Question

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 accelerations of the blocks are a₁, a₂ and a₃ as shown in figure. The system is released from rest with the spring in its unstretched state. The maximum extension of the spring is x₀. Which of the following option(s) is/are correct ? [g is the acceleration due to gravity. Neglect friction]

• A. $x_{0}=\frac{4 Mg}{k}$
• B. When spring achieves an extension of $\frac{x_{0}}{2}$ for the first time, the speed of the block connected to the spring is $3g\sqrt{\frac{M}{5k}}$
• C. a₂ - a₁ = a₁ - a₃
• D. 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 1

Answer: (C)

$T = \frac{2\left(2m\right)\left(m\right)}{3m}\left(g-a_{1}\right)$
$= \frac{4m}{3}\left(g-a_{1}\right)$
$\frac{8m}{3}\left(g-a_{1}\right)-kx = 2ma_{1}$
$\frac{8Mg}{3}-\frac{8Ma_{1}}{3}-kx = 2ma_{1}$
$\frac{8Mg}{3}-kx = \frac{14ma_{1}}{3}$
$\frac{8Mg-3kx}{14m} = a_{1}$
$a_{1} = \frac{8Mg-3kx}{14m}$
$\frac{vdv}{dx} = \left(\frac{8Mg}{14m}-\frac{3kx}{14m}\right)$
$\int vdv = \frac{1}{14m}\int\left(8Mg-3kx\right)dx$
for max elongation
$0 = \frac{1}{14m} \int\limits^{x_0}_{0}\left(8Mg-3kx\right)dx$
$= \frac{1}{14m}\left(8Mgx_{0}-\frac{3kx^{2}_{0}}{2}\right)$
$8Mgx_{0} = \frac{3kx^{2}_{0}}{2}$
$x_{0} = \frac{16Mg}{3k}$
$at \,x = \frac{x_{0}}{2}$
$\int\limits^{v}_{0} vdv = \frac{1}{14m}\int\limits^{x_0/2}_{0} \left(8Mg-3kx\right)dx$
$\frac{v^{2}}{2} = \frac{1}{14m}\left(\frac{8Mgx_{0}}{2}-\frac{3kx^{2}_{0}}{2\times4}\right)$
$v^{2} = \frac{1}{7m}\left(\frac{8Mg}{2}\times\frac{16Mg}{3x}-\frac{3x}{8}\times\frac{16M^{2}g^{2}}{3x\times3x}\right)$
$= \frac{1}{7m}\left(\frac{64M^{2}g^{2}}{3x}-\frac{2M^{2}g^{2}}{3x}\right)$
$v^{2} = \frac{62Mg^{2}}{21k}$
For acc. $2a_{1} = a_{2} + a_{3}$ therefore
$a_{2} - a_{1} = a_{1} - a_{3}$
$a_{1} = \frac{8Mg-3kx_{0}/4}{14m}$
$= \frac{8g}{14}-\frac{3kx_{0}}{14m\times4}$
$= \frac{8g}{14}-\frac{3x}{14m\times4}\times\frac{16Mg}{3x}$
$= \frac{8g}{14}-\frac{4g}{14}$
$= \frac{4g}{14} = \frac{2g}{7}$
OR
$\frac{8mg}{3}-\frac{8m}{3}a_{1}-kx = 2ma_{1}$
$\frac{14m}{3}a_{1} = -k\left[x-\frac{8mg}{3k}\right]$
$a_{1} = -\frac{3k}{14m}\left[x-\frac{8mg}{3k}\right]....\left(i\right)$
that means, block 2m (connected with the spring) will perform SHM about $x_{1} =\frac{8mg}{3k}$ therefore.
maximum elongation in the spring $x_{0} = 2x_{1} = \frac{16mg}{3k}$
on comparing equation (1) with
$a = \omega^{2}\left(x-x_{0}\right)$
$\omega = \sqrt{\frac{3k}{14m}}$
at $\left(\frac{x_{0}}{2}\right)$, block will be passing through its mean position therefore at mean position
$v_{0} = A\omega = \frac{8mg}{3k} \sqrt{\frac{3k}{14m}}$
At, $\frac{x_{0}}{4}\,\Rightarrow \,x = \frac{A}{2}$
$\therefore \,a_{cc} = -\frac{A}{2} \omega^{2}$
$= -\frac{4mg}{3k}. \frac{3h}{14m} = -\frac{2g}{7}$