Question

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 $\left(\frac{A_1}{A_2}\right)$ is

• A. ${\left(\frac{M+m}{M}\right)}^{1/2}$
• B. $\frac{M}{M+m}$
• C. $\frac{M+m}{M}$
• D. ${\left(\frac{M}{M+m}\right)}^{1/2}$

Answer 1

Answer: (A)

At mean position, initial velocity $=A,\omega$
New velocity $=\frac{M{A}_1{\omega }_1}{M+m}$
$\Rightarrow A_2{\omega }_2=\frac{M{A}_1{\omega }_1}{M+m}$
$\frac{A_2}{A_1}=\left(\frac{M}{M+m}\right)\frac{{\omega }_1}{{\omega }_2}$
${\omega }_1=\sqrt{\frac{k}{M}}$
${\omega }_2=\sqrt{\frac{k}{M+m}}$
$\Rightarrow \frac{A_2}{A_1}=\sqrt{\frac{M}{m+M}}$
or $\frac{A_1}{A_2}=\sqrt{\frac{m+M}{M}}$