The time period of simple harmonic motion of mass M in the given...

For series combination of the spring, equivalent spring constant is given by, $\dfrac{1}{{k}_{eq}} = \dfrac{1}{{k}_{1}} + \dfrac{1}{{k}_{2}}$.

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

Angular frequency of oscillation \left(\omega\right) = \sqrt{\dfrac{{k}_{eq}}{m}}$</p><p>$⇒ \left(\omega\right) = \sqrt{\dfrac{5k}{3m}}$</p><p>Period of oscillation \left(τ\right) = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{3m}{5k}}$

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

Therefore, $\alpha = 12$.