A block of mass $2M$ is attached to a massless spring with spring-constant $k$. This block is connected to two other blocks of masses $M$ and $2M$ using two massless pulleys and strings. The acceleration of the blocks are $a_1$, $a_2$ and $a_3$ as shown in the figure. The system is released from rest with the spring in its unstretched state. The maximum extension of the spring is $x_0$. Which of the following option(s) is/are correct? [$g$ is the acceleration due to gravity. Neglect friction]

Step 1: Establish the string constraint relation.

Let $x$ be the displacement of the block of mass $2M$ (attached to the spring) to the right. As this block moves right by $x$, the movable pulley $P_2$ moves down by $x$. Let $y_M$ and $y_{2M}$ be the downward displacements of the blocks of mass $M$ and $2M$ respectively.

The length of the string passing over $P_2$ is constant, which gives:

$$y_M + y_{2M} = 2x$$

Differentiating twice with respect to time:

$$a_2 + a_3 = 2a_1$$

Rearranging this relation:

$$a_2 - a_1 = a_1 - a_3$$

This shows that Option 4 is correct.

Step 2: Analyze the forces and tensions.

Let $T$ be the tension in the string connecting the blocks of mass $M$ and $2M$. Since the pulley $P_2$ is massless, the tension $T_u$ in the upper string connected to the block of mass $2M$ on the table is:

$$T_u = 2T$$

For the block of mass $2M$ on the table, the equation of motion is:

$$2M a_1 = T_u - kx$$ $$2M a_1 = 2T - kx$$

Step 3: Determine the tension $T$ using the frame of the movable pulley.

In the accelerating frame of pulley $P_2$ (which has downward acceleration $a_1$), the effective acceleration due to gravity is:

$$g_{\text{eff}} = g - a_1$$

The equations of motion for the hanging blocks of mass $M$ and $2M$ relative to the pulley are:

$$M(g - a_1) - T = M a_{\text{rel}}$$ $$2M(g - a_1) - T = -2M a_{\text{rel}}$$

Multiplying the first equation by 2 and adding it to the second equation to eliminate $a_{\text{rel}}$:

$$4M(g - a_1) - 3T = 0$$ $$T = \frac{4M}{3}(g - a_1)$$

Step 4: Find the acceleration $a_1$ as a function of $x$.

Substitute the expression for $T$ into the equation of motion for the block on the table:

$$2M a_1 = \frac{8M}{3}(g - a_1) - kx$$ $$\frac{14M}{3} a_1 = \frac{8Mg}{3} - kx$$ $$a_1 = \frac{4g}{7} - \frac{3k}{14M}x$$

Step 5: Determine the maximum extension $x_0$.

Rewrite the equation of motion in the standard form of simple harmonic motion:

$$a_1 = -\frac{3k}{14M}\left(x - \frac{8Mg}{3k}\right)$$

This represents simple harmonic motion with:

$$\omega^2 = \frac{3k}{14M}$$ $$x_{\text{eq}} = \frac{8Mg}{3k}$$

Since the system is released from rest at $x = 0$, the amplitude of oscillation is $A = x_{\text{eq}}$. The maximum extension $x_0$ is:

$$x_0 = 2x_{\text{eq}} = \frac{16Mg}{3k}$$

Since $\frac{16Mg}{3k} \neq \frac{4Mg}{k}$, Option 2 is incorrect.

Step 6: Evaluate the acceleration at $x = \frac{x_0}{4}$.

Substitute $x = \frac{x_0}{4} = \frac{4Mg}{3k}$ into the expression for $a_1$:

$$a_1 = \frac{4g}{7} - \frac{3k}{14M}\left(\frac{4Mg}{3k}\right)$$ $$a_1 = \frac{2g}{7}$$

Since $\frac{2g}{7} \neq \frac{3g}{10}$, Option 1 is incorrect.

Step 7: Evaluate the speed at $x = \frac{x_0}{2}$.

At $x = \frac{x_0}{2} = x_{\text{eq}}$, the block is at its equilibrium position, where its speed $v$ is maximum:

$$v = \omega x_{\text{eq}}$$ $$v = \sqrt{\frac{3k}{14M}} \cdot \frac{8Mg}{3k}$$ $$v = g\sqrt{\frac{32M}{21k}}$$

Since $g\sqrt{\frac{32M}{21k}} \neq 3g\sqrt{\frac{M}{5k}}$, Option 3 is incorrect.

Answer: The only correct option is $a_2 - a_1 = a_1 - a_3$.