If the period of oscillation of mass M suspended from a spring is one second, then the period of 9M will be

Options

A. 1/2 s
B. 1/4 s
C. 2 s
D. 4s

Explanation

To solve the problem, we need to understand the relationship between the mass attached to a spring and the period of oscillation. The formula for the period \( T \) of a mass-spring system is given by:

\[ T = 2\pi \sqrt{\frac{m}{k}} \]

where:
- \( T \) is the period of oscillation,
- \( m \) is the mass attached to the spring,
- \( k \) is the spring constant.

### Step-by-Step Solution:

1. **Identify the given information**:
- The period of oscillation for mass \( M \) is given as \( T = 1 \) second.

2. **Write the formula for the period with mass \( M \)**:
\[ T = 2\pi \sqrt{\frac{M}{k}} \]
Since \( T = 1 \) second, we can express this as:
\[ 1 = 2\pi \sqrt{\frac{M}{k}} \]

3. **Square both sides to eliminate the square root**:
\[ 1^2 = (2\pi)^2 \left(\frac{M}{k}\right) \]
This simplifies to:
\[ 1 = 4\pi^2 \frac{M}{k} \]

4. **Rearranging the equation to find \( k \)**:
\[ k = 4\pi^2 M \]

5. **Now consider the new mass \( 9M \)**:
- We need to find the new period \( T' \) when the mass is \( 9M \):
\[ T' = 2\pi \sqrt{\frac{9M}{k}} \]

6. **Substituting \( k \) from the previous step**:
\[ T' = 2\pi \sqrt{\frac{9M}{4\pi^2 M}} \]

7. **Simplifying the expression**:
\[ T' = 2\pi \sqrt{\frac{9}{4\pi^2}} = 2\pi \cdot \frac{3}{2\pi} = 3 \]

8. **Conclusion**:
- The period of oscillation for the mass \( 9M \) is \( T' = 3 \) seconds.

### Final Answer:
The period of oscillation of mass \( 9M \) will be **3 seconds**.

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