A horizontal spring-block system of mass $M$ executes simple harmonic motion. When the block is passing through its equilibrium position, an object of mass $m$ is put on it and the two move together. Find the new amplitude of vibration.

If $A$ is the initial amplitude of motion, then velocity of the block at mean position, then $v=\omega A=\sqrt{\frac{k}{M}}A$

Now by consevation of linear momentum, we have

$$\begin{gathered} Mv+0=\left(M+m\right)v\text{'} \\ \therefore v\text{'}=\left[\frac{Mv}{M+m}\right] \end{gathered}$$

If $A\text{'}$ be the new amplitude of motion, then

$$\begin{gathered} \frac{1}{2}\left(M+m\right){\left[\frac{Mv}{M+m}\right]}^2=\frac{1}{2}kA{\text{'}}^2 \\ A\text{'}=\sqrt{\frac{M^2}{k\left(M+m\right)}}v \\ = \sqrt{\frac{M^2}{k\left(M+m\right)}}\times \sqrt{\frac{k}{M}}A \\ =\sqrt{\frac{M}{\left(M+m\right)}}A \end{gathered}$$