Updated On: 27-06-2022

Updated On: 27-06-2022

Text Solution

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

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