A simple pendulum executing `SHM` in a straight line has zero velocity at 'A' and 'B' whose distances form 'O' in the same line `OAB` are 'a' and 'b'. If the velocity half way between them is 'v' then its time period is

Options

A. ` ( 2pi ( b - a ))/( v ) `
B. ` ( pi ( b - a )) / (v) `
C. ` ( pi ( b + a )) /( v ) `
D. None

Explanation

To solve the problem step by step, we will analyze the motion of a simple pendulum executing Simple Harmonic Motion (SHM) between points A and B.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a simple pendulum moving in a straight line along points O, A, and B.
- The distances from point O to point A and point B are given as 'a' and 'b', respectively.
- The pendulum has zero velocity at points A and B.

2. **Identifying the Amplitude**:
- The amplitude (A) of the pendulum's motion can be defined as the maximum displacement from the mean position (O).
- The midpoint M between A and B is where the velocity is maximum.
- The amplitude can be calculated as:
\[ A = \frac{b - a}{2} \]

3. **Velocity at the Midpoint**:
- It is given that the velocity halfway between A and B (at point M) is 'v'.
- In SHM, the maximum velocity (V_max) occurs at the mean position, which is given by:
\[ V_{max} = A \omega \]
- Here, \(\omega\) is the angular frequency.

4. **Relating Velocity to Time Period**:
- The angular frequency \(\omega\) can be expressed in terms of the time period (T):
\[ \omega = \frac{2\pi}{T} \]
- Substituting this into the equation for maximum velocity:
\[ V_{max} = A \cdot \frac{2\pi}{T} \]

5. **Substituting the Amplitude**:
- Now substituting the expression for amplitude:
\[ v = \left(\frac{b - a}{2}\right) \cdot \frac{2\pi}{T} \]

6. **Solving for Time Period (T)**:
- Rearranging the equation to find T:
\[ v = \frac{(b - a) \pi}{T} \]
- Therefore, we can express T as:
\[ T = \frac{(b - a) \pi}{v} \]

### Final Answer:
The time period \(T\) of the simple pendulum is given by:
\[ T = \frac{(b - a) \pi}{v} \]