Question

A mass \( M \), attached to a horizontal spring executes simple harmonic motion with amplitude \( A_1 \). When mass \( M \) passes mean position, then a smaller mass \( m \) is attached to it, and both of them together execute simple harmonic motion with amplitude \( A_2 \). Then the value of \( \frac{A_1}{A_2} \) is:

• A. \( \sqrt{\frac{m^2 + M^2}{M^2}} \)
• B. \( \sqrt{\frac{m+M}{M^2}} \)
• C. \( \sqrt{\frac{m+M}{M}} \)
• D. \( \frac{m+M}{M} \)

Hint:

When additional mass is attached at the mean position, the total energy remains conserved. The amplitude of oscillation decreases as the effective mass increases.

Answer 1

The Correct Option is C

Step 1: Understanding the SHM System When a mass \( M \) executes SHM with amplitude \( A_1 \), its energy is given by: \[ E_1 = \frac{1}{2} M \omega^2 A_1^2 \] When an additional mass \( m \) is attached, the new system's energy is: \[ E_2 = \frac{1}{2} (M + m) \omega^2 A_2^2 \]
Step 2: Apply Energy Conservation Since no external force acts on the system, mechanical energy is conserved: \[ \frac{1}{2} M \omega^2 A_1^2 = \frac{1}{2} (M+m) \omega^2 A_2^2 \] Solving for \( A_1/A_2 \), \[ \frac{A_1}{A_2} = \sqrt{\frac{M+m}{M}} \]