The time taken by a particle performing `SHM` on a straight line to pass from point `A` to `B` where its velocities are same is `2` seconds .After another `2` seconds it return to `B` The time period of oscillation is (in seconds)

Options

A. `2`
B. `4`
C. `6`
D. `8`

Explanation

To solve the problem, we need to analyze the motion of the particle performing Simple Harmonic Motion (SHM) between points A and B. Given that the time taken to pass from point A to point B, where the velocities are the same, is 2 seconds, and it takes another 2 seconds to return to point B, we can deduce the time period of the oscillation.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The particle moves from point A to point B in 2 seconds.
- The particle has the same velocity at points A and B, indicating that these points are symmetrically located about the mean position O.

2. **Time Taken for One Complete Cycle**:
- Since the time taken to go from A to B is 2 seconds, and the particle returns to B after another 2 seconds, the total time taken from A to B and back to B is \(2 + 2 = 4\) seconds.
- However, this only accounts for the motion from A to B and back to B.

3. **Completing the Full Oscillation**:
- To complete one full oscillation (from A to B, then back to A), we need to consider the time taken to go from B back to A.
- Since the motion is symmetrical, the time taken to go from B to A will also be 2 seconds.

4. **Calculating the Total Time Period**:
- Therefore, the total time for one complete cycle (from A to B, back to A) is:
\[ \text{Total Time Period} = \text{Time from A to B} + \text{Time from B to A} = 2 + 2 + 2 = 6 \text{ seconds} \]

5. **Final Calculation**:
- The total time period of the oscillation is thus \(6\) seconds.

### Conclusion:
The time period of oscillation is **6 seconds**.

---