Question

A mass \(m_1\) performs S.H.M. with amplitude A which is connected to horizontal spring. While mass \(m_1\) passing through mean position, another mass \(m_2\) (\(m_2 < m_1\)) is placed on it so that both masses move together with amplitude \(A_1\). The ratio of \(A/A_1\) is

• A. \([\frac{m_1+m_2}{m_1}]^{\frac{1}{2}}\)
• B. \([\frac{m_1}{m_1+m_2}]^{\frac{1}{2}}\)
• C. \(\frac{m_1}{m_1+m_2}\)
• D. \(\frac{m_2}{m_1+m_2}\)

Hint:

Momentum is conserved when the mass is dropped on at the mean position.

Answer 1

The Correct Option is A

Step 1: Energy view
At the mean position all energy is kinetic: $\frac12m_1v^2$. Momentum conservation gives KE after $=\frac{m_1}{m_1+m_2}\cdot\frac12m_1v^2$.

Step 2: Amplitudes
Energy is $\frac12kA^2$ before and $\frac12kA_1^2$ after, so $\frac{A_1^2}{A^2} = \frac{m_1}{m_1+m_2}$.

Step 3: Result
$\frac A{A_1} = \sqrt{\frac{m_1+m_2}{m_1}}$. Option (A).

Final Answer:
Option A. \[ \boxed{\text{(A)}\ \left[\frac{m_1+m_2}{m_1}\right]^{1/2}} \]