Question

A block of mass 2$M$ 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 accelerations 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]

• A. At an extension of $\frac{x_0}{4}$ of the spring. The magnitude of acceleration of the block connected to the spring is $\frac{3g}{10}$
• B. $x_0=\frac{4Mg}{k}$
• C. When spring achieves an extension of $\frac{x_0}{2}$ for the first time, the speed of the block connected to the spring is $3g\sqrt{\frac{M}{5k}}$
• D. $a_2-a_1=a_1-a_3$

Answer 1

The correct option is D

$a_2-a_1=a_1-a_3$

By constraint equation,
$2a_1=a_2+a_3$
$a_1-a_3=a_2-a_1$
for other options use $m$ equivalent



We know $\frac{2m_1{m}_2}{m_1{m}_2}=\frac{T}{g}$ for the system shown above.

So, $\frac{T}{g}=\frac{2\left(2m\right)\left(m\right)}{2m+m}=\frac{4m}{3}$

$⟹\frac{2T}{g}=\frac{8m}{3}$

hence $M_{eq.}=\frac{8m}{3}$



At maximum extension the velocity od system shown in the figure is zero. So energy stored in the spring is work done by gravity.

$\frac{1}{2}k{x}_0^2=\frac{8Mg}{3}{x}_0$

$x_0=\frac{16Mg}{3k}$
System will be in S.H.M so $\omega =\sqrt{\frac{k}{2M+\frac{8M}{3}}}$
Amplitude of oscillation $A=\frac{x_o}{2}=\frac{8Mg}{3K}$
At $\frac{x_o}{2}$, body is passing through mean position.
$\Rightarrow V=A\omega =8g\sqrt{\frac{M}{42K}}$
At $x=x_o/4$, $x=\frac{A}{2}$
$a=-\frac{{\omega }^{2}A}{2}=-\frac{2g}{7}$
So, only option D is the correct answer.