A mass $0.9$ kg, attached to a horizontal spring, executes SHM with an amplitude $A_1$. When this mass passes through its mean position, then a smaller mass of $124$ g is placed over it and both masses move together with amplitude $A_2$. If the ratio $\dfrac{A_1}{A_2}$ is $\dfrac{\alpha}{\alpha - 1}$, then the value of $\alpha$ will be ______ .
Numerical value type. Enter your answer.
Answer: 16
At the mean position momentum is conserved: $Mv = (M + m)v'$.
Amplitude $A = \dfrac{v_{\max}}{\omega}$ with $\omega = \sqrt{k/\text{mass}}$:
$$\frac{A_1}{A_2} = \frac{v}{v'}\cdot\frac{\omega'}{\omega} = \frac{M + m}{M}\sqrt{\frac{M}{M + m}} = \sqrt{\frac{M + m}{M}}$$
$$\frac{A_1}{A_2} = \sqrt{\frac{1.024}{0.9}} = \frac{32}{30} = \frac{16}{15}$$
$\dfrac{\alpha}{\alpha - 1} = \dfrac{16}{15} \Rightarrow \alpha = 16$.
Solution by Sreeraj P, M.Sc Physics