The total energy of a harmonic oscillator of mass 2kg is 9J. If its potential energy at mean position is 5J, its K.E. at the mean position will be :
A
9J
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B
14J
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C
4J
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D
11J
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Solution
The correct option is C4J Since the total energy of a harmonic oscillator remains conserved. So (T.E.)=(T.E.)atmean=9J. Now, (T.E.)atmean=(P.E.)atmean+(K.E.)atmean ⇒9=5+(K.E.)mean (K.E.)atmean=4J Hence, (c) is the correct answer.