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
\[x=\pm \,\frac{r}{\sqrt{2}}\]
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