A) Maximum potential energy is 100 J
B) Maximum K.E. is 100 J
C) Maximum P.E. is 160 J
D) Minimum P.E. is zero
Correct Answer: B
Solution :
Harmonic oscillator has some initial elastic potential energy and amplitude of harmonic variation of energy is \[\frac{1}{2}K{{a}^{2}}=\frac{1}{2}\times 2\times {{10}^{6}}{{(0.01)}^{2}}=100J\] This is the maximum kinetic energy of the oscillator. Thus \[{{K}_{\max }}=100J\]
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