A horizontal spring is connedted to a mass `M`. It exectues simple harmonic motion. When the mass `M` passes through its mean position, an object of mass `m` is put on it ans the two move together. The ratio of frequencies before and after will be-

Options

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

Explanation

To solve the problem, we need to analyze the situation before and after the mass \( m \) is added to the mass \( M \) connected to a spring. We will use the concepts of simple harmonic motion (SHM) and the properties of oscillating systems.

### Step-by-Step Solution:

1. **Understanding the System**:
- Initially, we have a mass \( M \) attached to a spring, which executes simple harmonic motion with a certain frequency.
- The frequency of oscillation \( f \) for a mass-spring system is given by the formula:
\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{M}} \]
where \( k \) is the spring constant.

2. **Frequency Before Adding Mass**:
- Let the frequency before adding the mass \( m \) be \( f_1 \):
\[ f_1 = \frac{1}{2\pi} \sqrt{\frac{k}{M}} \]

3. **Adding Mass \( m \)**:
- When the mass \( m \) is placed on the mass \( M \) at the mean position, the total mass becomes \( M + m \).

4. **Frequency After Adding Mass**:
- The new frequency \( f_2 \) after adding the mass \( m \) is given by:
\[ f_2 = \frac{1}{2\pi} \sqrt{\frac{k}{M + m}} \]

5. **Finding the Ratio of Frequencies**:
- We need to find the ratio of the frequencies before and after adding the mass:
\[ \frac{f_1}{f_2} = \frac{\frac{1}{2\pi} \sqrt{\frac{k}{M}}}{\frac{1}{2\pi} \sqrt{\frac{k}{M + m}}} \]
- Simplifying this gives:
\[ \frac{f_1}{f_2} = \frac{\sqrt{\frac{k}{M}}}{\sqrt{\frac{k}{M + m}}} = \sqrt{\frac{M + m}{M}} \]

6. **Final Result**:
- Therefore, the ratio of frequencies before and after adding the mass is:
\[ \frac{f_1}{f_2} = \sqrt{\frac{M + m}{M}} \]

### Conclusion:
The ratio of frequencies before and after adding the mass \( m \) to the mass \( M \) is \( \sqrt{\frac{M + m}{M}} \).