Show that for a particle in linear S.H.M., the average kinetic energy over a period of oscillation equals the average potential energy over the same period.

Explanation

Let us consider a particle of mass m executing simple harmonic motion. The displacement of particle at any time t is given by
`y = a sin omegat`
`v = "velocity" = (dy)/(dt) = a omegacos omegat`
K.E. `= (1)/(2)mv^(2) = (1)/(2)ma^(2)omega^(2)cos^(2)omegat`
Average K.E. over one complete cycle
`K_(av) = (1)/(T)int_(0)^(T)E_(k)dt`
` = (1)/(T) int_(0)^(T)momega^(2)a^(2)cos^(2)omegatdt`
`=(1)/(2T) ma^(2)omega^(2)cos^(2)omegatdt`
` = (ma^(2)omega^(2))/(2T)f((1 + cos 2 omegat)/(2))dt`
` = (ma^(2)omega^(2))/(4T) [t + (sin 2 omegat)/(2omega)]_(0)^(T)`
` = (ma^(2)omega^(2))/(4T) T = (ma^(2)omega^(2))/(4)`
Average potential energy over one complete cycle.
`E = (1)/(T) int_(0)^(T)E_(p)dt = (1)/(T) int_(0)^(T)(1)/(2)ky^(2)dt`
` = (1)/(T)int_(0)^()(1)/(2)momega^(2)a^(2)sin^(2)omegatdt`
` = (1)/(2T) momega^(2)a^(2)int_(0)^()sin^(2)omegatdt`
` = (momega^(2)a^(2))/(2T)int_(0)^(T)(1 + cos^(2)omegat)/(2)dt`
` (momega^(2)a^(2))/(4T)(|T-sin2omegat|^(T))/(|2omega|_(0))`
` = (momega^(2)a^(2))/(4T)(T) = (1)/(4)momega^(2)a^(2)`
`therefore` From (i) and (ii)
We get Average K.E. = Average P.E.