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 $\frac{A_{1}}{A_{2}}=\sqrt{\frac{m+M}{M}}$