### Q.22:- 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:-**The equation of displacement of a particle executing SHM at an instant

*t*is given as:x =

*A*sin ω*t*where,

*A*= Amplitude of oscillation

ω = Angular frequency = √

*k*/*M*The velocity of the particle is:*v* = *dx*/*dt *= Aωcosω*t*

The kinetic energy of the particle is:

E_{k} = 1/2 *Mv*^{2} = 1/2 *MA*^{2} ω^{2} cos^{2} ω*t*

The portential energy of the particle is:

E_{p} = 1/2 *kx*^{2} = 1/2 *M*^{2} ω^{2} A^{2} sin^{2} ω*t*

*
*For time period

*T*, the average kinetic energy over a single cycle is given as:

And, average potential energy over one cycle is given as:

It can be inferred from equations **(****i****)** and **(****ii****)** that the average kinetic energy for a given time period is equal to the average potential energy for the same time period.