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

Explanation

average kinetic energy = `lt` kinetic energy `gt=(1)/(T)int_(0)^(T)` (Kinetic energy) dt and average Potential energy = `lt` Potential energy `gt=(1)/(T)int_(0)^(T)` (Potential energy) dt]
Suppose a particle of mass m executes SHM of period T. The displacement of the particles at any instant t is given by `y=Asinomegat`
Velocity `v=(dy)/(dt)=omegaAcosomegat`
Kinetic energy `E_(K)=(1)/(2)mv^(2)=(1)/(2)momega^(2)A^(2)cos^(2)omegat`
Potential energy, `E_(P)=(1)/(2)momega^(2)y^(2)=(1)/(2)momega^(2)A^(2)sin^(2)omegat`
(a) Average K.E. over a period of oscillation,
`E_(K_(av))=(1)/(T)int_(0)^(T)E_(K)dt=(1)/(T)int_(0)^(T)(1)/(2)momega^(2)A^(2)cos^(2)omegatdt`
`=(1)/(2T)momega^(2)A^(2)int_(0)^(T)((1+cos2omegat)/(2))dt`
`=(1)/(4T)momega^(2)A^(2)[t+(sin2omegat)/(2omega)]_(0)^(T)=(1)/(4T)momega^(2)A^(2)T`
`E_(K_(av))=(1)/(4)momega^(2)A^(2)`
(b) Average P.E. over a period of oscillation
`E_(P_(av))=(1)/(T)int_(0)^(T)E_(P)dt=(1)/(T)int_(0)^(T)(1)/(2)momega^(2)A^(2)sin^(2)omegatdt`
`=(1)/(2T)momega^(2)A^(2)int_(0)^(T)((1-cos2omegat)/(2))dt`
`=(1)/(4T)momega^(2)A^(2)[t-(sin2omegat)/(2omega)]_(0)^(T)=(1)/(4T)momega^(2)A^(2)T`
`E_(P_(av))=(1)/(4)momega^(2)A^(2)`