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. Zero
B. `V_(max)`
C. `(V_(max))/(2)`
D. `(V_(max))/(sqrt(2))`

Answer

To solve the problem, we need to analyze the relationship between the time period of a particle executing simple harmonic motion (SHM) and the velocity it can achieve during this motion.

### Step-by-Step Solution:

1. **Understanding SHM**:
- A particle in SHM oscillates back and forth around an equilibrium position. The motion is periodic, and the time taken for one complete cycle is called the time period (T).

2. **Time Period of SHM**:
- The time period \( T \) of a particle in SHM is given by the formula:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
where \( m \) is the mass of the particle and \( k \) is the spring constant.

3. **Velocity in SHM**:
- The velocity \( v \) of a particle in SHM at a displacement \( x \) is given by:
\[ v = \pm \omega \sqrt{A^2 - x^2} \]
where \( \omega = 2\pi f \) (angular frequency), \( A \) is the amplitude, and \( x \) is the instantaneous displacement.

4. **Maximum Velocity**:
- The maximum velocity \( v_{max} \) occurs when \( x = 0 \) (at the equilibrium position) and is given by:
\[ v_{max} = \omega A \]

5. **Relating Time Period and Velocity**:
- The problem states that the time period \( T \) is equal to the smallest time interval in which the particle acquires a particular velocity \( v \). This means that we need to determine what possible values \( v \) can take relative to \( v_{max} \).

6. **Possible Values of Velocity**:
- The options given are:
1. 0
2. \( v_{max} \)
3. \( \frac{v_{max}}{2} \)
4. \( \frac{v_{max}}{\sqrt{2}} \)

7. **Analyzing Each Option**:
- **Option 1 (0)**: The particle can have zero velocity at the extreme points of its motion.
- **Option 2 (\( v_{max} \))**: This is the maximum velocity, which can be attained at the equilibrium position.
- **Option 3 (\( \frac{v_{max}}{2} \))**: This is a valid velocity that can be achieved at some point during the oscillation.
- **Option 4 (\( \frac{v_{max}}{\sqrt{2}} \))**: This corresponds to the velocity at \( x = \frac{A}{\sqrt{2}} \), which is a significant point in SHM.

8. **Conclusion**:
- Since the time period \( T \) is defined as the time taken to complete one full oscillation, the particle can achieve any of the velocities mentioned in the options during its motion. However, the maximum velocity and the velocity at specific points in the motion (like \( \frac{v_{max}}{2} \) and \( \frac{v_{max}}{\sqrt{2}} \)) are particularly relevant.
- Therefore, the possible magnitudes of \( v \) could be \( 0 \), \( v_{max} \), \( \frac{v_{max}}{2} \), or \( \frac{v_{max}}{\sqrt{2}} \).

### Final Answer:
The magnitude of \( \vec{v} \) may be: **0, \( v_{max} \), \( \frac{v_{max}}{2} \), or \( \frac{v_{max}}{\sqrt{2}} \)**.