1. Analyze the initial state of the system (particles 1 and 2)

The positions of particles 1 and 2 are given by: $x_1(t) = (x_0 + d) + a \sin(\omega t)$ $x_2(t) = (x_0 - d) - a \sin(\omega t)$

Their velocities are found by differentiating the positions with respect to time: $v_1(t) = \frac{dx_1}{dt} = a \omega \cos(\omega t)$ $v_2(t) = \frac{dx_2}{dt} = -a \omega \cos(\omega t)$

The center of mass (CM) of this two-particle system is initially at rest at $x_0$. The motion described is an oscillation about this CM. The amplitude of oscillation for each particle is $a$.

The total energy of this oscillation, $E_{\text{osc, initial}}$, can be calculated at the point of maximum kinetic energy (when potential energy is zero, at $t=0$). $E_{\text{osc, initial}} = K_{\text{max}} = \frac{1}{2} m v_1(0)^2 + \frac{1}{2} m v_2(0)^2 = \frac{1}{2} m (a\omega)^2 + \frac{1}{2} m (-a\omega)^2 = m a^2 \omega^2$. Alternatively, the energy can be calculated at maximum potential energy (when kinetic energy is zero). This occurs when $\cos(\omega t) = 0$, for instance at $t = \pi/(2\omega)$. At this point, the extension of the spring from its mean length is maximum.

2. Determine the state of the system just before the collision

The collision occurs at time $t_0 = \frac{\pi}{2\omega}$. At this time: $\sin(\omega t_0) = \sin(\frac{\pi}{2}) = 1$ $\cos(\omega t_0) = \cos(\frac{\pi}{2}) = 0$

The velocities of particles 1 and 2 just before the collision are: $v_1(t_0) = a \omega \cos(\frac{\pi}{2}) = 0$ $v_2(t_0) = -a \omega \cos(\frac{\pi}{2}) = 0$

Both particles are momentarily at rest at their maximum displacement. At this instant, the kinetic energy of the oscillation is zero, and the potential energy stored in the spring is at its maximum, equal to the total oscillation energy: $U_{\text{spring}}(t_0) = E_{\text{osc, initial}} = m a^2 \omega^2$.

3. Analyze the collision

Particle 3 (mass $m$) with velocity $v_3 = -u_0 = -a\omega/2$ (assuming it moves in the negative x-direction) collides elastically with particle 2 (mass $m$), which is at rest ($v_2 = 0$). For an elastic collision between two particles of equal mass, they exchange their velocities.

Let $v'_2$ and $v'_3$ be the velocities of particles 2 and 3 just after the collision. $v'_2 = v_3 = -a\omega/2$ $v'_3 = v_2 = 0$

The collision is instantaneous, so the velocity of particle 1 does not change. Its velocity just after the collision is $v'_1 = v_1(t_0) = 0$.

4. Determine the state of the system (1 and 2) just after the collision

Immediately after the collision, the velocities of particles 1 and 2 are: $v'_1 = 0$ $v'_2 = -a\omega/2$

The positions of the particles have not changed during the instantaneous collision, so the potential energy stored in the spring is the same as just before the collision: $U'_{\text{spring}} = U_{\text{spring}}(t_0) = m a^2 \omega^2$.

5. Calculate the new energy and momentum of the system (1 and 2)

The total kinetic energy of the system (1+2) just after the collision is: $K'_{\text{total}} = \frac{1}{2} m (v'_1)^2 + \frac{1}{2} m (v'_2)^2 = \frac{1}{2} m (0)^2 + \frac{1}{2} m (\frac{-a\omega}{2})^2 = \frac{1}{8} m a^2 \omega^2$.

The total energy of the system (1+2) is the sum of its kinetic and potential energies: $E'_{\text{total}} = K'_{\text{total}} + U'_{\text{spring}} = \frac{1}{8} m a^2 \omega^2 + m a^2 \omega^2 = \frac{9}{8} m a^2 \omega^2$.

The total momentum of the system (1+2) after the collision is: $P'_{\text{total}} = m v'_1 + m v'_2 = m(0) + m(\frac{-a\omega}{2}) = -\frac{1}{2} m a \omega$.

6. Analyze the new motion (CM motion and oscillation)

The system (1+2) now has a non-zero momentum, so its center of mass will move. The velocity of the new center of mass is: $v_{\text{cm}} = \frac{P'_{\text{total}}}{M_{\text{total}}} = \frac{-ma\omega/2}{2m} = -\frac{a\omega}{4}$.

The kinetic energy of the center of mass motion is: $K_{\text{cm}} = \frac{1}{2} M_{\text{total}} v_{\text{cm}}^2 = \frac{1}{2} (2m) (-\frac{a\omega}{4})^2 = m (\frac{a^2\omega^2}{16}) = \frac{1}{16} m a^2 \omega^2$.

The total energy of the system can be expressed as the sum of the kinetic energy of the CM and the energy of oscillation about the CM, $E'_{\text{osc}}$: $E'_{\text{total}} = K_{\text{cm}} + E'_{\text{osc}}$

So, the new oscillation energy is: $E'_{\text{osc}} = E'_{\text{total}} - K_{\text{cm}} = \frac{9}{8} m a^2 \omega^2 - \frac{1}{16} m a^2 \omega^2 = (\frac{18}{16} - \frac{1}{16}) m a^2 \omega^2 = \frac{17}{16} m a^2 \omega^2$.

7. Relate the new oscillation energy to the new amplitude $b$

The energy of oscillation is proportional to the square of the amplitude. For the new oscillation with amplitude $b$ and angular frequency $\omega$, the energy is: $E'_{\text{osc}} = m b^2 \omega^2$.

Equating the two expressions for the new oscillation energy: $m b^2 \omega^2 = \frac{17}{16} m a^2 \omega^2$ $b^2 = \frac{17}{16} a^2$ $\frac{b^2}{a^2} = \frac{17}{16}$.

8. Calculate the final value

The question asks for the value of $\frac{4b^2}{a^2}$. $\frac{4b^2}{a^2} = 4 \times \frac{17}{16} = \frac{17}{4} = 4.25$.