Blocks on table!

The question can be transformed to situation in which these blocks lie on the horizontal table and $Mg$ force acts on the right most block.

Let us consider these three blocks and spring as a system.

Net force acting on the system is $Mg$ . So the acceleration of the CoM of the system is $g/3$ .

Let us choose our reference which is moving with CoM. As it is a non inertial frame so we have to consider pseudo force.

Now work done by all the external force on the system is equal to change in Kinetic and potential energy of the system. In order to have maximum separation change in KE should be zero.

As we have to find work done by external forces so we don't have to worry about tension in the string and force acting because of spring.

$\frac { mgy }{ 3 } -\frac { mgx }{ 3 } +\left( mg-\frac { mgx }{ 3 } \right) =\frac { 1 }{ 2 } { k\left( x+y \right) }^{ 2 }$

On solving we get $(x+y)=2cm$ . So Maximum distance between the blocks are 3cm.

I defined the coordinates in the figure and applied Lagrangian mechanics:

$T=\frac{1}{2}2m\dot{x}^2+\frac{1}{2}2m(\dot{x}-\dot{l})^2 \quad \textcolor{#3D99F6}{ \text{kinetic energy}} \\ V=\frac{1}{2}k(l-l_0)^2 - mgx \quad \textcolor{#3D99F6} {\text{potencial energy} }$

$\dfrac{d}{dt}\dfrac{\partial T}{\partial \dot{l} } = - \dfrac{\partial V}{\partial l } \quad \implies \quad -m\ddot{x} + m\ddot{l} = -k(l-l_0) \phantom{mg} \quad (1)\\ \dfrac{d}{dt}\dfrac{\partial T}{\partial \dot{x} } = - \dfrac{\partial V}{\partial x } \quad \implies \quad 3m\ddot{x} - m\ddot{l} = mg \phantom{-k(l-l_0)} \quad (2)$

Combining $(1)$ and $(2)$ : $\quad \ddot{l}=-\dfrac{3k}{2m}(l-l_0)+\dfrac{g}{2} \quad (3)$

For the initial conditions $l(0)=0$ and $\dot{l}(0)=0$ , the solution for $(3)$ is: $\quad l(t)=A\sin(\omega t-\varphi_0) + l_0 + \frac{mg}{3k}$ , where $\omega=\sqrt{\frac{3k}{2m}}$ i $\varphi_0=\arcsin(\frac{mg}{3kA})$

Given that $\dot{l}(0)=0$ , we can determine the value of $A$ :

$\dot{l}(t)=\omega A\cos(\omega t-\varphi_0) \\ \dot{l}(0)=0 \implies \omega t-\varphi_0=\frac{\pi}{2} \implies A=\frac{mg}{3k}\approx 1 \text{ cm}$

The maximum value of $l$ is reached when $\sin(\omega t-\varphi_0)=1$ :

$l_{max}=A+l_0 + \dfrac{mg}{3k}=l_0 + \dfrac{2mg}{3k}\approx \boxed{3 \text{ cm}}$

Appreciating the other solutions , here's my solution( others must also have done by this ) , just two steps .

Analyse the problem -

The elongation can be maximum only when the right block on the table stops .

F.B.D of the given block justifies that T =kx

now since right block stops , the load will also stop ( connected through string )

hence , T= mg

Equating , mg = kx , x = mg/k , x = 3*10 / 1000 m = 3 cm

Must be the easiest !!