Spring Mass Pulley System Acceleration and Maximum Extension

To determine which option(s) is/are correct, we analyze the system using kinematics and dynamics.

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1. Kinematic Constraint (Option D)

Let $x_1$ be the horizontal displacement of the block of mass $2M$ connected to the spring.
Since string 1 is inextensible, the downward displacement of the lower (movable) pulley is equal to $x_1$. Thus, the downward acceleration of the lower pulley is $a_P = a_1$.

Let $y_2$ and $y_3$ be the downward displacements of the blocks of mass $M$ and $2M$ attached to the lower pulley, respectively. Their downward accelerations are $a_2$ and $a_3$.

The total length of string 2 passing over the lower pulley is: $L_2 = (y_2 - y_P) + (y_3 - y_P) = y_2 + y_3 - 2y_P$

Differentiating this constraint equation twice with respect to time: $\frac{d^2 y_2}{dt^2} + \frac{d^2 y_3}{dt^2} - 2\frac{d^2 y_P}{dt^2} = 0$ $a_2 + a_3 - 2a_1 = 0 \implies a_2 - a_1 = a_1 - a_3$

Thus, Option D is correct.

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2. Dynamic Equations and Maximum Extension (Option A)

Let $T_1$ be the tension in string 1 attached to the block of mass $2M$, and $T_2$ be the tension in string 2 holding masses $M$ and $2M$.

Since the lower pulley is massless: $T_1 = 2T_2$

The equations of motion for the suspended blocks are:

1. For mass $M$: $M g - T_2 = M a_2 = M(a_1 - a_r)$
2. For mass $2M$: $2M g - T_2 = 2M a_3 = 2M(a_1 + a_r)$

where $a_r$ is the acceleration of the blocks relative to the lower pulley.

Subtracting the equation for mass $M$ from the equation for mass $2M$: $M g = 2M(a_1 + a_r) - M(a_1 - a_r) = M a_1 + 3M a_r \implies a_r = \frac{g - a_1}{3}$

Substituting $a_r$ back into the force equation for mass $M$: $T_2 = M g - M(a_1 - a_r) = M(g - a_1) + M\left(\frac{g - a_1}{3}\right) = \frac{4}{3} M (g - a_1)$

Thus, tension $T_1$ is: $T_1 = 2T_2 = \frac{8}{3} M (g - a_1)$

Now, applying Newton's second law to the block of mass $2M$ on the horizontal surface (with displacement $x$ and acceleration $a_1 = \ddot{x}$): $T_1 - k x = 2M a_1$ $\frac{8}{3} M (g - a_1) - k x = 2M a_1$ $\frac{8}{3} M g - k x = \left(2M + \frac{8}{3}M\right) a_1 = \frac{14}{3} M a_1$

Solving for acceleration $a_1$: $a_1 = \frac{8 M g - 3 k x}{14 M}$

This equation represents Simple Harmonic Motion with an equilibrium extension $x_{\text{eq}}$ where $a_1 = 0$: $8 M g - 3 k x_{\text{eq}} = 0 \implies x_{\text{eq}} = \frac{8 M g}{3 k}$

Since the system is released from rest at $x = 0$, the maximum extension $x_0$ is: $x_0 = 2 x_{\text{eq}} = \frac{16 M g}{3 k}$

Thus, Option A is incorrect.

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3. Speed of the Block at $x = x_0 / 2$ (Option B)

At $x = \frac{x_0}{2} = x_{\text{eq}} = \frac{8 M g}{3 k}$, the speed $v_1$ of the block attached to the spring reaches its maximum value: $v_1 = \omega A$

where: $\omega = \sqrt{\frac{3 k}{14 M}} \quad \text{and} \quad A = x_{\text{eq}} = \frac{8 M g}{3 k}$

Substituting these values: $v_1 = \sqrt{\frac{3 k}{14 M}} \cdot \left(\frac{8 M g}{3 k}\right) = 4 g \sqrt{\frac{2 M}{21 k}}$

Thus, Option B is incorrect.

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4. Acceleration at $x = x_0 / 4$ (Option C)

At $x = \frac{x_0}{4} = \frac{4 M g}{3 k}$, the magnitude of the acceleration of the block connected to the spring is: $a_1 = \frac{8 M g - 3 k \left(\frac{4 M g}{3 k}\right)}{14 M} = \frac{8 M g - 4 M g}{14 M} = \frac{4 M g}{14 M} = \frac{2 g}{7}$

Thus, Option C is incorrect.

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Conclusion

The only correct option is D.