The time period of simple harmonic motion of mass $$M$$ in the given figure is $$\pi\sqrt{\frac{\alpha M}{5K}}$$, where the value of $$\alpha$$ is

We need equivalent spring constant first.

Left spring has spring constant

k

On right side, the two upper springs (k and k) are in parallel.

So their equivalent is

$$k_p=k+k=2k$$

This combination is in series with lower spring k.

So right-side equivalent:

$$\frac{1}{k_r}=\frac{1}{2k}+\frac{1}{k}$$

$$=\frac{1+2}{2k}$$$$=\frac{3}{2k}$$

Thus

$$k_r=\frac{2k}{3}$$

Now this right equivalent acts in parallel with left spring k.

Total effective spring constant:

$$k_{\text{eff}}=k+\frac{2k}{3}$$

$$=\frac{5k}{3}$$

Time period:

$$T=2\pi\sqrt{\frac{M}{k_{\text{eff}}}}$$

$$2\pi\sqrt{\frac{3M}{5k}}$$

$$\pi\sqrt{\frac{12M}{5k}}$$

Given

$$T=\pi\sqrt{\frac{\alpha M}{5k}}$$

Comparing,

$$α=12$$