Force on the particle is $F=-\frac{d V}{d x}$

$\Rightarrow \quad F=-\frac{d}{d x}\left(\frac{1}{2} k x^2-V_o \cos \left(\frac{x}{a}\right)\right)$

$=-\left(k x+\frac{V_o}{a} \sin \left(\frac{x}{a}\right)\right)$

As $\quad x << a, \Rightarrow \frac{x}{a} << 1$

So, $\sin \frac{x}{a} \approx \frac{x}{a}$

Hence, $F=-\left(k x+\frac{V_0}{a} \cdot \frac{x}{a}\right)=-\left(k+\frac{V_0}{a^2}\right) \cdot x$

Acceleration $A$ of the particle is

$A=\frac{F}{m}=-\frac{1}{m}\left(k+\frac{V_{0}}{a^2}\right) \cdot x$

As, acceleration $A=-\omega^2 x$, we have

$c 0=\sqrt{\frac{1}{m}\left(k+\frac{V_0}{a^2}\right)}=\sqrt{\left(\frac{k a^2+V_0}{m a^2}\right)}$

$\therefore$ Time period of oscillation is

$T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\left(\frac{m a^2}{k a^2+V_0}\right)}$