A mass `M` is suspended from a massless spring. An additional mass `m` stretches the spring further by a distance `x`. The combined mass will oscillate with a period

Options

A. `2pisqrt({((M+m)x)/(mg)})`
B. `2pisqrt({(mg)/((M+m)x)})`
C. `(pi)/(2)sqrt({(mg)/((M+m)x)})`
D. `2pisqrt({((M+m))/(mgx)})`

Explanation

To solve the problem, we need to determine the period of oscillation of a mass-spring system where a mass \( M \) is suspended from a massless spring and an additional mass \( m \) stretches the spring further by a distance \( x \).

### Step-by-Step Solution:

1. **Understanding the System**:
- A mass \( M \) is hanging from a spring, and when an additional mass \( m \) is added, it stretches the spring by a distance \( x \).
- The system reaches a new equilibrium position due to the additional mass.

2. **Applying Hooke's Law**:
- The force exerted by the spring when it is stretched by distance \( x \) is given by Hooke's Law:
\[ F_{\text{spring}} = kx \]
- Here, \( k \) is the spring constant.

3. **Balancing Forces**:
- At equilibrium, the gravitational force acting on the additional mass \( m \) is balanced by the spring force:
\[ mg = kx \]
- Rearranging this gives us the spring constant \( k \):
\[ k = \frac{mg}{x} \]

4. **Finding the Total Mass**:
- The total mass that will oscillate is the sum of the two masses:
\[ M_{\text{total}} = M + m \]

5. **Using the Formula for Period of Oscillation**:
- The period \( T \) of a mass-spring system is given by:
\[ T = 2\pi \sqrt{\frac{M_{\text{total}}}{k}} \]
- Substituting \( M_{\text{total}} = M + m \) and \( k = \frac{mg}{x} \) into the formula:
\[ T = 2\pi \sqrt{\frac{M + m}{\frac{mg}{x}}} \]

6. **Simplifying the Expression**:
- This simplifies to:
\[ T = 2\pi \sqrt{\frac{(M + m)x}{mg}} \]

7. **Final Result**:
- Therefore, the period of oscillation of the combined mass-spring system is:
\[ T = 2\pi \sqrt{\frac{(M + m)x}{mg}} \]

### Conclusion:
The correct expression for the period of oscillation is \( T = 2\pi \sqrt{\frac{(M + m)x}{mg}} \).