Show that for a particle executing S.H.M. average value of kinetic and potential energy is the same and each i (2023)

Solution

Average kinetic energy.
Kinetic energy at any instant =12mv2=12m(dydt)2
For a S.H.M. y=asin(ωt+ϕ)
and dydt=aωcos(ω+ϕ)
∴ Instantaneous K.E. =12ma2ω2cos2(ωt+θ)
If T is the time period, then
Average K.E. =<K.E.≥1T∫T012ma2ω2cos2(ωt+ϕ)dt
=ma2ω22T∫T0cos2(ωt+ϕ)dt=ma2ω22T∫T012[1+cos2(ωt+ϕ)]dt
=ma2ω24T[∫T0cos2(ωt+ϕ)dt]=ma2ω24TT [∴∫T0cos(2ωt+ϕ)dt=0]
=14ma2ω2
Average potential energy. Instantaneous potential energy is given by
P.E.=12ky2=12ka2sin2(ωt+ϕ)
∴ Average P.E.=1T∫T012ka2sin2(ωt+ϕ)dt
=ka22T∫T012[1−cos2(ωt+ϕ)dt]=ka24T[∫T0dt−∫T0cos2(ωt+ϕ)dt]
=ka24TT=14ka2 [∴∫T0cos2(ωt+ϕ)dt=0]
But ωt=km ∴k=mω2
∴ Average P.E. =14ma2ω2
Total Energy. The total energy =12ma2ω2
Thus it is clear from equations (i), (ii), and (iii) that the average kinetic energy of a harmonicoscillator is equal to the average potential energy and is equal to half the total energy i.e.,
<K.E.≥<P.E.≥12Etotal