Pulleys and masses connected to a spring

JEE Advanced 2019 Paper 2, Question 2

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 $2 M$ 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]

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$x_{0}=\frac{4 M g}{k}$
When spring achieves an extension of $\frac{x_{0}}{2}$ for the first time, the speed of the block connected to the spring is $3 g \sqrt{\frac{M}{5 k}}$
At an extension of $\frac{x_{0}}{4}$ of the spring, the magnitude of acceleration of the block connected to the spring is $\frac{3 g}{10}$
$a_{2}-a_{1}=a_{1}-a_{3}$

Solution

WARNING: I am unsure whether I’ve made some mistake in my solutions, so please try this problem yourself and let me know if you get a different answer.

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From the force diagram, we have

(1) $\begin{equation*} \begin{aligned} T_1 - kx &= 2M a_1 \\ Mg - T &= M a_2 \\ 2Mg - T &= 2M a_3 \\ T_1 &= 2 T \end{aligned} \end{equation*}$

To relate the accelerations we use the constraint that the lengths of the strings connecting the masses are constant. Then,

(2) $\begin{equation*} (y_p - y_2) + (y_p - y_3) + \pi R = l_2 , \end{equation*}$

where $R$ is the radius of the second pulley and $l_2$ is the length of the string around it. Differentiating this expression twice w.r.t. we get

(3) $\begin{equation*} \ddot{y}_p - \ddot{y}_2 - \ddot{y}_3 = 0 . \end{equation*}$

We’ve also marked the initial position of the block attached to the spring by $0$ with its displacement $x$ at some later time. Then,

(4) $\begin{equation*} x + y_p + l_1 = {\rm constant} \implies \ddot x + \ddot{y}_p = 0 , \end{equation*}$

Next, we note that $\ddot{x} = a_1, \ddot{y}_2 = -a_2$ and $\ddot{y}_3 = -a_3$ for the directions indicated. Thus, from (3) and (4),

(5) $\begin{equation*} \boxed{2 a_1 = a_2 + a_3} \implies \text{option (D) is correct.} \end{equation*}$

Using the above relation we can eliminate $a_2$ from the equations (1),

(6) $\begin{equation*} \begin{rcases} Mg - T &= M a_2 \\ 2Mg - T &= 2M a_3 \end{rcases} \implies 4 Mg - 3 T = 2 M (a_2 + a_3) \implies T = \frac{4M}{3} (g-a_1) . \end{equation*}$

Keeping in mind that $\ddot{x} = a_1$ , we obtain

(7) $\begin{equation*} 2T - kx = 2M a_1 \implies x = \frac{8Mg}{3k} - \frac{14 M}{3k} \ddot x \end{equation*}$

This equation represents simple harmonic motion about the equilibrium point $x=\frac{8Mg}{3k}$ , where the acceleration $\ddot{x}=0$ . Since the oscillations were started at $x=0$ with zero initial velocity, the solution to the differential equation (7) is

(8) $\begin{equation*} x(t) = \frac{8Mg}{3k} - \frac{8Mg}{3k} \cos (\omega t) , \end{equation*}$

with $\omega = \frac{3k}{14M}$ . This is the blue line in the figure below. The maximum extension $x_0$ is corresponds to $\cos(\omega t) = -1$ ,

(9) $\begin{equation*} \boxed{x_0 = \frac{16Mg}{3k}} \implies \text{option (A) is incorrect.} \end{equation*}$

The point $x_0/2$ is just the equilibrium point, where the velocity reaches its maximum possible value $v_{\rm max} = \omega \times {\rm amplitude}$ , which is

(10) $\begin{equation*} \boxed{ v_{\rm max} = \frac{8Mg}{3k} \sqrt{\frac{3k}{14M}} } \implies \text{option (B) is incorrect.} \end{equation*}$

Finally, the acceleration at $x_0/4$ can be calculated directly from (7) by setting $x=x_0/4$ ,

(11) $\begin{equation*} \frac{4Mg}{3k} = \frac{8Mg}{3k} - \frac{14 M}{3k} \ddot x \implies \boxed{\ddot{x} = \frac{2g}{7}} \implies \text{option (C) is incorrect.} \end{equation*}$

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