A mass `M`, attached to a horizontal spring, excutes `SHM` with a amplitude `A_(1)`. When the mass `M` passes through its mean position then a smaller mass `m` is placed over it and both of them move together with amplitude `A_(2)`, the ratio of `((A_(1))/(A_(2)))` is :

Options

A. `M/M+m`
B. `M+m/M`
C. `(M/M+m)^(1//2)`
D. `(M+m/M)^(1//2)`

Explanation

To solve the problem, we will analyze the situation step by step, using the principles of conservation of momentum and the properties of simple harmonic motion (SHM).

### Step-by-Step Solution:

1. **Understanding the Initial Conditions:**
- A mass \( M \) is attached to a spring and is executing simple harmonic motion (SHM) with an amplitude \( A_1 \).
- The maximum velocity \( V_1 \) of the mass \( M \) at the mean position can be expressed as:
\[ V_1 = \omega_1 A_1 \]
where \( \omega_1 \) is the angular frequency of the motion.

2. **Introducing the Smaller Mass:**
- When the mass \( M \ passes through the mean position, a smaller mass \( m \) is placed on it. The total mass now becomes \( M + m \).
- The new maximum velocity \( V_2 \) of the combined mass at the mean position can be expressed as:
\[ V_2 = \omega_2 A_2 \]
where \( A_2 \) is the new amplitude after mass \( m \) is added, and \( \omega_2 \) is the new angular frequency.

3. **Applying Conservation of Momentum:**
- According to the conservation of momentum, the initial momentum must equal the final momentum:
\[ M V_1 = (M + m) V_2 \]
Substituting the expressions for \( V_1 \) and \( V_2 \):
\[ M (\omega_1 A_1) = (M + m) (\omega_2 A_2) \]

4. **Expressing Angular Frequencies:**
- The angular frequencies can be expressed in terms of the spring constant \( k \):
\[ \omega_1 = \sqrt{\frac{k}{M}} \quad \text{and} \quad \omega_2 = \sqrt{\frac{k}{M + m}} \]

5. **Substituting Angular Frequencies:**
- Substitute \( \omega_1 \) and \( \omega_2 \) into the momentum equation:
\[ M \left(\sqrt{\frac{k}{M}} A_1\right) = (M + m) \left(\sqrt{\frac{k}{M + m}} A_2\right) \]

6. **Simplifying the Equation:**
- Cancel \( \sqrt{k} \) from both sides:
\[ M \sqrt{M + m} A_1 = (M + m) \sqrt{M} A_2 \]
- Rearranging gives:
\[ \frac{A_1}{A_2} = \frac{(M + m) \sqrt{M}}{M \sqrt{M + m}} \]

7. **Final Ratio of Amplitudes:**
- Thus, the ratio of the amplitudes is:
\[ \frac{A_1}{A_2} = \sqrt{\frac{(M + m)}{M}} \]

### Conclusion:
The final result for the ratio of the amplitudes \( \frac{A_1}{A_2} \) is:
\[ \frac{A_1}{A_2} = \sqrt{\frac{M + m}{M}} \]