Question

A mass 'M' attached to a horizontal spring executes S.H.M. of amplitude $A_{1}$. When the mass M passes through its mean position, a smaller mass 'm' is placed over it and both move together with amplitude $A_{2}$. The ratio $(A_{1}/A_{2})$ is}

• A. $\frac{M+m}{M}$
• B. $\frac{M}{M+m}$
• C. $(\frac{M+m}{M})^{1/2}$
• D. $(\frac{M}{M+m})^{1/2}$

Hint:

Conservation of momentum is key when masses collide or are added at the mean position.

Answer 1

The Correct Option is C

Step 1: Concept
At the mean position, all energy is kinetic. When mass 'm' is added, momentum is conserved.

Step 2: Analysis
- Initial velocity $v_1 = \omega_1 A_1 = \sqrt{\frac{k}{M}} A_1$.
- Conservation of momentum: $M v_1 = (M+m) v_2$.
- New velocity $v_2 = \frac{M}{M+m} v_1$.

Step 3: Calculation
Also, $v_2 = \omega_2 A_2 = \sqrt{\frac{k}{M+m}} A_2$.
$\sqrt{\frac{k}{M+m}} A_2 = \frac{M}{M+m} \sqrt{\frac{k}{M}} A_1$
$\frac{A_1}{A_2} = \frac{\sqrt{k/(M+m)}}{\frac{M}{M+m} \sqrt{k/M}} = \frac{M+m}{M} \sqrt{\frac{M}{M+m}} = \sqrt{\frac{M+m}{M}}$.

Step 4: Conclusion
Hence, the ratio is $(\frac{M+m}{M})^{1/2}$. Final Answer: (C)