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Q.
A particle is executing SHM at mid point of mean position and extremity. What is the potential energy in terms of total energy
ManipalManipal 2010Oscillations
Solution:
If a particle executes SHM, its kinetic energy is given by
$ KE=\frac{1}{2}m{{\omega }^{2}}({{A}^{2}}-{{x}^{2}}) $
Or $ KE=\frac{1}{2}k({{A}^{2}}-{{x}^{2}}) $
where, $ k=m{{\omega }^{2}}= $ constant
Its potential energy is given by
$ PE=\frac{1}{2}m{{\omega }^{2}}{{x}^{2}} $
$ =\frac{1}{2}k{{x}^{2}} $
Thus, total energy of particle
$ E=KE+PE $
$ =\frac{1}{2}k({{A}^{2}}-{{x}^{2}})+\frac{1}{2}k{{x}^{2}} $
$ =\frac{1}{2}k{{A}^{2}} $
Hence, $ PE=\frac{1}{2}k{{x}^{2}}=\frac{1}{2}k{{\left( \frac{A}{2} \right)}^{2}} $
$ (\because x=A/2) $
$ =\frac{1}{4}\left( \frac{1}{2}k{{A}^{2}} \right) $
$ =\frac{1}{4}E $
Hence, potential energy is one-fourth of total energy.