The variation of kinetic energy (KE) of a particle executing simple...

The formula to calculate the kinetic energy of a particle executing SHM is given by

$K = \dfrac{1}{2}m{\omega}^{2}\left({A}^{2} - {x}^{2}\right) . . . \left(1\right)$

Substitute $0$ for $x$ into equation (1) to obtain the kinetic energy at the mean position.

$\begin{cases}K & = & \dfrac{1}{2}m{\omega}^{2}\left({A}^{2} - {0}^{2}\right) \\ = & \dfrac{1}{2}m{\omega}^{2}{A}^{2}\end{cases}$

Substitute $A$for $x$ into equation (1) to obtain the kinetic energy at the end points.

$\begin{cases}K & = & \dfrac{1}{2}m{\omega}^{2}\left({A}^{2} - {A}^{2}\right) \\ & = & 0\end{cases}$

Hence, the variation can be shown as follows-