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.

(N/A) The displacement of a particle executing $SHM$ at an instant $t$ is given by $x = A \sin \omega t$,where $A$ is the amplitude and $\omega$ is the angular frequency.
The velocity of the particle is $v = \frac{dx}{dt} = A \omega \cos \omega t$.
The kinetic energy is $E_k = \frac{1}{2} M v^2 = \frac{1}{2} M A^2 \omega^2 \cos^2 \omega t$.
The potential energy is $E_p = \frac{1}{2} k x^2 = \frac{1}{2} M \omega^2 A^2 \sin^2 \omega t$.
The average kinetic energy over a period $T$ is $\langle E_k \rangle = \frac{1}{T} \int_0^T E_k dt = \frac{1}{T} \int_0^T \frac{1}{2} M A^2 \omega^2 \cos^2 \omega t dt$.
Using $\cos^2 \omega t = \frac{1 + \cos 2 \omega t}{2}$,we get $\langle E_k \rangle = \frac{M A^2 \omega^2}{2T} \int_0^T \frac{1 + \cos 2 \omega t}{2} dt = \frac{M A^2 \omega^2}{4T} [t + \frac{\sin 2 \omega t}{2 \omega}]_0^T = \frac{1}{4} M A^2 \omega^2 \dots (i)$.
The average potential energy over a period $T$ is $\langle E_p \rangle = \frac{1}{T} \int_0^T E_p dt = \frac{1}{T} \int_0^T \frac{1}{2} M \omega^2 A^2 \sin^2 \omega t dt$.
Using $\sin^2 \omega t = \frac{1 - \cos 2 \omega t}{2}$,we get $\langle E_p \rangle = \frac{M \omega^2 A^2}{2T} \int_0^T \frac{1 - \cos 2 \omega t}{2} dt = \frac{M \omega^2 A^2}{4T} [t - \frac{\sin 2 \omega t}{2 \omega}]_0^T = \frac{1}{4} M A^2 \omega^2 \dots (ii)$.
Comparing $(i)$ and $(ii)$,we see that $\langle E_k \rangle = \langle E_p \rangle$.