Question

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

Answer 1

The displacement for the particle executingsimple harmonic motion is given as,

x=Asinωt

By differentiating the above equation to determine the particle velocity, we get

d dt ( x )= d dt ( Asinωt ) v=Aωcosωt

The average kinetic energy of the particle over one oscillation is,

K av = 1 T ∫ 0 T 1 2 m v 2 = 1 T ∫ 0 T 1 2 m ( Aω cos 2 ωt ) 2 dt = m A 2 ω 2 2T ∫ 0 T ( 1+cos2ωt ) 2 dt = m A 2 ω 2 4T ∫ 0 T ( 1+cos2ωt )dt

By simplifying the above equation further, we get

K av = m A 2 ω 2 4T [ t+ sin2ωt 2ω ] 0 T = m A 2 ω 2 4T [ T+( sin2ωT 2ω ) ] = m A 2 ω 2 4T [ T+( sin4π 2ω ) ]           ( ω= 2π T ) = 1 4 m A 2 ω 2 …… (1)

The angular frequency in terms of spring constant k is given as,

ω= k m k=m ω 2

The average value of the potential energy over one oscillation is,

U av = 1 T ∫ 1 2 k x 2

By substituting the values of k and x in the above equation, we get

U av = 1 T ∫ 0 T ( 1 2 m ω 2 A 2 sin 2 ωt )dt = m ω 2 A 2 2T ∫ 0 T ( 1−cos2ωt ) 2 dt = m ω 2 A 2 4T ∫ 0 T ( 1−cos2ωt )dt

By simplifying the above equation further, we get

U av = m ω 2 A 2 4T [ t− sin2ωt 2ω ] 0 T = m ω 2 A 2 4T [ T− sin2ωT 2ω ] = m ω 2 A 2 4T [ T− sin4π 2ω ]           ( ω= 2π T ) = 1 4 m ω 2 A 2 …… (2)

From equations (1) and (2), we can say that the average kinetic energy over a period of oscillation equals the average potential over the same period.