The relation between T and g is given by

Options

A. `T prop g`
B. `T prop g^(2)`
C. `T^(2) prop g`
D. `T prop sqrt(1/g)`

Answer

To find the relation between the time period \( T \) of a simple pendulum and the acceleration due to gravity \( g \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Pendulum**: A simple pendulum consists of a mass (bob) attached to a string or rod of fixed length \( L \). When the pendulum swings, it undergoes oscillatory motion.

2. **Defining Time Period**: The time period \( T \) is defined as the time taken for the pendulum to complete one full oscillation, which includes moving from the mean position to the extreme position and back.

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

4. **Analyzing the Relation**: From the formula, we can see that:
- \( T \) is proportional to the square root of the length \( L \) and inversely proportional to the square root of \( g \).
- This means that if \( g \) increases, \( T \) decreases, and vice versa.

5. **Expressing the Inverse Relationship**: We can express the relationship between \( T \) and \( g \) as:
\[ T \propto \frac{1}{\sqrt{g}} \]
This indicates that the time period \( T \) is inversely proportional to the square root of the acceleration due to gravity \( g \).

6. **Alternative Form**: If we square both sides of the equation, we can also express it as:
\[ T^2 \propto \frac{1}{g} \]
This shows that the square of the time period \( T^2 \) is directly proportional to the inverse of \( g \).

7. **Conclusion**: Therefore, the relation between the time period \( T \) and the acceleration due to gravity \( g \) can be summarized as:
- \( T \propto \frac{1}{\sqrt{g}} \) or \( T^2 \propto \frac{1}{g} \).

### Final Answer:
The relation between \( T \) and \( g \) is given by:
- \( T \propto \frac{1}{\sqrt{g}} \) (or equivalently, \( T^2 \propto \frac{1}{g} \)).