A particle of mass m executes SHM with amplitude ‘a’ and frequency f, the average kinetic energy during motion from the position of equilibrium to the end is
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a
2π2mA2f
b
π2mA2f2
c
14mA2f2
d
4π2mA2f2
answer is B.
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Detailed Solution
average K.E=KEmax+02AverageKE=12KEmax=12×12mw2A2 here w=angular frequency = 2πf =14m2πf2A2 =14m4π2f2A2 KE=π2mA2f2