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

Let the particle executing SHM starts oscillating with initial phase `phi =0`, its displacement at time t,
`x= A sin omega t`
Particle velocity `v= A omega cos omega t`
Instantaneous kinetic energy, `K = (1)/(2) mv^(2)`
`= (1)/(2)mA^(2) omega^(2) cos^(2) omega t`
Average value of kinetic energy over one complete cycle,
`lt K gt = (1)/(T) int_(0)^(T) (1)/(2) mA^(2) omega^(2) cos^(2) omega t dt`
`=(mA^(2)omega^(2))/(2T) int_(0)^(T) ((1+cos 2omega t)/(2))dt`
`=(mA^(2)omega^(2))/(4T) int_(0)^(T) (1+cos 2omega t)dt`
`=(mA^(2)omega^(2))/(4T) [t+ (sin 2omega t)/(2omega)]_(0)^(T)`
`=(mA^(2)omega^(2))/(4T) [(T-0) +(sin 2omega T- sin 0)/(2omega)]`
`=(mA^(2)omega^(2))/(4T) [T +(sin 4pi - sin 0)/(2omega)]`
`=(mA^(2)omega^(2))/(4T) T+0" "[therefore sin 4pi = sin0 =0]`
`lt K gt = (1)/(4) mA^(2)omega^(2)" ""........"(1)`
Now instantaneous potential energy, `U= (1)/(2) kx^(2)`
`=(1)/(2)m omega^(2) (A^(2) sin^(2) omega t)`
`=(1)/(2) m omega^(2) A^(2) sin^(2) omega t`
Average value of potential energy over one complete cycle,
`lt U gt = (1)/(T) int_(0)^(T) (1)/(2) m omega^(2) A^(2) sin^(2) omega t dt`
`=(m omega^(2)A^(2))/(2T) int_(0)^(T) sin^(2) omega t dt`
`=(mA^(2)omega^(2))/(2T) int_(0)^(T) ((1-cos 2omega t))/(2)dt`
`=(mA^(2)omega^(2))/(4T) [int_(0)^(T) (1- cos 2omega t)dt ]`
`=(mA^(2)omega^(2))/(4T) [t- (sin 2omega t)/(2omega)]_(0)^(T)`
`=(mA^(2)omega^(2))/(4T) [(T-0) -(sin 2omega T- sin 0)/(2omega)]`
`=(mA^(2)omega^(2))/(4T) [T -(sin 4pi - sin 0)/(2omega)]`
`=(mA^(2)omega^(2))/(4) " "[therefore sin 4pi = 0, sin0 =0]`
`=(mA^(2)omega^(2)T)/(T)+0`
`lt U gt = (1)/(4) mA^(2)omega^(2)" ""........"(2)`
From equation (1) and (2),
`lt K gt = lt U gt = (1)/(4)m A^(2) omega^(2)`.