For a particle executing simple harmonic motion, the kinetic energy K is given by, K=K0cos2ωt. The maximum value of potential energy is?
A
K0
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B
zero
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C
K0/2
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D
not obtainable
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Solution
The correct option is DK0 Kmax=K0=total energy As total energy remains conserved in SHM hence when U is maximum in SHM, K=0, i.e., is also equal to Umax, i.e., Umax=E=K0.