Question

A horizontal spring executes S.H.M. with amplitude \(A_{1}\) when mass \(m\) is attached to it. When it passes through mean position another mass \(m\) is placed on it. Both masses move together with amplitude \(A_{2}\). The ratio \(\frac{A_{1}}{A_{2}}\) is

• A. \(\frac{1}{\sqrt{2}}\)
• B. \(\sqrt{2}\)
• C. \(2\)
• D. \(\frac{1}{2}\)

Hint:

At the mean position momentum is conserved, so the speed halves while the angular frequency also changes.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
At the mean position the speed is maximum: \(v=A\omega\) with \(\omega=\sqrt{\dfrac{k}{m}}\).

Step 2: Before and after
Before: \(v_1=A_1\omega_1\).
After adding a mass \(m\), momentum is conserved: \(mv_1=2mv_2\), so \(v_2=\dfrac{v_1}{2}\).
New angular frequency \(\omega_2=\sqrt{\dfrac{k}{2m}}=\dfrac{\omega_1}{\sqrt{2}}\).

Step 3: New amplitude
\(A_2=\dfrac{v_2}{\omega_2}=\dfrac{A_1\omega_1/2}{\omega_1/\sqrt{2}}=\dfrac{A_1}{\sqrt{2}}\).

Step 4: Ratio
\(\dfrac{A_1}{A_2}=\sqrt{2}\), option (B).

Final Answer:
The ratio A1 to A2 is root 2. \[ \boxed{\text{(B) }\sqrt{2}} \]