**JEE Advanced 2019 Paper 2, Question 2**

A block of mass is attached to a massless spring with spring-constant . This block is connected to two other blocks of masses and using two massless pulleys and strings. The accelerations of the blocks are and 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 . Which of the following option(s) is/are correct?

[ is the acceleration due to gravity. Neglect friction]

- When spring achieves an extension of for the first time, the speed of the block connected to the spring is
- At an extension of of the spring, the magnitude of acceleration of the block connected to the spring is

**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.

From the force diagram, we have

(1)

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

(2)

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

(3)

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

(4)

Next, we note that and for the directions indicated. Thus, from (3) and (4),

(5)

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

(6)

Keeping in mind that , we obtain

(7)

This equation represents simple harmonic motion about the equilibrium point , where the acceleration . Since the oscillations were started at with zero initial velocity, the solution to the differential equation (7) is

(8)

with . This is the blue line in the figure below. The maximum extension is corresponds to ,

(9)

The point is just the equilibrium point, where the velocity reaches its maximum possible value , which is

(10)

Finally, the acceleration at can be calculated directly from (7) by setting ,

(11)