A particle is executing simple harmonic motion with a time period of 2 second. At t = 0, the particle was in its extreme position. Then the time after which the potential energy of the particle will be equal to 50% of its total energy is
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a
0.125 S
b
1 S
c
0.25 S
d
0.5 S
answer is C.
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Detailed Solution
If E be the total energy and V be the potential energy at time t, then V=E2(1+cos2ωt), where ω= angular frequency of SHM.when V=E/2, we get, E2=E2(1+cos2ωt)⇒2ωt=π2⇒t=πT42π=T8t=28sec=0.25sec