If the collision occurs at time $$t_0 = \pi/(2\omega)$$, then the value of $$4b^2/a^2$$ will be ________.

At any instant the coordinates of the two identical masses are given as
$$x_1(t)=\bigl(x_0+d\bigr)+a\sin\omega t,\qquad x_2(t)=\bigl(x_0-d\bigr)-a\sin\omega t$$

Hence their centre-of-mass (C-O-M) position is
$$x_{cm}=\frac{x_1+x_2}{2}=x_0$$ which is fixed (initially the C-O-M is at rest).

For the system “two masses $$m$$ joined by a mass-less spring of force constant $$k$$” the oscillation (stretch-compress) mode has angular frequency
$$\omega=\sqrt{\frac{2k}{m}}\;.\qquad -(1)$$

At the instant of collision, $$t_0=\dfrac{\pi}{2\omega}\;,$$ we have
$$\sin\omega t_0 =\sin\frac{\pi}{2}=1,\qquad \cos\omega t_0 =0.$$

Therefore, just before collision (denote by the superscript “−”)
$$v_1^-=\dot x_1= a\omega\cos\omega t_0 =0,$$ $$v_2^-=\dot x_2=-a\omega\cos\omega t_0 =0.$$ Both particles of the spring system are instantaneously at rest.

A third particle, also of mass $$m$$, approaches particle 2 with speed
$$u_0=\frac{a\omega}{2}.$$

Because the collision between 3 and 2 is perfectly elastic and the two masses are equal, the standard 1-D result gives
$$v_3^+=0,\qquad v_2^+=u_0$$ immediately after collision (superscript “+”). Particle 1 is untouched, so $$v_1^+=0$$.

Hence the C-O-M velocity acquired by the spring system (masses 1 & 2) is
$$v_{cm}=\frac{v_1^+ + v_2^+}{2} =\frac{u_0}{2} =\frac{a\omega}{4}.$$

The internal (relative) motion is analysed in the C-O-M frame. Let the extension of the spring from its natural length be $$x = (x_1-x_2)-2d.$$ Initially, the oscillation had amplitude $$a$$ for each mass, so the relative amplitude was $$2a$$. At $$t=t_0$$ the spring is at its extreme extension:
$$x_0 = 2a,\qquad \dot x^- = 0.$$

Immediately after collision the relative velocity becomes $$\dot x^+ = v_1^+ - v_2^+ = 0 - u_0 = -u_0 = -\frac{a\omega}{2}.$$ (The negative sign shows the spring will start to compress.)

Total internal energy in the C-O-M frame just after collision is the sum of
(i) spring potential energy and (ii) kinetic energy of relative motion.
For the two-body system the reduced mass is $$\mu=\dfrac{m}{2}$$, and with $$k$$ related to $$\omega$$ by (1): $$k=\dfrac{m\omega^2}{2}$$.

Thus
$$E'=\frac12 kx_0^{\,2}+\frac12\mu\dot x^{+2} =\frac12\left(\frac{m\omega^2}{2}\right)(2a)^2 +\frac12\left(\frac{m}{2}\right)\left(\frac{a\omega}{2}\right)^2 =2k a^2+\frac{mu_0^{\,2}}{4}\;.$$

After the collision the two masses again execute simple harmonic motion (same $$\omega$$) but with a larger amplitude $$b$$. At an extreme, kinetic energy is zero, so
$$E'=\frac12 k(2b)^2 = 2k b^2.$$

Equating the two expressions for $$E'$$:
$$2k b^2 = 2k a^2 + \frac{m u_0^{\,2}}{4}.$$

Divide by $$2k$$ and substitute $$u_0=\dfrac{a\omega}{2}$$ and $$k=\dfrac{m\omega^2}{2}$$:
$$b^2 = a^2 + \frac{m(a\omega/2)^2}{4}\,\frac{1}{2k} = a^2 + \frac{a^2\omega^2/4}{4\omega^2} = a^2 + \frac{a^2}{16} = a^2\left(1 + \frac1{16}\right) = a^2\frac{17}{16}.$$

Therefore
$$\frac{4b^2}{a^2} = 4\left(\frac{17}{16}\right) = \frac{17}{4}=4.25.$$

Hence, the required value is 4.25.