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Simple hormonic motion
Question

 A particle of mass m executes simple harmonic motion with amplitude 'a' and frequency v. The average kinetic energy during its motion from the
position of equilibrium to the end is

Difficult
Solution

 The kinetic energy is given by
K.E.=12mω2a2sin2⁡ωt
The average kinetic energy during its motion from the position of equilibrium to the end is given by
(K.E.)Average =∫0T/4 12mω2a2sin2⁡ωtdt =π2ma2v2 (∵ω=2πv)

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