A particle performs simple hormonic motion with a period of 2 seconds. The time taken by it to cover a displacement equal to half of its amplitude from the mea n position is

Options

A. `1//2` s
B. `1//3` s
C. `1//4` s
D. `1//6` s

Explanation

To solve the problem, we need to determine the time taken by a particle performing simple harmonic motion (SHM) to cover a displacement equal to half of its amplitude from the mean position.

### Step-by-Step Solution:

1. **Understand the Parameters**:
- The period of the motion (T) is given as 2 seconds.
- The amplitude (A) is not provided directly, but we will use it symbolically.

2. **Identify the Displacement**:
- The displacement we are interested in is half of the amplitude, which is \( y = \frac{A}{2} \).

3. **Use the SHM Displacement Formula**:
- The displacement in SHM can be expressed as:
\[ y = A \sin(\omega t) \]
- Where \( \omega \) is the angular frequency.

4. **Calculate Angular Frequency**:
- The angular frequency \( \omega \) is related to the period \( T \) by the formula:
\[ \omega = \frac{2\pi}{T} \]
- Substituting \( T = 2 \) seconds:
\[ \omega = \frac{2\pi}{2} = \pi \text{ rad/s} \]

5. **Set Up the Equation**:
- Now, substituting \( y = \frac{A}{2} \) into the displacement equation:
\[ \frac{A}{2} = A \sin(\pi t) \]
- Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ \frac{1}{2} = \sin(\pi t) \]

6. **Solve for Time (t)**:
- The value of \( \sin(\theta) = \frac{1}{2} \) corresponds to angles \( \theta = \frac{\pi}{6} \) and \( \theta = \frac{5\pi}{6} \). Therefore:
\[ \pi t = \frac{\pi}{6} \quad \text{or} \quad \pi t = \frac{5\pi}{6} \]
- Solving for \( t \):
- From \( \pi t = \frac{\pi}{6} \):
\[ t = \frac{1}{6} \text{ seconds} \]
- From \( \pi t = \frac{5\pi}{6} \):
\[ t = \frac{5}{6} \text{ seconds} \]

7. **Determine the Relevant Time**:
- Since we want the time taken to reach the first instance of \( \frac{A}{2} \) from the mean position, we take:
\[ t = \frac{1}{6} \text{ seconds} \]

### Final Answer:
The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is \( \frac{1}{6} \) seconds.