A particle of mass $m$ is located in a unidimensional potential field where the potential energy of the particle depends on the coordinate $x$ as $U\left(x\right)=U_0(1-\cos cx)$; $U_0$ and $c$ are constants. Find the period of small oscillation that the particle performs about the equilibrium position.

Given, $U\left(x\right)=U_0(1-\cos cx)$

We know that the field force is given by

$$\begin{gathered} F=-\frac{dU}{DX} \\ =-\frac{d}{dx}\left[U_0\right(1-\cos cx\left)\right] \\ =-U_{0}c \sin cx \end{gathered}$$

For small value of $x$,$\sin cx=cx$

$F=-U_{0}c\left(cx\right)=-U_0{c}^{2}x$

Acceleration $a=\frac{F}{m}=\frac{U_0{c}^2}{m}(-x)$

As ,$a\propto (-x)$ so above equation represents $SHM$. Compare this with standard equation of $SHM$, we get

$$\begin{gathered} \omega =\sqrt{\frac{U_0{c}^2}{M}} \\ T=\frac{2\mathrm{\pi }}{\omega }=2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{{\mathrm{U}}_0{\mathrm{c}}^2}} \end{gathered}$$