A simple pendulum performs simple harmonic motion about `x=0` with an amplitude a ans time period T. The speed of the pendulum at `x = (a)/(2)` will be

Options

A. `(pi a sqrt(3))/(2T)`
B. `(pi a)/(T)`
C. `(3pi^(2) a)/(T)`
D. `(pi a sqrt(3))/(T)`

Answer

To find the speed of a simple pendulum at a position \( x = \frac{a}{2} \), we can use the formula for the speed of a particle in simple harmonic motion (SHM):

### Step-by-Step Solution:

1. **Understand the parameters**:
- Amplitude \( a \): The maximum displacement from the mean position (which is \( x = 0 \)).
- Time period \( T \): The time taken for one complete cycle of motion.
- Displacement \( x \): The position at which we want to find the speed, given as \( x = \frac{a}{2} \).

2. **Use the formula for speed in SHM**:
The speed \( v \) of a particle in SHM can be expressed as:
\[ v = \omega \sqrt{a^2 - x^2} \]
where \( \omega \) is the angular frequency.

3. **Calculate the angular frequency \( \omega \)**:
The angular frequency \( \omega \) is related to the time period \( T \) by:
\[ \omega = \frac{2\pi}{T} \]

4. **Substitute \( x = \frac{a}{2} \)** into the speed formula**:
Substitute \( x \) into the speed formula:
\[ v = \omega \sqrt{a^2 - \left(\frac{a}{2}\right)^2} \]
Simplifying the expression inside the square root:
\[ v = \omega \sqrt{a^2 - \frac{a^2}{4}} = \omega \sqrt{\frac{4a^2}{4} - \frac{a^2}{4}} = \omega \sqrt{\frac{3a^2}{4}} = \omega \cdot \frac{a\sqrt{3}}{2} \]

5. **Substitute \( \omega \) into the speed equation**:
Now substituting \( \omega = \frac{2\pi}{T} \):
\[ v = \frac{2\pi}{T} \cdot \frac{a\sqrt{3}}{2} \]
Simplifying gives:
\[ v = \frac{\pi a \sqrt{3}}{T} \]

6. **Final result**:
Therefore, the speed of the pendulum at \( x = \frac{a}{2} \) is:
\[ v = \frac{\pi a \sqrt{3}}{T} \]