JEE Advanced 2024 — Paper 2 — Question 42

Two particles, 1 and 2 , each of mass $m$, are connected by a massless spring, and are on a horizontal frictionless plane, as shown in the figure. Initially, the two particles, with their center of mass at $x_{0}$, are oscillating with amplitude a and angular frequency $\omega$. Thus, their positions at time $t$ are given by $x_{1}(t)=\left(x_{0}+d\right)+a \sin \omega t$ and $x_{2}(t)=\left(x_{0}-d\right)-a \sin \omega t$, respectively, where $d>2 a$. Particle 3 of mass $m$ moves towards this system with speed $u_{0}=a \omega / 2$, and undergoes instantaneous elastic collision with particle 2, at time $t_{0}$. Finally, particles 1 and 2 acquire a center of mass speed $\mathrm{v}_{\mathrm{cm}}$ and oscillate with amplitude b and the same angular frequency $\omega$.

If the collision occurs at time $t_{0}=\pi /(2 \omega)$, then the value of $4 b^{2} / a^{2}$ will be