If the mass of the the pulleys shown in figure are small and the cord is inextensible, the angular frequency of oscillation of the system is

Let T be the tension in the cord and x_a and x_b the displacements of pulleys A and B respectively. Now assume that pulley B is fixed ; then extension of spring $x_b=\frac{x}{2}\text{ or }x=2x_b$. $\therefore x=2x_a+2x_b\dots .\left(1\right)$

From free body diagram of pulleys, $2\mathrm{T}={\mathrm{k}}_{\mathrm{b}}{\mathrm{x}}_{\mathrm{b}}\dots .\left(2\right)\text{ or }2T=k_{\mathrm{b}}{x}_a\dots .\text{ (3) }$

If K_eq denotes equivalent spring constant, $\frac{\mathrm{T}}{{\mathrm{k}}_{\mathrm{eq}}}=\mathrm{x}=2{\mathrm{x}}_{\mathrm{a}}+2{\mathrm{x}}_{\mathrm{b}}$ From equations (2) and (3)${\mathrm{x}}_{\mathrm{a}}=\frac{2\mathrm{T}}{{\mathrm{k}}_{\mathrm{a}}}\text{ and }{\mathrm{x}}_{\mathrm{b}}=\frac{2\mathrm{T}}{{\mathrm{k}}_{\mathrm{b}}}\text{ i.e., }{\mathrm{k}}_{\mathrm{eq}}=\frac{1}{4\left(\frac{1}{{\mathrm{k}}_{\mathrm{a}}}+\frac{1}{{\mathrm{k}}_{\mathrm{b}}}\right)}$

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

2nd Method