If the collision occurs at time $$t_0 = 0$$, the value of $$v_{cm}/(a\omega)$$ will be ________.

The positions of the two identical masses just before the collision are

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

Differentiating, their velocities are

$$v_1(t)=a\omega\cos\omega t,\qquad v_2(t)=-a\omega\cos\omega t.$$

The collision takes place at $$t_0=0$$, so

$$v_{1i}=v_1(0)=a\omega,\qquad v_{2i}=v_2(0)=-a\omega.$$

Particle 3 (mass $$m$$) approaches from the left with speed

$$u_0=\tfrac{a\omega}{2}$$

towards the right (+ve $$x$$ direction). All three masses are equal, so a one-dimensional perfectly elastic collision simply makes the two colliding particles exchange their velocities.

Hence, immediately after the collision,

$$v_{3f}=-a\omega,\qquad v_{2f}=u_0=\tfrac{a\omega}{2},\qquad v_{1f}=v_{1i}=a\omega \;(\text{particle 1 is untouched}).$$

The centre-of-mass velocity of particles 1 and 2 after the collision is

$$v_{cm}=\frac{v_{1f}+v_{2f}}{2}=\frac{a\omega+\tfrac{a\omega}{2}}{2} =\frac{3a\omega/2}{2}= \frac{3}{4}\,a\omega.$$

Therefore,

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

The required value is 0.75.