Resolving tension $T$, we have Restoring force on bead $=2 T \cos \theta$

Here, $\cos \theta=\frac{x}{\sqrt{x^2+\left(\frac{l}{2}\right)^2}}$

As $l \gg x, \frac{l^2}{4}+x^2 \approx \frac{l^2}{4}$

So, $\quad \cos \theta=\frac{2 x}{l}$

Now, if $a=$ acceleration of bead. Then,

Restoring force $=$ Mass $\times$ Acceleration

$\Rightarrow \quad 2 T \cos \theta=-m a$

$\Rightarrow \frac{2 T(2 x)}{l} =-m a$

$\Rightarrow a =-\left(\frac{4 T}{m l}\right) \cdot x$

Hence, $\omega^2=\frac{4 T}{m l}$

$\therefore$ Time period of oscillation is

$T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m l}{4 T}}$

Hence, frequency of oscillation is

$f=\frac{1}{T}=\frac{1}{2 \pi} \sqrt{\frac{4 T}{m l}}$