$\therefore \quad M v_{1}=(M+m) v_{2}$
$M A_{1} \sqrt{\frac{k}{M}}=(M+m) A_{2} \sqrt{\frac{k}{m+M}} \therefore \quad(V=A \sqrt{\frac{k}{M}})$
$A_{1} \sqrt{M}=A_{2} \sqrt{M+m} \quad \therefore \frac{A_{1}}{A_{2}}=\sqrt{\frac{m+M}{M}}$
If the position and velocity of the particle at $t=0\, {s}$ are $2\, {cm}$ and $2\, \omega \,{cm} \,{s}^{-1}$ respectively, then its amplitude is $x \sqrt{2} \,{cm}$ where the value of $x$ is ..... .