P.E. is maximum then $K.E$. is zero and vice versa.