The bob of a simple pendulum of length L is released at time t = 0 from a position of small angular displacement ` theta _ 0 ` . Its linear displacement at time t is given by :

Options

A. `l theta_(0)cossqrt((g)/(l))t`
B. `lsqrt((g)/(l))t costheta_(0)`
C. `l gsintheta_(0)`
D. `l theta_(0)sinsqrt((g)/(l))t`

Answer

To solve the problem of finding the linear displacement of a simple pendulum bob released from a small angular displacement, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Pendulum Setup**:
- A simple pendulum consists of a mass (bob) attached to a string of length \( L \).
- When the bob is displaced to an angle \( \theta_0 \) and released, it will undergo simple harmonic motion (SHM).

2. **Equation of Motion**:
- The angular displacement \( \theta \) of the pendulum as a function of time \( t \) can be expressed as:
\[ \theta(t) = \theta_0 \cos(\omega t) \]
- Here, \( \theta_0 \) is the initial angular displacement, and \( \omega \) is the angular frequency.

3. **Determine the Angular Frequency**:
- The angular frequency \( \omega \) for a simple pendulum is given by:
\[ \omega = \sqrt{\frac{g}{L}} \]
- Where \( g \) is the acceleration due to gravity.

4. **Linear Displacement**:
- The linear displacement \( x \) of the bob from the vertical position can be related to the angular displacement by:
\[ x = L \theta \]
- Substituting the expression for \( \theta(t) \):
\[ x(t) = L \theta_0 \cos(\omega t) \]

5. **Final Expression for Linear Displacement**:
- Now substituting \( \omega \) into the equation:
\[ x(t) = L \theta_0 \cos\left(\sqrt{\frac{g}{L}} t\right) \]

### Conclusion:
The linear displacement of the pendulum bob at time \( t \) is given by:
\[ x(t) = L \theta_0 \cos\left(\sqrt{\frac{g}{L}} t\right) \]