A linear harmonic oscillator of force constant 2xx10^6 Nm^(-1) and amplitude 0.01 m has a total mechanical energy of 160 J. Then find maximum P.E and maximum KE?

A linear harmonic oscillator of force constant 2xx10^6 Nm^(-1) and amp

1.	A linear harmonic oscillator of force constant 2xx10^6 Nm^(-1) and amplitude 0.01 m has a total mechanical energy of 160 J. Then find maximum P.E and maximum KE?
Answer» Solution :Maximum elasticP.E =`1/2 KA^2 = 1/2 xx 2 xx 10^6 xx (0.01)^2 = 100 J` As the oscillator goes from the EXTREME to mean position, this is CONVERTED into K.E. So, maximum K.E is 100 J. Since total energy is 160 J, maximum P.E is 160J. From this it is understood that at the mean position potential energy of the simple harmonic oscillator is MINIMUM which NEED not be zero.