A mass $$M$$, attached to a horizontal spring, executes S.H.M. with amplitude $$A_1$$. When the mass $$M$$ passes through its mean position then a smaller mass $$m$$ is placed over it and both of them move together with amplitude $$A_2$$. The ratio of $$\left(\dfrac{A_1}{A_2}\right)$$ is:

The spring (force constant $$k$$) carries only the large mass $$M$$ at first, so its natural angular frequency is
$$\omega_1 = \sqrt{\dfrac{k}{M}}$$

In simple-harmonic motion the maximum speed is attained at the mean position and equals
$$v_{\max} = \omega_1 A_1 = \sqrt{\dfrac{k}{M}}\;A_1$$

Exactly at this instant the small mass $$m$$ is gently placed on $$M$$ and sticks to it. The process is so quick that the external spring force has no time to change the horizontal momentum, hence

Conservation of linear momentum (just before and just after contact):
$$M\,v_{\max} = (M+m)\,v'$$ $$\Rightarrow \; v' = \dfrac{M}{M+m}\;\sqrt{\dfrac{k}{M}}\;A_1$$

After sticking, the combined mass $$(M+m)$$ executes SHM with the same spring. Its new angular frequency is
$$\omega_2 = \sqrt{\dfrac{k}{M+m}}$$

This new SHM has amplitude $$A_2$$, and its maximum speed is
$$v' = \omega_2 A_2 = \sqrt{\dfrac{k}{M+m}}\;A_2$$

Equate the two expressions for $$v'$$:

$$\sqrt{\dfrac{k}{M+m}}\;A_2 = \dfrac{M}{M+m}\;\sqrt{\dfrac{k}{M}}\;A_1$$

Cancel $$\sqrt{k}$$ from both sides and rearrange:

$$A_2 = \dfrac{M}{M+m}\;A_1\;\sqrt{\dfrac{M+m}{M}} = A_1\;\sqrt{\dfrac{M}{M+m}}$$

Hence the required ratio of the initial to the final amplitude is

$$\dfrac{A_1}{A_2} = \sqrt{\dfrac{M+m}{M}}$$

Therefore the correct choice is:
Option C which is: $$\left(\dfrac{M+m}{M}\right)^{1/2}$$