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 the particle executing SHM starts oscillating from its mean position. Then displacement equation is
`x=A sin omegat ` ltbRgt `:. ` Particle velocity, `v=A omega cos omega t`
`:.` Istantaneous K.E. `K=1/2 mv^(2)=1/2 mA^(2) omega^(2) cos^(2) omega t`
`:.` Average value of K.E. over one complete cylce
`K_(av)=1/T int_(0)^(T) 1/2 mA^(2)omega^(2)cos^(2)omegat dt=(mA^(2)omega^(2))/(2T) int_(0)^(T) cos^(2) omega t dt `
`=(mA^(2) omega^(2))/(2T) int_(0)^(T) ((1+cos 2 omegat))/2 dt`
`=(mA^(2)omega^(2))/(4T) [t+(sin 2 omegat)/(2 omega)]_(0)^(T)` ltBrgt `=(mA^(2)omega^(2))/(4T) [(T-0)+((sin 2 omegat- sin 0)/(2omega))]`
`=1/4 mA^(2) omega^(2)`
Again instanteous P.E. =`1/2 kx^(2)=1/2 momega^(2) x^(2)=1/2 momega^(2)A^(2)sin^(2)omegat`
`:.` average value P.E. over one complete cycle
`U_(av)=1/T in_(0)^(T) 1/2 momega^(2) A^(2)sin^(2) omega t=(momega^(2)A^(2))/(2T)int_(0)^(T) sin^(2)omega t dt`
`=(momega^(2)A^(2))/(2T) int_(0)^(T) ((1-2 cos omegat))/2 dt`
`=(m omega^(2)A^(2))/(4T)[t-(sin2 omegat)/(2omega)]_(0)^(T)`
`=(momega^(2)A^(2))/(4T)[(T-0)-((sin2 omegat-sin0))/(2 omega)]`
`=1/4 momega^(2)A^(2)......(ii)`
Simple comparison (i) and (ii) , shows that
`K_(av)=U_(av)=1/4momega^(2)A^(2)`