A block of mass $2\mathrm{M}$ is attached to a massless spring with spring-constant $\mathrm{k}.$ This block is connected to two other blocks of masses $\mathrm{M}$ and $2\mathrm{M}$ using two massless pulleys and strings. The accelerations of the blocks are ${\mathrm{a}}_1,{\mathrm{a}}_2$ and ${\mathrm{a}}_3$ as shown in figure. The system is released from rest with the spring in its unstretched state. The maximum extension of the spring is ${\mathrm{x}}_0.$ Which of the following option(s) is/are correct?
[ $\mathrm{g}$ is the acceleration due to gravity. Neglect friction]

• A

  ${\mathrm{x}}_0=\frac{4\mathrm{M}\mathrm{g}}{\mathrm{k}}$
• B

  When spring achieves an extension of $\frac{{\mathrm{x}}_0}{2}$ for the first time, the speed of the block connected to the spring is $3\mathrm{g}\sqrt{\frac{\mathrm{M}}{5\mathrm{k}}}$
• Correct answer
  C

  ${\mathrm{a}}_2-{\mathrm{a}}_1={\mathrm{a}}_1-{\mathrm{a}}_3$
• D

  At an extension of $\frac{{\mathrm{x}}_0}{4}$ of the spring, the magnitude of acceleration of the block connected to the spring is $\frac{3\mathrm{g}}{10}$

Solution

Using string constraint
$a_1=\frac{a_2+a_3}{2}$
$\Rightarrow 2a_1=a_2+a_3$
$a_1-a_3=a_2-a_1\Rightarrow \left(\mathrm{C}\right)$ is correct.
for other options use $m$ equivalent

Equivalent mass
Or Reduced mass
$m_{eq}=\frac{2m_1\times m_2}{m_1+m_2}$ (for a two-mass system)
$=\frac{2\left(2m\right)\times m}{2m+m}=\frac{4m}{3}$
$\Rightarrow T=\left(\frac{4m}{3}\right).g$
$\therefore 2T=2.\frac{4mg}{3}=\left(\frac{8m}{3}\right)g$
$\Rightarrow m_{eq}=\frac{8m}{3}$

Using mechanical energy conservation
$\Rightarrow \frac{1}{2}k{x}_0^2=\frac{8mg}{3}{x}_0$
$x_0=\frac{16mg}{3k}\Rightarrow \left(\mathrm{A}\right)$ is incorrect
Option $\left(\mathrm{B}\right)\Rightarrow V_{\frac{x_0}{2}}=V_{\mathrm{m}\mathrm{a}\mathrm{x}}=\frac{x_0}{2}\omega =\frac{x_0}{2}\sqrt{\frac{k}{2m+\frac{8m}{3}}}=\frac{x_0}{2}\sqrt{\frac{3k}{14m}}=g\sqrt{\frac{32}{21k}}$
Option $\left(\mathrm{D}\right)\Rightarrow a_{\frac{x_0}{4}}=\frac{x_0}{4}{\omega }^2=\frac{x_0}{4}\frac{3k}{14m}=\frac{3k{x}_0}{42m}=\frac{8g}{21}$
For a simple harmonic motion $x_0\Rightarrow$ maximum displacement from extreme to extreme
$\Rightarrow x_0=2A\Rightarrow A=\frac{x_0}{2}$
and $\omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{3k}{14m}}$
$\therefore v_{\mathrm{m}\mathrm{a}\mathrm{x}}=A.\omega$
and $a={\omega }^2.x$

Hence, the correct answer is option: 'C'.