The length of a pendulum is doubled and the mass of its bob is halved. Its time period would

Options

A. become double
B. become half
C. become `sqrt(2)` times
D. remains the same

Answer

To solve the problem, we need to determine how the time period of a pendulum changes when its length is doubled and the mass of its bob is halved.

### Step-by-Step Solution:

1. **Understand the Formula for Time Period**:
The time period \( T \) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

2. **Identify the Changes**:
- The length of the pendulum is doubled: \( L' = 2L \)
- The mass of the bob is halved: \( m' = \frac{m}{2} \)

3. **Substitute the New Length into the Formula**:
We need to find the new time period \( T' \) when the length is doubled:
\[ T' = 2\pi \sqrt{\frac{L'}{g}} = 2\pi \sqrt{\frac{2L}{g}} \]

4. **Simplify the Expression**:
We can factor out the square root:
\[ T' = 2\pi \sqrt{2} \sqrt{\frac{L}{g}} = \sqrt{2} \cdot (2\pi \sqrt{\frac{L}{g}}) \]
Thus, we can express \( T' \) in terms of the original time period \( T \):
\[ T' = \sqrt{2} \cdot T \]

5. **Conclusion**:
The new time period \( T' \) is \( \sqrt{2} \) times the original time period \( T \). Therefore, the time period increases by a factor of \( \sqrt{2} \) when the length is doubled, and the mass of the bob does not affect the time period.

### Final Answer:
The time period would become \( \sqrt{2} \) times the original time period.

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