A particle executing linear `SHM`. Its time period is equal to the smallest time interval in which paricle acquires a particular velcity `overset(vec)(v)`, the magnitude of `overset(vec)(v)` may be :

Options

A. `v_(max)`
B. 0
C. between 0 and `v_(max)`
D. between and 0 and `-v_(max)`

Answer

To solve the problem, we need to understand the relationship between the time period of a particle in Simple Harmonic Motion (SHM) and the velocity it acquires during this motion.

### Step-by-Step Solution:

1. **Understanding SHM**:
- In SHM, a particle oscillates back and forth about an equilibrium position. The motion can be described by a sinusoidal function.

2. **Time Period (T)**:
- The time period (T) of SHM is the time taken for the particle to complete one full oscillation. It is a constant for a given system and is independent of the amplitude of the motion.

3. **Velocity in SHM**:
- The velocity of a particle in SHM varies throughout its motion. The maximum velocity (Vmax) occurs as the particle passes through the equilibrium position.

4. **Acquiring a Particular Velocity**:
- The problem states that the time period is equal to the smallest time interval in which the particle acquires a particular velocity (v). In SHM, the smallest time interval to reach a specific velocity corresponds to the maximum velocity.

5. **Conclusion**:
- Since the particle can achieve its maximum velocity (Vmax) in the shortest time interval, we conclude that the magnitude of the velocity (v) mentioned in the question must be equal to the maximum velocity of the particle in SHM.

Thus, the answer is:
**The magnitude of \( \vec{v} \) may be \( V_{max} \)**.