14.22 Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.

TheequationofdisplacementofaparticleexecutingSHM:
x=Asinωt
where,
A=Amplitudeofoscillation
ω=Angularfrequency=kM
Thevelocityoftheparticleis:
v=dxdt=Aωcosωt
Thekineticenergyoftheparticleis:
Ek=12Mv2=12MA2ω2cos2ωt
Thepotentialenergyoftheparticleis:
EP=12kx2=122A2sin2ωt
FortimeperiodT,theaveragekineticenergyoverasingle
cycle:
(Ek)Avg=1T0TEkdt=1T0T12MA2ω2cos2ωtdt=12TMA2ω20T(1+cos2ωt)2dt
=14TMA2ω2[t+sin2ωt2ω]0T=14TMA2ω2(T)=14MA2ω2...(i)
And,averagepotentialenergyoveronecycle:
(EP)Avg=1T0TEpdt=1T0T122A2sin2ωtdt=12T2A20T(1cos2ωt)2dt
=14T2A2[tsin2ωt2ω]0T=14T2A2(T)=2A24...ii

 

So the average kinetic energy for a given time period is equal to the average potential energy for the same time period.