A simple pendulum of length L has a period T. If length is changed by `Delta L`, the change in period `Delta T` is proportional to

Options

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

Explanation

To solve the problem, we need to analyze how the period \( T \) of a simple pendulum changes when the length \( L \) is altered by a small amount \( \Delta L \).

### Step-by-Step Solution:

1. **Understand the Formula for the Period of a Pendulum**:
The 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. **Differentiate the Period with Respect to Length**:
To find how the period changes with a change in length, we differentiate \( T \) with respect to \( L \):
\[ \frac{dT}{dL} = 2\pi \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{g}} \cdot \frac{1}{\sqrt{L}} = \frac{\pi}{\sqrt{gL}} \]

3. **Express the Change in Period**:
The change in period \( \Delta T \) when the length changes by \( \Delta L \) can be expressed using the derivative:
\[ \Delta T \approx \frac{dT}{dL} \cdot \Delta L \]
Substituting the expression for \( \frac{dT}{dL} \):
\[ \Delta T \approx \frac{\pi}{\sqrt{gL}} \cdot \Delta L \]

4. **Identify the Proportionality**:
From the equation \( \Delta T \approx \frac{\pi}{\sqrt{gL}} \cdot \Delta L \), we can see that the change in period \( \Delta T \) is proportional to \( \Delta L \) and inversely proportional to \( \sqrt{L} \).

5. **Final Expression**:
Thus, we can conclude that:
\[ \Delta T \propto \frac{\Delta L}{\sqrt{L}} \]
This indicates that the change in period \( \Delta T \) is proportional to the change in length \( \Delta L \) divided by the square root of the original length \( L \).

### Conclusion:
The change in period \( \Delta T \) is proportional to \( \frac{\Delta L}{\sqrt{L}} \).