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 and minimum values of P.E and K.E?

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

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 and minimum values of P.E and K.E?
Answer» Solution :MAXIMUM `K.E= (1)/(2)KA^(2)= (1)/(2)xx2xx10^(6)xx(0.01)^(2)= 100j` Since total energy is 160J, 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. `PE_("min")= TE-KE_("max")= 160-100= 60J= KE_("min")= 0`