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

Hey mate,

◆ Proof-
Consider a particle of mass m moving in SHM.
The position of particle is given by-
x = Asin(ωt+δ)

Now, we can configure velocity,
v = dx/dt
v = Aωcos(ωt+δ)

a) Kinetic energy is
K.E. = 1/2 mv^2
K.E. = 1/2 m [Aωcos(ωt+δ)]^2
K.E. = 1/2 m A^2 ω^2 cos^2(ωt+δ)

Average K.E. over full cycle is
KE(avg) = 1/T ∫K.E.dt
KE(avg) = 1/T ∫[1/2 m A^2 ω^2 cos^2(ωt+δ)]dt
KE(avg) = mA^2ω^2/2T ∫cos^2(ωt+δ)dt
KE(avg) = mA^2ω^2/4 ...(1)

b) Potential energy is
P.E. = 1/2 kx^2
P.E. = 1/2 m ω^2 [Asin(ωt+δ)]^2
P.E. = 1/2 m A^2 ω^2 sin^2(ωt+δ)

Average P.E. over full cycle is
PE(avg) = 1/T ∫P.E.dt
PE(avg) = 1/T ∫[1/2 m A^2 ω^2 sin^2(ωt+δ)]dt
PE(avg) = mA^2ω^2/2T ∫sin^2(ωt+δ)dt
PE(avg) = mA^2ω^2/4 ...(2)

From (1) & (2),
KE(avg) = PE(avg)


Hope this helps you...