Calculating distances might not be easy

Taking forces on the whole system,

$\begin{aligned} mg &= 3ma\\ a&=\frac{g}{3} \end{aligned}$

W.R.T to the block at the leftmost end, forces acting on the other block on the table are $kx$ , $kx$ and $\frac{2mg}{3}$ .

Thus conserving energy in the block frame,

$\begin{aligned} 0 &= -kx^2 + \dfrac{2mgx}{3}\\ x&= \dfrac{2mg}{3k} \end{aligned}$

$\boxed {\therefore x = \dfrac{2mg}{3k}}$

Let blocks 1, 2 and 3 be numbered from left to right. The equations of motion for each block are

$mg-T=ma_3$

$T-kx=ma_2$

$kx=ma_1$ ,

where $x$ is the elongation of the string and blocks 2 and 3 have the same acceleration (say, $a_2$ ) because they are equally spaced by the string at all times. Let $x_1$ and $x_2$ be the positions of blocks 1 and 2 with respect to some reference frame. We then have

$x_2-x_1-L=x$ or

${\displaystyle a_{2}-a_{1}=\frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}}}$ .

Finding $a_1$ and $a_2$ from the equations of motion and replacing them on this last one, we get the differential equation for $x$

${\displaystyle \frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}}+\frac{3k}{2m}x=\frac{g}{2}}$ ,

which solution is

${\displaystyle x=-\frac{mg}{3k}\cos(\omega t)+\frac{mg}{3k}}$ .

Given this, the maximum value of $x$ is when $\cos(\omega t)=-1$ , i.e. ${\displaystyle \frac{2mg}{3k}}$ . Hence, the maximum distance between blocks 1 and 2 should be

$L+{\displaystyle \frac{2mg}{3k}}=3$ cm,

given $g=10m/s^2$ .

First analyze block falling down. mg - T= ma
Now understand that we can associate the spring block system as that of a block of mass 2m. So we analyze than T=2ma
mg-2ma=ma ; ma+2ma = mg ; 3ma= mg a = g/3
NOW PLUG BACK IN T=2ma ; T=2mg/3 Now anaylaze system in depth.

T-kx=0 ; T=kx ; 100 * 2mg/3k = x (* 100; to convert to centimeters) x+1 = 3

TADA!

The acceleration of centre of mass will be g/3.(Mg=3Ma) Now,let us view everything from the centre of mass frame.When the spring is at it's maximum extension the velocity of all three blocks with respect to centre of mass will be 0.(This statement can be proved using Proof by Contradiction) Now writing kinematic equation of velocity for the leftmost block. V^{2}=integral of 2 a dx. The value of a is the acceleration of the leftmost block with respect to centre of mass frame. Hence a= (g/3)-(kx/m) Now,if we make the integral 0 we get value of x as 2cm.This is maximum extension in spring.Adding the initial distance between the blocks to it.We get maximum distance between them as 3cm.