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Question

A linear harmonic oscillator of force constant 2×106 N/m and amplitude 0.01 m has a total mechanical energy of 160 Joules. Which of the following is true?

A
Maximum potential energy is 100 J
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B
Maximum K.E. is 160 J
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C
Maximum P.E. is 160 J
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D
Minimum P.E. is zero
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Solution

The correct option is C Maximum P.E. is 160 J
Harmonic oscillator has some initial elastic potential energy PE0 and amplitude of harmonic variation of energy is
12kA2=12×2×106×(0.01)2=100 J
This is the maximum kinetic energy of the ocsillator. Thus, KEmax=100 J.
But total mechanical energy is 160 J
As total mechanical energy is conserved
PE0+KEmax=160 J (at mean position)
PE0=PEmin=60 J
We know during oscillation, at some instant entire KE is coverted to PE
PEmax=PE0+KEmax
PEmax=60 J+100 J=160 J (At extreme position)

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