1. Set up the situation

The oscillator of mass $M$ moves in the potential $V=\frac12 k(x-X)^2,$ so its equilibrium position is at $x=X$.

Each time it crosses equilibrium, a particle of mass $m$ comes from the right with speed $u$ and sticks to it completely inelastically.

Given: $M=10,\quad m=5,\quad u=1,\quad k=1.$

Initially, the oscillator is at rest at equilibrium.

2. First collision

Initially, mass $M$ is at rest. A particle of mass $m$ with velocity $-u$ (coming from the right toward the left) collides and sticks.

By conservation of momentum: $m(-u)+M(0)=(M+m)v_1$ $v_1=-\frac{mu}{M+m}=-\frac{5\cdot 1}{10+5}=-\frac13.$

So just after the first collision, the combined mass is $M_1=M+m=15,$ and it is at equilibrium with speed magnitude $\frac13$.

For SHM, if a particle starts from equilibrium with speed $v$, then amplitude is $A=\frac{v}{\omega},\qquad \omega=\sqrt{\frac{k}{M_1}}.$

Thus $A_1=\frac{1/3}{\sqrt{1/15}}=\frac{\sqrt{15}}{3}=\sqrt{\frac53}.$

This is not yet one of the options, so clearly the process of repeated collisions matters.

3. Velocity when it returns to equilibrium

After half a cycle, the mass $15$ returns to equilibrium from the left toward the right with the same speed magnitude: $v=+\frac13.$

At that instant, another particle of mass $m=5$ comes from the right with velocity $-1$ and sticks.

4. Recurrence relation for velocity after each collision

Let $M_n=M+n m$ be the total mass after the $n$-th collision, and let $v_n$ be the velocity just after the $n$-th collision.

Before the $(n+1)$-th collision, the oscillator crosses equilibrium with velocity $-v_n$ or $+v_n$ depending on phase. Since the incoming particle always comes from the right, the sign alternates. It is more convenient to track the speed magnitude.

Let $s_n=|v_n|$.

At each equilibrium crossing, just before the next collision the oscillator speed magnitude is still $s_n$.

Now momentum conservation at the next collision gives: $M_n(\pm s_n)+m(-u)=M_{n+1}v_{n+1}.$

Because the direction alternates, one finds that after every collision the motion continues toward the left/right alternately, but the speed magnitude satisfies the simple relation $s_{n+1}=\left|\frac{M_n s_n-mu}{M_n+m}\right|.$

Let us compute explicitly from the actual sequence.

5. Compute successive collisions

After 1st collision

$M_1=15,\qquad v_1=-\frac13.$

Just before 2nd collision

At equilibrium again, $v=+\frac13.$

Second collision: $15\left(\frac13\right)+5(-1)=20v_2$ $5-5=20v_2$ $v_2=0.$

This is the key point: after the second collision, the system comes to rest exactly at equilibrium.

Therefore amplitude after the second collision is $A_2=0.$

6. What happens afterwards?

Since after the second collision the oscillator is at rest at equilibrium, it will not move again.

Hence it will never cross the equilibrium position again on its own, so no further collisions can occur dynamically.

Thus the amplitude remains $A=0.$

So after 13 collisions, the physical outcome implied by the given data is still $A=0.$

7. Check against options

The computed amplitude is $0,$ which is not present among the options.

Therefore the stored answer $\text{A} = \frac{1}{\sqrt3}$ does not agree with the mechanics of the stated problem.

It appears there is likely an error in the question/options or in the stored answer.