A particle of mass m is attached to a network of four springs as shown in the figure. The entire system is on a horizontal smooth surface. The frequency of oscillation of the particle along Y-direction is

Let a force dF is applied on the particle along the Y-direction.
Let the displacement of the particle is dy. Deformation of each upper spring is $d{\mathcal{l}}_1$ and that of each lower spring is $d{\mathcal{l}}_2$ (say). Let the static deformation of the system us dy.

 

From the figure, $x^2+y^2={\mathcal{l}}_1^2$
or  $2xdx+2ydy=2{\mathcal{l}}_{1}d{\mathcal{l}}_1$
x is constant.So, $dx=0$

Thus,  $ydy={\mathcal{l}}_{1}d{\mathcal{l}}_1$

$$\begin{gathered} d{\mathcal{l}}_1=\frac{y}{{\mathcal{l}}_1}dy=\sin {30}^{0}dy \\ or d{\mathcal{l}}_1=\frac{dy}{2} \\ Similarly, d{\mathcal{l}}_2=\sin {60}^{0}dy \\ \Rightarrow d{\mathcal{l}}_2=\frac{\sqrt{3}}{2}dy \end{gathered}$$

Net spring force in equilibrium
$=2[d{F}_{s_1}.\cos {60}^0]+2[d{F}_{s_2}.\cos {30}^0]=2[kd{\mathcal{l}}_1.\frac{1}{2}]+2[2kd{\mathcal{l}}_2.\frac{\sqrt{3}}{2}]=\frac{7}{2}kdy$ 
This force is in downward direction and is equal in magnitude to force applied dF.
$$\begin{gathered} \therefore dF=\frac{7}{2}kdy \\ \Rightarrow \frac{dF}{dy}=\frac{7}{2}k \\ So, k_{eff}=\frac{7}{2}k \\ \omega =\sqrt{\frac{k_{eff}}{m}}=\sqrt{\frac{7k}{2m}}\Rightarrow f=\frac{1}{2\pi }\sqrt{\frac{7k}{2m}} \end{gathered}$$