question_answer 56)

                  Find the displacement of a simple harmonic oscillator at which its P.E. is half of the maximum energy of the oscillator.

Answer:

                  Maximum energy of a particle executing S.H.M is given by                 \[E=\frac{1}{2}M,\,{{\omega }^{2}}{{r}^{2}}\] where r is the amplitude.                 P.E. of a particle executing S.H.M. at a distance \[x\] from mean position is given by                 \[P.E.=\,\frac{1}{2}\,M\,{{\omega }^{2}}\,{{x}^{2}}\]                 Since \[P.E.\,=\frac{1}{2}\,E\]                 \[\therefore \] \[\frac{1}{2}\,M\,{{\omega }^{2}}\,{{x}^{2}}\,=\frac{1}{2}\,M\,{{\omega }^{2}}\,{{r}^{2}}\]                 or  \[x=\pm \,\frac{r}{\sqrt{2}}\]