Show that for a particle executing simple harmonic motion average value of kinetic energy is equal to the average value of potential energy.

Explanation

Suppose a particle of mass m exectues SHM of time period T. The displacement of the particle at any instant t is given by
`y=A sin omega t" "....(1)`
Velocity `v=(dy)/(dt)=(d)/(dt)(A sin omega t)`
`=A omega cos omega t= omega A cos omega t`
`:. {:("Kinetic"),("energy"):}}E_(k)=(1)/(2)mv^(2)`
`=(1)/(2) m omega^(2)A^(2) cos^(2) omega t" "....(2)`
`:. {:("Potential"),("energy"):}}E_(P)=(1)/(2)m^(2) v^(2)`
`=(1)/(2) m omega^(2)A^(2) sin^(2) omega t" "....(3)`
`:.` Average K.E for one period of oscillation,
`E_(k_(av))=(1)/(T_(0)) E_(k)dt`
`=(1)/(T) overset(T) underset(0)int (1)/(2) m omega^(2)A^(2)cos ^(2) omega t dt`
`=(1)/(2T)m omega^(2)A^(2) overset(T) underset(0)int ((1-cos2 omega t))/(2)dt`
`K_(k_(av))=(1)/(4T)m omega^(2)A^(2)[t+(sinj 2 omega t)/(2 Omega)]_(0)^(T)`
`K_(k_(av))=(1)/(4T)m omega^(2)A^(2)[T+0]`
`k_(k_(av))=(1)/(4T)m omega^(2)A^(2)(T)`
`E_(K_(av))=(1)/(4)m omega^(2)A^(2) " "...(4)`
Average potential energy over a period of oscillation is
`E_(p_(av))=(1)/(T)overset(T) underset(0)int E_(rho)dt`
`=(1)/(T)overset(T) underset(0)int m omega^(2)A^(2)sin^(2) omega t dt`
`=(1)/(2)m omega^(2)A^(2)A^(2)overset(T) underset(0)int ((1- cos 2 omega t))/(2)dt`
`E_(p_(av))=(1)/(4) m omega^(2)A^(2)[-t-(sin 2 omegat)/(2 omega)]_(0)^(T)`
`=(1)/(4T) m omega^(2)A^(2)[T-0]`
`=(1)/(4T)m omega^(2)A^(2)(T)`
`E_(p_(av))=(1)/(4T) m omega^(2)A^(2) " "....(5)`
Clearly from equation (4) and (5)
`E_(K_(av))=E_(p_(av))`
`:. {:("Average"),("value of K.E"):}}= {{:("Average"),("value of P.E"):}`