JEE Main & Advanced Physics Simple Harmonic Motion Average Value of P.E. and K.E.

The average value of potential energy for complete cycle is given by 

\[{{U}_{average}}=\frac{1}{T}\int_{\,0}^{\,T}{U\,dt}=\frac{1}{T}\int_{\,0}^{\,T}{\frac{1}{2}m\,{{\omega }^{2}}{{a}^{2}}{{\sin }^{2}}(\omega \,t+\varphi )}\]\[=\frac{1}{4}m{{\omega }^{2}}{{a}^{2}}\]

The average value of kinetic energy for complete cycle \[{{K}_{average}}=\frac{1}{T}\int_{\,0}^{\,T}{K\,dt}\]\[=\frac{1}{T}\int_{\,0}^{\,T}{\frac{1}{2}m\,{{\omega }^{2}}{{a}^{2}}{{\cos }^{2}}\omega \,t\,dt}=\frac{1}{4}m{{\omega }^{2}}{{a}^{2}}\]

Thus average values of kinetic energy and potential energy of harmonic oscillator are equal and each equal to half of the total energy \[{{K}_{average}}={{U}_{average}}\]\[=\frac{1}{2}E=\frac{1}{4}m{{\omega }^{2}}{{a}^{2}}\].