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 _______.

For series combination of the spring, equivalent spring constant is given by, $\frac{1}{k_{eq}}=\frac{1}{k_1}+\frac{1}{k_2}$.

Therefore, for combined system net spring constant $k_{eq}=\frac{2k\times k}{2k+k}+k=\frac{5k}{3}$

Angular frequency of oscillation $\left(\omega \right)=\sqrt{\frac{k_{eq}}{m}}$

$\Rightarrow \left(\omega \right)=\sqrt{\frac{5k}{3m}}$

Period of oscillation $\left(\tau \right)=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{3m}{5k}}$

$=\pi \sqrt{\frac{12m}{5k}}$

Therefore, $\alpha =12$.