Q.

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

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a

4π2ma2v2

b

2π2ma2v2

c

π2ma2v2

d

14ma2v2

answer is C.

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Detailed Solution

The kinetic energy is given byK.E.=12mω2a2sin2⁡ωtThe 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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