Here, for calculation of time period we have neglected acceleration of gravity. Because, it is constant at all places and doesn't affect on resultant restoring force.

Let in the equilibrium position, the spring has extended by an amount $x_{0}$. Let the mass be pulled through a distance $x$ and then released. But, string is inextensible, hence the spring alone will contribute the total extension $x+x=2 x$. So net extension in the spring will be $2 x+x_{0}$.

When the extension of spring is $x_{0},(\mathrm{M}$ is not suspended) the restoring force in spring $\mathrm{F}=2 k x_{0}, \ldots$ (1) $\left[\because \mathrm{F}=\mathrm{T}+\mathrm{T}\right.$ and $\left.\mathrm{T}=k x_{0}\right]$

When mass is suspended, the net extension in spring $2 x+x_{0}$ and restoring force in spring, $\mathrm{F}^{\prime}=2 k\left(2 x+x_{0}\right)=4 k x+2 k x_{0} \quad \ldots$ (2)

Restoring force on system,

$f^{\prime} =-\left(\mathrm{F}^{\prime}-\mathrm{F}\right)$

$=\mathrm{F}-\mathrm{F}^{\prime}$

$=2 k x_{0}-4 k x-2 k x_{0} \quad \text { [From equ. (1) and (2)] }$

$=-4 k x$