The period of oscilation of a mass `M` suspended a spring negligible mass is `T` if along with it and there mass `M` is also suspended the period of oscillation now be

Options

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

Explanation

To solve the problem, we need to find the new period of oscillation \( T' \) when a second mass \( M \) is added to the original mass \( M \) suspended from a spring.

### Step-by-Step Solution:

1. **Understand the Formula for the Period of Oscillation:**
The period of oscillation \( T \) for a mass-spring system is given by the formula:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
where \( m \) is the mass attached to the spring and \( k \) is the spring constant.

2. **Identify the Initial Mass:**
Initially, we have a mass \( M \) suspended from the spring. Thus, the period of oscillation is:
\[ T = 2\pi \sqrt{\frac{M}{k}} \]

3. **Determine the New Mass:**
When another mass \( M \) is added, the total mass \( m' \) becomes:
\[ m' = M + M = 2M \]

4. **Calculate the New Period of Oscillation:**
Now, we need to find the new period of oscillation \( T' \) with the total mass \( m' \):
\[ T' = 2\pi \sqrt{\frac{m'}{k}} = 2\pi \sqrt{\frac{2M}{k}} \]

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

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

### Conclusion:
The new period of oscillation is \( \sqrt{2} \) times the original period \( T \).

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