The period of oscillation of mass M suspended from a spring of negligible mass is T. If along with it another mass M is also suspended, the period of oscillation will now be

Options

A. T
B. `T//sqrt(2)`
C. `2T`
D. `sqrt(2)T`

Answer

To solve the problem, we need to determine the new period of oscillation when an additional mass is added to a mass-spring system.

### Step-by-Step Solution

1. **Understanding the Original System**:
The original system consists of a mass \( M \) suspended from a spring with a spring constant \( k \). The formula for the period of oscillation \( T \) of a mass-spring system is given by:
\[ T = 2\pi \sqrt{\frac{M}{k}} \]

2. **Adding Another Mass**:
When we add another mass \( M \) to the system, the total mass becomes \( 2M \). The spring constant \( k \) remains the same.

3. **Calculating the New Period**:
We denote the new period of oscillation as \( T' \). The formula for the new period with the total mass \( 2M \) is:
\[ T' = 2\pi \sqrt{\frac{2M}{k}} \]

4. **Relating the New Period to the Original Period**:
We can express \( T' \) in terms of the original period \( T \):
\[ T' = 2\pi \sqrt{\frac{2M}{k}} = 2\pi \sqrt{2} \sqrt{\frac{M}{k}} = \sqrt{2} \cdot (2\pi \sqrt{\frac{M}{k}}) = \sqrt{2} \cdot T \]

5. **Final Result**:
Therefore, the new period of oscillation when another mass \( M \) is added is:
\[ T' = \sqrt{2} \cdot T \]

### Conclusion
The period of oscillation when an additional mass \( M \) is suspended from the spring will be \( \sqrt{2} T \).