As given that, the potential energy (associated with the field) for a body of mass $m$.
$$
U(x)=U_0(1-\cos \alpha x)
$$
$\because$ for conservative force $f$, we can write $f=\frac{-d u}{d x}$
We have assumed the field to be conservative
$$
\begin{aligned}
F &=-\frac{d}{d x}\left(U_0-U_0 \cos \alpha x\right) \\
&=-U_0 \alpha \sin \alpha x
\end{aligned}
$$
For SHM, $\alpha x$ is small so $\sin \alpha x$ i.e. $\sin \alpha x=\alpha x$ so,
$$
\begin{aligned}
&F=-U_0 \alpha^2 x \\
&\therefore F \propto(-x)
\end{aligned}
$$
As, $U_0, \alpha$ being constant.
$\therefore$ So motion is SHM for small oscillations.
Standard equation for SHM,
$$
F=-m \omega^2 x
$$
Comparing Eqs. (ii) and (iii), then
$$
\begin{aligned}
&-m \omega^2 x=-U_0 \alpha^2 x \\
&m \omega^2=U_0 \alpha^2 \\
&\omega^2=\frac{U_0 \alpha^2}{m} \text { or } \omega=\sqrt{\frac{U_0 \alpha^2}{m}} \\
&\therefore \text { Time period } T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m}{U_0 \alpha^2}}
\end{aligned}
$$
From (i) this period is valid for small angle $\alpha x$.
$$
T=\frac{2 \pi}{\alpha} \sqrt{\frac{m}{U_0}}
$$