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.

The equation of displacement of a particle executing SHM:
x=Asinωt
where,
A=Amplitude of oscillation 
ω=Angular frequency=kM
The velocity of the particle is:
v=dxdt=Aωcosωt
The kinetic energy of the particle is:
Ek=12Mv2=12MA2ω2cos2ωt
The potential energy of the particle is:
EP=12kx2=122A2sin2ωt
For time period T, the average kinetic energy over a single 
cycle:
(Ek)Avg=1T0TEkdt=1T0T12MA2ω2cos2ωtdt   =12TMA2ω20T(1+cos2ωt)2dt
=14TMA2ω2[t+sin2ωt2ω]0T=14TMA2ω2(T)=14MA2ω2                                     ...(i)
And, average potential energy over one cycle:
(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.