Question

A mass \(M\) attached to a horizontal spring executes S.H.M. of amplitude \(A_1\). When the mass passes through its mean position, a smaller mass \(m\) is placed over it and both move together with amplitude \(A_2\). The ratio \( \left(\dfrac{A_1}{A_2}\right) \) is

• A. \( \dfrac{M+m}{M} \)
• B. \( \left(\dfrac{M}{M+m}\right)^{1/2} \)
• C. \( \left(\dfrac{M+m}{M}\right)^{1/2} \)
• D. \( \dfrac{M}{M+m} \)

Hint:

When mass changes suddenly at mean position, use conservation of momentum to relate amplitudes.

Answer 1

The Correct Option is C

Step 1: Velocity at mean position.
At mean position, the velocity of the mass is maximum and given by \[ v = \omega A_1 \] where \[ \omega = \sqrt{\frac{k}{M}} \]

Step 2: Apply conservation of momentum.
When mass \(m\) is placed on \(M\), total mass becomes \(M+m\). Using conservation of momentum at mean position, \[ Mv = (M+m)v' \]

Step 3: Relate new velocity and amplitude.
New angular frequency \[ \omega' = \sqrt{\frac{k}{M+m}} \] and \[ v' = \omega' A_2 \]

Step 4: Solve for amplitude ratio.
\[ M\omega A_1 = (M+m)\omega' A_2 \] \[ \frac{A_1}{A_2} = \sqrt{\frac{M+m}{M}} \]