Answer
$$
\tau=\pi \sqrt{\frac{m}{k}}
$$
Work Step by Step
$$
\begin{gathered}
T+V=\text { const. } \\
T=\frac{1}{2} m(\dot{y})^2 \\
V=m g y+\frac{1}{2}(4 k)(\Delta s-y)^2 \\
T+V=\frac{1}{2} m(\dot{y})^2+m g y+\frac{1}{2}(4 k)(\Delta s-y)^2 \\
m \ddot{y} \ddot{y}+m g \dot{y}-4 k(\Delta s-y) \dot{y}=0 \\
m \ddot{y}+m g+4 k y-4 k \Delta s=0
\end{gathered}
$$
Since $\Delta s=\frac{m g}{4 k}$
Then
$$
\begin{gathered}
m \ddot{y}+4 k y=0 \\
y+\frac{4 k}{m} y=0 \\
\omega_n=\sqrt{\frac{4 k}{m}} \\
\tau=\frac{2 \pi}{\omega_n}=\pi \sqrt{\frac{m}{k}}
\end{gathered}
$$
