1. Initial motion of particles 1 and 2

Given $x_1(t)=(x_0+d)+a\sin \omega t,$ $x_2(t)=(x_0-d)-a\sin \omega t.$

At $t=0$, $x_1(0)=x_0+d, \qquad x_2(0)=x_0-d.$

Their velocities are $v_1(t)=\dot x_1=a\omega \cos \omega t,$ $v_2(t)=\dot x_2=-a\omega \cos \omega t.$

So at $t_0=0$, $v_1(0)=+a\omega, \qquad v_2(0)=-a\omega.$

Thus particle 2 is moving left with speed $a\omega$.

2. Velocity of particle 3 before collision

Particle 3 moves towards the system with speed $u_0=\frac{a\omega}{2}.$

Since it collides with particle 2 from the left/right? For collision with particle 2 while moving towards the two-particle system, particle 3 must be approaching from the left and moving rightward.

Hence before collision, $v_3=+\frac{a\omega}{2}.$

Particle 2 has $v_2=-a\omega.$

3. Elastic collision between equal masses 2 and 3

For a 1D elastic collision between equal masses, the velocities are exchanged.

Therefore after collision, $v_2'=+\frac{a\omega}{2}, \qquad v_3'=-a\omega.$

Particle 1 is unaffected instantaneously, so $v_1'=a\omega.$

4. Center of mass velocity of particles 1 and 2 after collision

Now consider only particles 1 and 2 as the spring system after collision.

Their total momentum just after collision is $P=m v_1'+m v_2'=m(a\omega)+m\left(\frac{a\omega}{2}\right)=\frac{3}{2}ma\omega.$

Total mass of particles 1 and 2 is $2m$, so their center of mass speed is $v_{\mathrm{cm}}=\frac{P}{2m}=\frac{\frac{3}{2}ma\omega}{2m}=\frac{3}{4}a\omega.$

Hence $\frac{v_{\mathrm{cm}}}{a\omega}=\frac{3}{4}=0.75.$

5. Comparison with stored answer

Derived answer: $\boxed{0.75}$

This matches the stored correct answer.