JEE Advanced 2019 — Paper 2 — Question 2

In the frame of pulley B, the hanging masses have accelerations :

$M \rightarrow\left(a_{2}-a_{1}\right), 2 M \rightarrow\left(a_{3}-a_{1}\right):$ downward.

$\therefore\left(\mathrm{a}_{2}-\mathrm{a}_{1}\right)=-\left(\mathrm{a}_{3}-\mathrm{a}_{1}\right)$ [constant]

Assuming that the extension of the spring is x

We consider the FBD of A :

2M. $\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=2 \mathrm{~T}-\mathrm{kx}$

where $\mathrm{a}_{1} \equiv \frac{\mathrm{~d}^{2} \mathrm{x}}{\mathrm{dt}^{2}} …(i)$

and the FBD of the rest of the system in the frame of pulley B :

Upward acceleration of block $M$ w.r.t. the pulley $B=$ Downward acceleration of block 2 M w.r.t the pulley

$\frac{T-M\left(g-a_{1}\right)}{M}=\frac{2 M\left(g-a_{1}\right)-T}{2 M}$

$$\begin{gathered} \Rightarrow \mathrm{T}=\frac{4 \mathrm{M}}{3}\left(\mathrm{~g}-\mathrm{a}_{1}\right) …(ii) \end{gathered}$$

Substituting in equation (i), we get

$$2 \mathrm{M} . \mathrm{a}_{1}=\frac{8 \mathrm{M}}{3}\left(\mathrm{~g}-\mathrm{a}_{1}\right)-\mathrm{kx}$$

or $\quad \frac{14 \mathrm{M}}{3} \mathrm{a}_{1}=\frac{8 \mathrm{Mg}}{3}-k x …(iii)$

This is the equation of SHM Maximum extension $=2 \times$ Amplitude i.e.

$x_{0}=2 \times \frac{8 \mathrm{Mg}}{3 \mathrm{k}}$

At $\frac{\mathrm{x}_{0}}{4}$, acceleration is easily found from equation (iii):

$\frac{14 M}{3} a_{1}=\frac{8 M g}{3}-\frac{4 M g}{3}$

$a_{1}=\frac{2 g}{7}$

At $\frac{x_{0}}{2}$, speed of the block $(2 M)=\omega \times$ amplitude

$=\sqrt{\frac{3 \mathrm{k}}{14 \mathrm{M}}} \times \frac{8 \mathrm{Mg}}{3 \mathrm{k}}$

$\therefore$ option (A) is correct