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
4Ï2ma2v2
b
2Ï2ma2v2
c
Ï2ma2v2
d
14ma2v2
answer is C.
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Detailed Solution
The kinetic energy is given byK.E.=12mÏ2a2sin2â¡ÏtThe average kinetic energy during its motion from the position of equilibrium to the end is given by(K.E.)Average =â«0T/4â12mÏ2a2sin2â¡Ïtdt =Ï2ma2v2 (âµÏ=2Ïv)
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