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

Average Value of P.E. and K.E.

Category : JEE Main & Advanced

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}}$.

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