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 $a_{1}, a_{2} a n d a_{3}$ 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]

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When spring achieves an extension of $\frac{x_{0}}{2}$ for the first time, the speed of the block connected to the spring is 3g $\sqrt{\frac{M}{5 k}}$

$a_{2} - a_{1} = a_{1} - a_{3}$

$x_{0} = \frac{4 M g}{k}$

At an extension of $\frac{x_{0}}{4}$ of the spring, the magnitude of acceleration of the block connected to the spring is $\frac{3 g}{10}$

answer is D.

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Detailed Solution

Using constraint relation

$a_{2} - a_{1} = a_{1} - a_{3}$

Also,

$\begin{array}{l} 2 Mg - T = 2 {Ma}_{3} \\ M g - T = M a_{2} \\ 2 T - k X = 2 M a_{1} \\ Solving, \\ a_{1} = \frac{4 g}{7} - \frac{3 k x}{14 M} \\ \Rightarrow a_{1} = - \frac{3 k}{14 M} [x - \frac{8 M g}{3 k}] \end{array}$

$This equation represents SHM of block 2M attached with spring.$

${Comparing with a=-ω}^{2} (x - x_{m e a n})$

$\begin{array}{l} \Rightarrow ω = \sqrt{\frac{3 k}{14 M}} \\ and x_{mean} = \frac{8 M g}{3 k} \\ Since, a_{1} = 0, at two points x = 0 and x = x_{0}, \\ \Rightarrow Amplitude = \frac{x_{0}}{2} \\ \Rightarrow \frac{x_{0}}{2} = \frac{8 M g}{3 k} \\ \Rightarrow x_{0} = \frac{16 Mg}{3 k} \end{array}$

$\begin{array}{l} Now, put x = \frac{x_{0}}{4} = \frac{4 M g}{3 k} \\ \Rightarrow a_{1} = \frac{4 g}{7} - \frac{3 k}{14 M} \cdot \frac{4 M g}{3 k} \\ \Rightarrow a_{1} = \frac{2 g}{7} \end{array}$