Show that for a particle executing simple harmonic motion
a. the average value of kinetic energy is equal to the average value of potential energy.
b. average potential energy = average kinetic energy `=1/2` (total energy) (Hint : average kinetic energy = < kinetic energy >
`=1/Tint_(0)^(T)` (Kinetic energy) dt and
average Potential energy =
`=1/Tint_(0)^(T)` (potential energy) dt

Explanation

a). `(E_(k))"ave"=1/Tint_(0)^(T)E_(k)dt=1/Tint_(0)^(T)(1)/(2)mA^(2)omega^(2)cos^(2)omegatdt`
`=1/(4T)mA^(2)square^(2).T" "...(1)`
`(E_(p))"ave"=1/Tint_(0)^(T)E_(p)dt=1/Tint_(0)^(T)(1)/(2)mA^(2)omega^(2)sin^(2)omegatdt`
`=1/(2T)momega^(2)A^(2)int_(0)^(2)((1-cos2omegat))/(2).dt`
`=1/(4T).momega^(2)A^(2)[t-(sin2omegat)/(2omega)]_(0)^(T)`
`(E_(p))_("ave")=1/(4T)msquare^(2)A^(2)T" "..(2)`
`=msquare^(2)A^(2)//4`
From equations (1) and (2) it can show that the average K.E. is equal to average P.E.
(Total Energy) `TE=1/2msquare^(2)A^(2)`
`(E_(k))_("ave")=(E_(p))_("ave")=1/2.((1)/(2)momega^(2)A^(2))=1/2(TE)`
b. average kinetic energy =
`=1/Tint_(0)^(T)` (Kinetic energy) dt
and
average Potential energy =
`=1/Tint_(0)^(T)` (Potential energy) dt