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: Before
At the mean position the speed is \(v = A\omega_1\), with \(\omega_1 = \sqrt{\frac{k}{m_1}}\).

Step 2: Momentum
After placing \(m_2\), \(m_1v = (m_1+m_2)v'\), so \(v' = \frac{m_1v}{m_1+m_2}\).

Step 3: New amplitude
New angular frequency \(\omega_2 = \sqrt{\frac{k}{m_1+m_2}}\) and \(A_1 = \frac{v'}{\omega_2}\).

Step 4: Ratio
\(A_1 = \frac{m_1}{m_1+m_2}\cdot A\frac{\omega_1}{\omega_2} = \frac{m_1}{m_1+m_2}A\sqrt{\frac{m_1+m_2}{m_1}} = A\sqrt{\frac{m_1}{m_1+m_2}}\). So \(\frac{A}{A_1} = \left(\frac{m_1+m_2}{m_1}\right)^{1/2}\). Option (A).

Final Answer:
A/A1 equals root of (m1+m2)/m1. \[ \boxed{\text{(A)}\ \left[\frac{m_1+m_2}{m_1}\right]^{1/2}} \]