A linear harmonic oscillator of force constant 2×106N/m and amplitude 0.01m has a total mechanical energy of 160 Joules. Which of the following is true?
A
Maximum potential energy is 100J
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B
Maximum K.E. is160J
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C
Maximum P.E. is 160J
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D
Minimum P.E. is zero
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Solution
The correct option is CMaximum P.E. is 160J Harmonic oscillator has some initial elastic potential energy PE0 and amplitude of harmonic variation of energy is 12kA2=12×2×106×(0.01)2=100J
This is the maximum kinetic energy of the ocsillator. Thus, KEmax=100J.
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)