Answer:
\[U(x)=\text{
}{{U}_{0}}(1\text{ }\cos \,\,\alpha x)\]
\[\therefore
\] \[F=-\frac{dU}{dx}\,=-\frac{d}{dx}\,[{{U}_{0}}-{{U}_{0}}\,\cos
\,\alpha x]\]
\[=0-{{U}_{0}}\,\alpha
\,\sin \,\alpha x\]
If \[(\alpha
x)\] is small, then \[\sin
\,\alpha x=\alpha x\]
\[\therefore
\] \[F=-\,{{U}_{0}}\,{{\alpha
}^{2}}x\]
and
acceleration, \[a=\,\frac{F}{m}\,=-\left( \frac{{{U}_{0}}{{\alpha }^{2}}}{m}
\right)\,x\]
Hence,
motion is S.H.M.
\[\therefore
\] \[\,T=2\pi
\,\sqrt{\frac{m}{{{U}_{0}}{{\alpha }^{2}}}}\]
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