If the collision occurs at time 𝑡₀ = 𝜋/(2𝜔), then the value of 4𝑏²/𝑎² will be ________.

The correct answer is 4.25.

Step-by-step Derivation:

1. Velocities of the oscillating particles before the collision:
The position functions of the two particles (each of mass m) connected by a spring are given by:
$x_1\left(t\right)=(x_0+d)+asin\left(\omega t\right)$
$x_2\left(t\right)=(x_0-d)-asin\left(\omega t\right)$

Differentiating these equations with respect to time t gives the velocities of the two particles before the collision:
$v_1\left(t\right)=a\omega cos\left(\omega t\right)$
$v_2\left(t\right)=-a\omega cos\left(\omega t\right)$

2. State of the system at the collision time $t_0=\frac{\pi }{2\omega }$:
Substitute the given collision time $t_0$ into the velocity equations:
$v_1\left(t_0\right)=a\omega cos\left(\frac{\pi }{2}\right)=0$
$v_2\left(t_0\right)=-a\omega cos\left(\frac{\mathrm{��}}{2}\right)=0$

Thus, immediately before the collision, both particles are momentarily at rest in the center-of-mass frame, and the oscillation energy is entirely stored as potential energy in the stretched spring. The total oscillation energy relative to the center of mass is:
$E_{\text{rel}}=m{a}^2{\omega }^2$

3. State of the system immediately after the elastic collision:
A third particle of mass m moving with velocity $u_0=\frac{a\omega }{2}$ collides elastically with particle 2. Since both particles have equal mass m, they completely exchange their velocities upon collision:
- Particle 3 comes to rest.
- Particle 2 gains the velocity of particle 3: $v_2^{\text{'}}=u_0=\frac{a\omega }{2}$.
- Particle 1 is unaffected during the instantaneous collision: $v_1^{\text{'}}=0$.

4. Analyzing the new oscillation:
The new center-of-mass velocity of the two-particle system (particles 1 and 2) is:
$v_{\text{cm}}^{\text{'}}=\frac{v_1^{\text{'}}+v_2^{\text{'}}}{2}=\frac{0+\frac{a\omega }{2}}{2}=\frac{a\omega }{4}$

The relative velocities of the particles in the new center-of-mass frame are:
$v_{\text{1,rel}}^{\text{'}}=0-\frac{a\omega }{4}=-\frac{a\omega }{4}$
$v_{\text{2,rel}}^{\text{'}}=\frac{a\omega }{2}-\frac{a\omega }{4}=\frac{a\omega }{4}$

The relative kinetic energy in the center-of-mass frame immediately after the collision is:
$K_{\text{rel}}^{\text{'}}=\frac{1}{2}m{\left(-\frac{a\omega }{4}\right)}^2+\frac{1}{2}m{\left(\frac{a\omega }{4}\right)}^2=m\frac{a^2{\omega }^2}{16}$

The potential energy of the spring at the instant of collision remains unchanged:
$U^{\text{'}}=E_{\text{rel}}=m{a}^2{\omega }^2$

Thus, the total energy of the new relative oscillation is:
$E_{\text{rel}}^{\text{'}}=K_{\text{rel}}^{\text{'}}+U^{\text{'}}=m\frac{a^2{\omega }^2}{16}+m{a}^2{\omega }^2=\frac{17}{16}m{a}^2{\omega }^2$

Since the new oscillation has an amplitude b, its total oscillation energy is:
$E_{\text{rel}}^{\text{'}}=m{b}^2{\omega }^2$

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

Multiplying by 4 to find the required ratio:
$\frac{4b^2}{a^2}=4\times \frac{17}{16}=\frac{17}{4}=4.25$