A block of mass $m$ is kept on smooth horizontal surface and connected with two springs as shown in figure. Initially springs are in their natural length. Time period of small horizontal oscillation of the block is:

When block is displaced slightly by $x$ slightly towards right then $x\text{'}=x\cos 30°$

The restoring force

$$\begin{gathered} F=-\left(kx+4kx\text{'}\cos 30°\right) \\ =-\left[kx+4k\left(x\cos 30°\right)\cos 30°\right] \\ =-\left[kx+3kx\right] \\ =- 4kx \\ a=\frac{F}{m}=\frac{4k}{m}\left(-x\right) \end{gathered}$$

On comparing with $a=-{\omega }^{2}x$, we get

$\omega =\sqrt{\frac{4k}{m}}$ and $\mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{4\mathrm{k}}}=\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$