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]

Rendered by QuickLaTeX.com

  1. x_{0}=\frac{4 M g}{k}
  2. 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}}
  3. 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}
  4. a_{2}-a_{1}=a_{1}-a_{3}


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.

Rendered by QuickLaTeX.com

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*}

Rendered by QuickLaTeX.com

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*}

Leave a Reply