Oscillations
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Question : 22
Total: 25
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.
Solution:
Consider a particle of mass m executing S.H.M. with period T. The displacement of the particle at an instant t, when time period is noted from the mean position is given by
y = A sin ω t
∴ Velocity,v =
= A ω cos ω t
K . E . , E k =
m v 2 =
m A 2 ω 2 cos 2 ω t
P.E.,E p =
k y 2 =
m A 2 ω 2 sin 2 ω t ( ∵ k = m ω 2 )
∴ Average K.E. over one cycleE k a v =
E k d t =
m A 2 ω 2 cos 2 ω t d t =
m A 2 ω 2
d t
=
m A 2 ω 2 [ t +
] 0 T
=
m A 2 ω 2 ( T ) =
m A 2 ω 2
Average P.E. over one cycleE p a v =
E p d t =
m A 2 ω 2 sin 2 ω t d t =
m ω 2 A 2
d t
=
m ω 2 A 2 [ 1 −
] 0 T
=
m ω 2 A 2 [ T ] =
m A 2 ω 2
From (i) and (ii),E k a v = E p a v
∴ Velocity,
P.E.,
∴ Average K.E. over one cycle
Average P.E. over one cycle
From (i) and (ii),
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