b
In first collision \(mu\) momentum will be imparted to system, in second collision when momentum of \((\mathrm{M}+\mathrm{m})\) is in opposite direction \(mu\) momentum of particle will make its momentum zero.

On \(13^{\text {th }}\) collision,

\(\mathrm{m} \rightarrow {\mathrm{M}+12} ; \quad \mathrm{M}+13 \mathrm{m} \rightarrow \mathrm{V}\)

\(\mathrm{mu}=(\mathrm{M}+13 \mathrm{m}) \mathrm{v} \Rightarrow \mathrm{v}=\frac{\mathrm{mu}}{\mathrm{M}+13 \mathrm{m}}=\frac{\mathrm{u}}{15}\)

\(v=\omega A \Rightarrow \frac{u}{15}=\sqrt{\frac{K}{M-13 m}} \times A\)

Putting value of \(M, m, u\) and \(K\) we get amplitude

\(A=\frac{1}{15} \sqrt{\frac{75}{1}}=\frac{1}{\sqrt{3}}\)