A disc of mass 3 m and a disc of mass m are connected by a massless spring of stiffness k. The heavier disc is placed on the ground with the spring vertical and lighter disc on top. From its equilibrium position, the upper disc is pushed down by a distance $\delta$ and released. Then,

$k{\delta }_0=mg$

where, ${\delta }_0=$ compression of the spring due to the weight of the upper block.

After the removal of the external force on the upper block ,the upper block starts moving up and when the elongation of the spring is y , the upper block stops and the lower block is on the verge of loosing contact with the ground.

Conserving energy ,  

      $$\begin{gathered} △K+△U = 0 \\ \Rightarrow mg(y+{\delta }_0+\delta ) +\frac{1}{2}k[y^2-{\left({\delta }_0+\delta \right)}^2] =0 \\ \Rightarrow \delta =\frac{4mg}{k} \end{gathered}$$

Again when $\delta =\frac{2mg}{k},normal reaction =k({\delta }_0+\delta )+3mg =6mg$