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

##### Updated On: 27-06-2022

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.1TT012ma2ω2cos2(ωt+ϕ)dt
=ma2ω22TT0cos2(ωt+ϕ)dt=ma2ω22TT012[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.=1TT012ka2sin2(ωt+ϕ)dt
=ka22TT012[1cos2(ωt+ϕ)dt]=ka24T[T0dtT0cos2(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

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