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

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

1.	A linear harmonic oscillator of force constant 2xx10^(-6)Nm^(-1) andamplitude 0.01 m has a total mechanical energy of 160J. Then find maximum and minilmum values o P.E and K.E?
Answer» Solution :MAXIMUM `K.E.=1/2KA^(2)=1/2xx2xx10^(6)xx(0.01)^(2)=100J` 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. `PE_("min")=TE-KE_("max")=160-100=60J=KE_("min")=0`