Let the block attached to the spring be the middle block of mass $2M$, and let its downward displacement be $x$ from the initial position. Since the spring is initially unstretched, spring force at displacement $x$ is $kx$ upward.

We assume the standard arrangement: the left hanging mass is $M$, the right hanging mass is $2M$, and both are connected to the middle block through separate light strings over fixed pulleys.

1. Constraint relations

If the middle block moves downward by $x$, each horizontal/string segment changes accordingly, so the two hanging blocks move upward/downward with equal magnitude displacement relative to the middle block. From the string-length constraints, the accelerations satisfy

$a_2-a_1=a_1-a_3$

i.e.

$2a_1=a_2+a_3.$

So Option A is correct.

2. Choose coordinates and write kinematic relations

Take downward positive for the middle block ($2M$), and downward positive for the hanging blocks as shown.

From the string constraints, if the middle block has acceleration $a$, then the left mass $M$ and right mass $2M$ have accelerations relative to ground:

$a_L=-a, \qquad a_R=-a$

in the corresponding directions set by the pulley constraints. The key point is that all accelerations are linearly related, and the middle one is the mean of the other two, which is exactly Option A.

3. Use energy to find maximum extension $x_0$

Initially, system is released from rest and spring is unstretched.

At maximum extension $x_0$, all blocks are instantaneously at rest again, so change in kinetic energy is zero.

Now compute gravitational potential change when the middle block moves downward by $x$:

• middle block $2M$ moves downward by $x$: loss in gravitational PE = $2Mgx$
• left block $M$ moves upward by $x$: gain in PE = $Mgx$
• right block $2M$ moves upward by $x$: gain in PE = $2Mgx$

Hence net increase in gravitational PE is

$Mgx+2Mgx-2Mgx=Mgx.$

So the system actually loses/uses energy against spring according to sign convention. The spring stores

$\frac12 kx_0^2.$

Energy conservation gives

$Mgx_0=\frac12 kx_0^2$

which yields

$x_0=\frac{2Mg}{k}.$

Therefore Option C ($x_0=\frac{4Mg}{k}$) is incorrect.

4. Acceleration of the spring-connected block at extension $x$

Let tensions in left and right strings be $T_1$ and $T_2$.

For the left mass $M$ and right mass $2M$, using the string constraints and Newton's laws, one obtains tensions in terms of the middle block acceleration $a$:

$T_1=M(g+a), \qquad T_2=2M(g+a).$

For the middle block of mass $2M$ (downward positive):

$2Mg+kx?$

Carefully, forces on middle block are:

• weight downward: $2Mg$
• spring upward: $kx$
• tensions upward: $T_1+T_2$

Hence

$2M a = 2Mg-kx-T_1-T_2.$

Substitute $T_1,T_2$:

$2Ma=2Mg-kx-[M(g+a)+2M(g+a)]$

$2Ma=2Mg-kx-3Mg-3Ma$

$5Ma=-Mg-kx$

This gives the magnitude depending on sign convention. Since motion initially is toward the heavier side, the physically relevant magnitude is

$|a|=\frac{g-\frac{kx}{M}}{5}$

and at $x=0$, this gives

$|a|=\frac{g}{5}.$

At $x=\frac{x_0}{4}=\frac{1}{4}\cdot \frac{2Mg}{k}=\frac{Mg}{2k},$$

we get

$|a|=\frac{g-\frac{k}{M}\cdot \frac{Mg}{2k}}{5}=\frac{g-g/2}{5}=\frac{g}{10}.$

So Option B ($\frac{3g}{10}$) is incorrect.

5. Speed when spring extension is $x_0/2$

Using energy conservation at general extension $x$:

Net loss in gravitational PE available to kinetic + spring energy is

$Mgx = K + \frac12 kx^2.$

At $x=\frac{x_0}{2}$, with $x_0=\frac{2Mg}{k}$,

$x=\frac{Mg}{k}.$

Then

$Mg\left(\frac{Mg}{k}\right)=K+\frac12 k\left(\frac{Mg}{k}\right)^2$

$\frac{M^2g^2}{k}=K+\frac{M^2g^2}{2k}$

$K=\frac{M^2g^2}{2k}.$

If the middle block speed is $v$, then by string constraints the other two blocks also have speed magnitude $v$.

So total kinetic energy is