A mass M is attached to a horizontal spring of force constant k fixed on one side to a rigid support as shown in figure. The mass oscillates on a frictionless surface with time period T and amplitude A. When the mass is in equilibrium position, another mass m is gently placed on it. What will be the new amplitude of oscillations

When a mass m is placed on mass M, the new system is of 
Mass = (M+m), attached to the spring. New time period of oscillation,

Let v = velocity of mass M while passing through the mean position.
v'= velocity of mass (M+m), while passing through the mean position.
According to law of conservation of linear momentum
$Mv=(M+m)v\text{'}$
At mean position,  $v=A\omega =andv\text{'}=A\text{'}\omega \text{'}$
 $\begin{array}{l}MA\omega =(M+m)A\text{'}\omega \text{'}\\ A\text{'}=\left(\frac{M}{M+m}\right)\frac{\omega }{\omega \text{'}}A=\frac{M}{M+m}\times \frac{T\text{'}}{T}\times A\\ =\left(\frac{M}{M+m}\right)\times \sqrt{\frac{M+m}{M}}\times A\\ =A\sqrt{\frac{M}{M+m}}\end{array}$