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 x₀ , are oscillating with amplitude a and angular frequency ω . Thus, their positions at time t are given by x₁(t) = (x₀ + d) + a sin ωt and x₂(t) = (x₀ − d) − a sin ωt , respectively, where d > 2a . Particle 3 of mass m moves towards this system with speed u₀ = aω/2 , and undergoes instantaneous elastic collision with particle 2, at time t₀ . Finally, particles 1 and 2 acquire a center of mass speed v_cm and oscillate with amplitude b and the same angular frequency ω .

If the collision occurs at time t₀ = π/(2ω) , then the value of 4b ² /a ² will be

The correct answer is 4.25 (or 17/4).

Analysis of the Given Data:
From the problem description and the attached diagram, we have three identical particles of mass m on a horizontal frictionless surface:
• Particle 1 and Particle 2 are connected by a spring.
• Particle 3 moves to the right with velocity $u_0=\frac{a\omega }{2}$ towards particle 2.
• Before collision, the equations of motion for particles 1 and 2 are:
$x_1\left(t\right)=(x_0+d)+a\sin \left(\omega t\right)$
$x_2\left(t\right)=(x_0-d)-a\sin \left(\omega t\right)$

Step 1: Velocities of particles just before collision
Differentiating the positions with respect to time t, we get the velocities:
$v_1\left(t\right)=\frac{d{x}_1}{dt}=a\omega \cos \left(\omega t\right)$
$v_2\left(t\right)=\frac{d{x}_2}{dt}=-a\omega \cos \left(\omega t\right)$

At the time of collision $t_0=\frac{\pi }{2\omega }$:
$\omega t_0=\frac{\pi }{2}$
Therefore,
$\cos \left(\omega t_0\right)=\cos \left(\frac{\pi }{2}\right)=0$
$\sin \left(\omega t_0\right)=\sin \left(\frac{\pi }{2}\right)=1$

Hence, just before the collision at $t=t_0$:
• $v_1=0$
• $v_2=0$
• The compression/elongation of the spring is at maximum amplitude, meaning the stored potential energy in the spring is $U=\frac{1}{2}k{\left(2a\right)}^2$ or in terms of reduced mass $\mu =\frac{m}{2}$, the oscillation energy is $E_{\text{osc}}=\frac{1}{2}\mu {\omega }^2{\left(2a\right)}^2=m{\omega }^2{a}^2$.

Step 2: Analysis of the Elastic Collision
Particle 3 of mass m traveling at velocity $u_0=\frac{a\omega }{2}$ collides head-on elastically with particle 2 (which is at rest at $t=t_0$).
Since particles 2 and 3 have equal masses, an instantaneous elastic collision causes them to exchange their velocities.
• Velocity of particle 3 after collision: $v_3^{\text{'}}=0$
• Velocity of particle 2 immediately after collision: $v_2^{\text{'}}=u_0=\frac{a\omega }{2}$
• Velocity of particle 1 remains unchanged: $v_1^{\text{'}}=0$

Step 3: Post-collision motion of system (Particles 1 and 2)
Immediately after the collision, the new velocities of particles 1 and 2 are:
$v_1^{\text{'}}=0$, and $v_2^{\text{'}}=\frac{a\omega }{2}$

The velocity of the center of mass of particles 1 and 2 after collision is:
$v_{\text{cm}}=\frac{m{v}_1^{\text{'}}+m{v}_2^{\text{'}}}{2m}=\frac{0+\frac{a\omega }{2}}{2}=\frac{a\omega }{4}$

Step 4: Finding the new amplitude of oscillation b
The total energy of the 2-particle system immediately after collision consists of center of mass kinetic energy and internal oscillation energy:
$E_{\text{total}}=E_{\text{cm}}+E_{\text{osc, new}}$

Alternatively, calculating kinetic energy + spring potential energy directly:
• Total Kinetic Energy after collision:
$K=\frac{1}{2}m{\left(v_1^{\text{'}}\right)}^2+\frac{1}{2}m{\left(v_2^{\text{'}}\right)}^2=0+\frac{1}{2}m{\left(\frac{a\omega }{2}\right)}^2=\frac{1}{8}m{a}^2{\omega }^2$

• Potential Energy stored in the spring at $t=t_0$:
Since the relative position between the two particles at $t=t_0$ is unchanged by the instantaneous collision, the potential energy remains:
$U=m{a}^2{\omega }^2$

• Total internal oscillation energy after collision:
$E_{\text{osc, new}}=K+U-E_{\text{cm}}$
Where $E_{\text{cm}}=\frac{1}{2}\left(2m\right)v_{\text{cm}}^2=m{\left(\frac{a\omega }{4}\right)}^2=\frac{1}{16}m{a}^2{\omega }^2$

Substituting into the expression for new oscillation energy:
$E_{\text{osc, new}}=\frac{1}{8}m{a}^2{\omega }^2+m{a}^2{\omega }^2-\frac{1}{16}m{a}^2{\omega }^2=\left(\frac{1}{8}+1-\frac{1}{16}\right)m{a}^2{\omega }^2=\frac{17}{16}m{a}^2{\omega }^2$

The new amplitude of oscillation of each particle is b (so relative oscillation amplitude is 2b):
$E_{\text{osc, new}}=m{b}^2{\omega }^2$

Equating the two expressions for $E_{\text{osc, new}}$:
$m{b}^2{\omega }^2=\frac{17}{16}m{a}^2{\omega }^2$
$\frac{b^2}{a^2}=\frac{17}{16}$

Multiplying by 4 on both sides gives:
$\frac{4b^2}{a^2}=4\times \frac{17}{16}=\frac{17}{4}=4.25$

Thus, the required value of $\frac{4b^2}{a^2}$ is 4.25.