Question

A horizontal spring executes S.H.M. with amplitude \( A_1 \), when mass \( m_1 \) is attached to it. When it passes through mean position, another mass \( m_2 \) is placed on it. Both masses move together with amplitude \( A_2 \). Therefore \( A_2 : A_1 \) is

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

Hint:

In S.H.M., the amplitude decreases as the total mass increases. This relationship is critical in problems involving mass and amplitude in harmonic motion.

Answer 1

The Correct Option is C

Step 1: Understanding S.H.M. with two masses.
When mass \( m_1 \) is attached to the spring, the amplitude is \( A_1 \). When mass \( m_2 \) is added, the two masses move together with amplitude \( A_2 \). The amplitude of S.H.M. is inversely proportional to the total mass attached to the spring. Step 2: Using the relationship between amplitude and mass.
Using conservation of energy and the fact that the system's effective mass increases when both masses move together, the new amplitude is related to the total mass. The amplitude ratio is: \[ A_2 : A_1 = \left( \frac{m_1}{m_1 + m_2} \right)^{\frac{1}{2}} \] Thus, the correct answer is (C) \( \left( \frac{m_1}{m_1 + m_2} \right)^{\frac{1{2}} \)}.