A particle executes SHM with periodic time of 6 seconds. The time taken for traversing a distance of half the amplitude from mean position is :

Options

A. 3 sec.
B. 2 sec.
C. 1 sec.
D. 1/2 sec.

Answer

To solve the problem, we need to determine the time taken for a particle in Simple Harmonic Motion (SHM) to traverse a distance of half the amplitude from its mean position.

### Step-by-Step Solution:

1. **Understand the Parameters of SHM**:
- The periodic time (T) of the particle is given as 6 seconds.
- The amplitude (A) is not provided, but we will denote it as A for our calculations.

2. **Determine the Angular Frequency**:
- The angular frequency (ω) can be calculated using the formula:
\[ \omega = \frac{2\pi}{T} \]
- Substituting the value of T:
\[ \omega = \frac{2\pi}{6} = \frac{\pi}{3} \text{ radians/second} \]

3. **Calculate the Distance to be Traversed**:
- We need to find the time taken to traverse a distance of half the amplitude from the mean position. This distance is:
\[ \text{Distance} = \frac{A}{2} \]

4. **Use the SHM Equation**:
- The displacement (x) in SHM can be described by the equation:
\[ x = A \sin(\omega t) \]
- To find the time taken to reach \( x = \frac{A}{2} \):
\[ \frac{A}{2} = A \sin(\omega t) \]
- Dividing both sides by A (assuming A ≠ 0):
\[ \frac{1}{2} = \sin(\omega t) \]

5. **Find the Angle Corresponding to \(\sin(\theta) = \frac{1}{2}\)**:
- The angle for which \(\sin(\theta) = \frac{1}{2}\) is:
\[ \omega t = \frac{\pi}{6} \quad \text{(first quadrant)} \]
- Therefore, we can solve for t:
\[ t = \frac{\frac{\pi}{6}}{\omega} = \frac{\frac{\pi}{6}}{\frac{\pi}{3}} = \frac{1}{2} \text{ seconds} \]

6. **Conclusion**:
- The time taken for the particle to traverse a distance of half the amplitude from the mean position is **0.5 seconds**.

### Final Answer:
The time taken for traversing a distance of half the amplitude from the mean position is **0.5 seconds**.