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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 $\frac{{{A_1}}}{{{A_2}}}$ is
$\frac{M}{{M + m}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$
$\;\frac{{M + m}}{M}$
${\left( {\;\frac{M}{{M + m}}} \right)^{\frac{1}{2}}}$
${\left( {\;\frac{{M + m}}{M}} \right)^{\frac{1}{2}}}$
Solution
The net force becomes zero at the mean point. Therefore, linear momentum must be conserved.
$\therefore \quad M v_{1}=(M+m) v_{2}$
$M A_{1} \sqrt{\frac{k}{M}}=(M+m) A_{2} \sqrt{\frac{k}{m+M}} \therefore \quad(V=A \sqrt{\frac{k}{M}})$
$A_{1} \sqrt{M}=A_{2} \sqrt{M+m} \quad \therefore \frac{A_{1}}{A_{2}}=\sqrt{\frac{m+M}{M}}$
