A particle performing `SHM` takes time equal to `T` (time period of `SHM`) in consecutive appearances at a perticular point. This point is :

Options

A. the mean position
B. an extreme position
C. between the mean position and the positie extreme
D. between the mean position and the negative extreme.

Answer

To solve the question, we need to analyze the motion of a particle performing Simple Harmonic Motion (SHM) and determine the point at which it takes time equal to the time period \( T \) to appear consecutively.

### Step-by-Step Solution:

1. **Understanding SHM**:
- In SHM, a particle oscillates back and forth around a mean position. The motion is periodic, meaning it repeats itself after a fixed interval of time known as the time period \( T \).

2. **Identifying the Motion**:
- The particle moves from the mean position to the extreme position (maximum displacement) on one side, then returns to the mean position, continues to the extreme position on the opposite side, and finally returns back to the mean position again.

3. **Time Period \( T \)**:
- The time period \( T \) is the total time taken for the particle to complete one full cycle of motion, which includes moving from the mean position to one extreme, back to the mean, to the other extreme, and back to the mean.

4. **Consecutive Appearances**:
- The question asks for the time taken for the particle to appear consecutively at a particular point. The only point where the particle can appear consecutively in a time interval of \( T \) is at the mean position.

5. **Conclusion**:
- Therefore, the point at which the particle takes time equal to \( T \) in consecutive appearances is the **mean position**.

### Final Answer:
The point is the **mean position** of the particle performing SHM.

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