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.

Explanation

Consider a particle of mass m executing S.H.M. with period T . The displacement of the particle at an instant t, when time period is noted from the mean position is given by y`=A sin omega t`
`therefore ` Velocity , `V= (dy)/(dt) = A omega cos omega t`,
Kinetic energy `E_K = 1/2 mV^2 = 1/2 m A^2 omega^2 cos^2 omega t`
Potential energy `E_P =1/2 ky^2 = 1/2 m omega^2 A^2 sin^2 omega t`
`(because K = m omega^2) therefore ` Average K.E. over one cycle.
`E_(k_(av)) = 1/T int_0^T E_k dt = 1/T int_0^2 1/2 mA^2 omega^2 cos^2 omega t dt`
`=1/(2T) m omega^2 A^2 int_0^T ((1+ cos 2 omegat))/2 dt = 1/(4T) m A^2 omega^2 [k+(sin 2omega t)/(2omega)]_0^T = 1/(4T) mA^2 omega^2 (T) = 1/4 mA^2 omega^2`
Average P.E. over one cycle is `E_(k_(av)) = 1/T int_0^T E_k dt = 1/T * 1/T int_0^T 1/2 mA^2 omega^2 sin^2 omega t dt`
`=1/(2T) m omega^2 A int_0^T ((1-cos 2omegat))/2 = 1/(4T) m omega^2 A^2 [1-(sin2 omega t)/(2omega)]_0^T`
`=1/(4T) m omega^2 A^2 [T] =1/4 A^2 omega^2`
From (i) and (ii) , `E_(K_(aÅ))= E_(P_(aÅ))`