$A$ body of mass $m$ is situated in a potential field $U(x) = U_0 (1 - \cos \alpha x)$,where $U_0$ and $\alpha$ are constants. Find the time period of small oscillations.

(D) The potential energy associated with the field is given by $U(x) = U_0 (1 - \cos \alpha x)$.
The force $F$ is related to potential energy by $F = -\frac{dU}{dx}$.
$\frac{dU}{dx} = \frac{d}{dx} [U_0 (1 - \cos \alpha x)] = U_0 \alpha \sin \alpha x$.
Therefore,$F = -U_0 \alpha \sin \alpha x$.
For small oscillations,$\alpha x$ is very small,so $\sin \alpha x \approx \alpha x$.
Thus,$F \approx -U_0 \alpha (\alpha x) = -U_0 \alpha^2 x$.
This is the equation of Simple Harmonic Motion $(SHM)$ of the form $F = -k x$,where the effective spring constant $k = U_0 \alpha^2$.
For $SHM$,the angular frequency $\omega$ is given by $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{U_0 \alpha^2}{m}} = \alpha \sqrt{\frac{U_0}{m}}$.
The time period $T$ is given by $T = \frac{2 \pi}{\omega}$.
Substituting $\omega$,we get $T = \frac{2 \pi}{\alpha \sqrt{\frac{U_0}{m}}} = 2 \pi \sqrt{\frac{m}{U_0 \alpha^2}}$.