Answer is (A)
P.E of the oscillating mass is given by
$E=\frac{1}{2} m \omega x^{2}=\frac{1}{2} k x^{2} \quad \text { Where } k=m \omega^{2}$
When only black is oscillating, at $x=A$
$E=E_{\max }=\frac{1}{2} m \omega A^{2}$ and at mean position i.e. $x=0$
$\therefore \quad A \propto \frac{1}{\sqrt{\mathrm{m}}}$
$E=0$
When mass $\mathrm{m}_{2}$ is placed on top of the mass $\mathrm{m}_{1}$
Then, total mass is $\left(m_{1}+m_{2}\right)$ and $E=0$ at this point, as $x=0$.
When the combination reaches $x=A$,
Then $A_{1} \propto \frac{1}{\sqrt{m_{1}+m_{2}}}$
$\therefore \frac{\mathrm{A}_{1}}{\mathrm{~A}}=\sqrt{\frac{\mathrm{m}_{1}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}}$