A mass $M$ attached to a horizontal spring executes simple harmonic motion with amplitude $A_1$. When mass…
A mass $M$ attached to a horizontal spring executes simple harmonic motion with amplitude $A_1$. When mass $M$ passes mean position then a smaller mass $m$ is attached to it and both of them together executing simple harmonic motion with amplitude $A_2$. Then the value of $\frac{A_1}{A_2}$ is
$\sqrt{\frac{m^2+M^2}{M^2}}$
$\sqrt{\frac{m+M}{M^2}}$
$\sqrt{\frac{m+M}{M}}$
$\frac{m+M}{M}$
Solution
For SHM , at mean position, $\mathrm{F}_{\text {net }}=0$
$\therefore$ By conservation of momentum,
$\begin{aligned}
& \mathrm{Mv}_1=(M+m) v_2 \\
& \Rightarrow M w_1 A_1=(m+m) w_2 A_2 \\
& \Rightarrow M\left(\sqrt{\frac{k}{M}}\right) A_1=(M+m)\left(\sqrt{\frac{k}{M+m}}\right) A_2 \\
& \therefore \frac{A_1}{A_2}=\sqrt{\frac{M+m}{M}}
\end{aligned}$