Q: A linear harmonic oscillator of force constant 2 × 106 Nm-1 and amplitude 0.01 m has a total mechanical energy of 160 J. Then find maximum and minimum values of PE and KE.
Sol: K.Emax = (1/2) KA2
= (1/2)× 2×106×(0.01)2 = 100 J
Since total energy is 160 J. Maximum P.E is 160 J. From this it is understood that at the mean position potential energy of the simple harmonic oscillator is minimum which need not be zero.
PEmin = TE-KEmax
= 160 -100 = 60 J
KEmin = 0