Let x₀ be the extension in the spring in equilibrium. Then equilibrium of A and B give,

                         T = kx₀ + mg sin θ                   ...(i)

and                  2T = mg                                   ...(ii)

Here, T is the tension in the string. Now, suppose A is further displaced by a distance x from its mean position and v be its speed at this moment. Then B lowers by $\frac{x}{2}$ and speed of B at this instant will be $\frac{v}{2}$ Total energy of the system in this position will be,

$\text{E}=\frac{1}{2}\text{k}{\left(x+x_0\right)}^2+\frac{1}{2}{\text{m}}_{\text{A}}{\text{v}}^2+\frac{1}{2}{\text{m}}_{\text{B}}{\left(\frac{\text{v}}{2}\right)}^2+{\text{m}}_{\text{A}}{\text{gh}}_{\text{A}}-{\text{m}}_{\text{B}}{\text{gh}}_{\text{B}}$

$\text{or}\text{E}=\frac{1}{2}\text{k}{\left(x+x_0\right)}^2+\frac{1}{2}\text{m}{\text{v}}^2+\frac{1}{8}{\text{mv}}^2+\text{mgx sin}\text{θ}-\text{mg}\frac{x}{2}$

$\text{or}\text{E}=\frac{1}{2}\text{k}{\left(x+x_0\right)}^2+\frac{5}{8}\text{m}{\text{v}}^2+\text{mgx sin}\text{θ}-\text{mg}\frac{x}{2}$

Since, E is constant,

$\frac{\text{dE}}{\text{dt}}=0$

$\text{or}0=\text{k}\left(x+x_0\right)\frac{\text{d}x}{\text{dt}}+\frac{5}{4}\text{mv}\left(\frac{\text{dv}}{\text{dt}}\right)+\text{mg}\left(\text{sin}\text{θ}\right)\left(\frac{\text{d}x}{\text{dt}}\right)-\frac{\text{mg}}{2}\left(\frac{\text{d}x}{\text{dt}}\right)$

$\text{Substituting,}\frac{\text{d}x}{\text{dt}}=\text{v}$

$\frac{\text{d}\text{v}}{\text{dt}}=a$

$\text{and}\text{k}{x}_2+\text{mg sin}\text{θ}=\frac{\text{mg}}{2}\left[\text{From equations (i) and (ii)}\right]$

$\text{We get,}\frac{5}{4}\text{m}a=-\text{k}x$

$\text{Since,}a\propto -x$

Motion is simple harmonic, time period of which is,

$\text{T}=2\text{π}\sqrt{\left|\frac{x}{a}\right|}=2\text{π}\sqrt{\frac{\text{5m}}{\text{4k}}}$

$\text{ω}=\frac{2\text{π}}{\text{T}}=\sqrt{\frac{\text{4k}}{\text{5m}}}$