If the mass of the pulley shown in figure is small and the cord is inextensible, the angular l frequency of oscillation of the system is:

Free body diagram

Let ‘T’ be the tension in the cord and$x_{\text{a}}\&x_{\text{b}}$  are the displacements of pulleys A & B respectively. Now assume that pulley B is fixed, then extension of spring

$x_{\text{b}}=\frac{x}{2}\Rightarrow x=2x_{\text{b}}$

Similarly if we imagine that pulley A is fixed$x=2x_{\text{a}}$ ,

But neither

Pulley B nor pulley A is fixed

$\therefore 2x_{\text{a}}+2x_{\text{b}}\to \left(1\right)$

F. B. D for pulleys:

$\text{2T}={\text{K}}_{\text{b}}{x}_{\text{b}}\to \left(2\right)$

                $2\text{T}={\text{K}}_{\text{a}}{x}_{\text{a}}\to \left(3\right)$

If ${\text{K}}_{\text{eq}}$ denotes equivalent spring constant

$\frac{\text{T}}{{\text{K}}_{\text{eq}}}=x=2x_{\text{a}}+2x_{\text{b}}\to \left(4\right)$

From equivalent (2), (3) & (4)

${\text{K}}_{\text{eq}}=\frac{1}{4\left(\frac{1}{{\text{K}}_{\text{a}}}+\frac{1}{{\text{K}}_{\text{b}}}\right)}$

$\therefore \omega =\sqrt{\frac{{\text{K}}_{\text{eq}}}{\text{m}}}=\sqrt{\frac{{\text{K}}_{\text{a}}{\text{K}}_{\text{b}}}{4\text{m}\left({\text{K}}_{\text{a}}+{\text{K}}_{\text{b}}\right)}}$