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. $x_0=\frac{4Mg}{k}$
• 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 $\frac{3g}{10}$

Answer 1

Answer: (C)

$T=\frac{2\left(2m\right)\left(m\right)}{3m}\left(g-a_1\right)$
$=\frac{4m}{3}\left(g-a_1\right)$
$\frac{8m}{3}\left(g-a_1\right)-kx=2m{a}_1$
$\frac{8Mg}{3}-\frac{8M{a}_1}{3}-kx=2m{a}_1$
$\frac{8Mg}{3}-kx=\frac{14m{a}_1}{3}$
$\frac{8Mg-3kx}{14m}=a_1$
$a_1=\frac{8Mg-3kx}{14m}$
$\frac{vdv}{dx}=\left(\frac{8Mg}{14m}-\frac{3kx}{14m}\right)$
$\int vdv=\frac{1}{14m}\int \left(8Mg-3kx\right)dx$
for max elongation
$0=\frac{1}{14m}\underset{0}{\overset{x_0}{\int }}\left(8Mg-3kx\right)dx$
$=\frac{1}{14m}\left(8Mg{x}_0-\frac{3k{x}_0^2}{2}\right)$
$8Mg{x}_0=\frac{3k{x}_0^2}{2}$
$x_0=\frac{16Mg}{3k}$
$atx=\frac{x_0}{2}$
$\underset{0}{\overset{v}{\int }}vdv=\frac{1}{14m}\underset{0}{\overset{x_0/2}{\int }}\left(8Mg-3kx\right)dx$
$\frac{v^2}{2}=\frac{1}{14m}\left(\frac{8Mg{x}_0}{2}-\frac{3k{x}_0^2}{2\times 4}\right)$
$v^2=\frac{1}{7m}\left(\frac{8Mg}{2}\times \frac{16Mg}{3x}-\frac{3x}{8}\times \frac{16M^2{g}^2}{3x\times 3x}\right)$
$=\frac{1}{7m}\left(\frac{64M^2{g}^2}{3x}-\frac{2M^2{g}^2}{3x}\right)$
$v^2=\frac{62M{g}^2}{21k}$
For acc. $2a_1=a_2+a_3$ therefore
$a_2-a_1=a_1-a_3$
$a_1=\frac{8Mg-3k{x}_0/4}{14m}$
$=\frac{8g}{14}-\frac{3k{x}_0}{14m\times 4}$
$=\frac{8g}{14}-\frac{3x}{14m\times 4}\times \frac{16Mg}{3x}$
$=\frac{8g}{14}-\frac{4g}{14}$
$=\frac{4g}{14}=\frac{2g}{7}$
OR
$\frac{8mg}{3}-\frac{8m}{3}{a}_1-kx=2m{a}_1$
$\frac{14m}{3}{a}_1=-k\left[x-\frac{8mg}{3k}\right]$
$a_1=-\frac{3k}{14m}\left[x-\frac{8mg}{3k}\right]....\left(i\right)$
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, $\frac{x_0}{4}\Rightarrow x=\frac{A}{2}$
$\therefore a_{cc}=-\frac{A}{2}{\omega }^2$
$=-\frac{4mg}{3k}.\frac{3h}{14m}=-\frac{2g}{7}$