Question

Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.

Answer 1

Let $x=Asin\omega t$

Then, $v=dx/dt=A\omega \cos \left(\omega t\right)$

$K.E.=\frac{1}{2}{mv}^2$

Average kinetic energy is $\frac{1}{T}{\int }_0^T\frac{1}{2}m{v}^{2}dt=\frac{1}{4}m{A}^2{\omega }^2$

Average potential energy is $\frac{1}{T}{\int }_0^T\frac{1}{2}k{x}^{2}dt=\frac{1}{4}m{A}^2{\omega }^2$