A particle executes SHM along a straight line so that its period is 12 s. The time it takes in traversing a distance equal to half its amplitude from its equilibrium position is

Options

A. 6s
B. 4s
C. 2s
D. 1s

Explanation

To solve the problem of finding the time it takes for a particle executing Simple Harmonic Motion (SHM) to traverse a distance equal to half its amplitude from its equilibrium position, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the parameters of SHM:**
- The period \( T \) of the SHM is given as 12 seconds.

2. **Calculate the angular frequency \( \omega \):**
- The angular frequency \( \omega \) is calculated using the formula:
\[ \omega = \frac{2\pi}{T} \]
- Substituting \( T = 12 \) seconds:
\[ \omega = \frac{2\pi}{12} = \frac{\pi}{6} \text{ rad/s} \]

3. **Set up the equation for SHM:**
- The displacement \( x \) in SHM can be expressed as:
\[ x = A \sin(\omega t + \phi) \]
- For simplicity, we can assume \( \phi = 0 \) (the particle starts from the equilibrium position):
\[ x = A \sin(\omega t) \]

4. **Determine the displacement for half the amplitude:**
- We want to find the time when the particle is at a distance equal to half its amplitude:
\[ x = \frac{A}{2} \]

5. **Set up the equation:**
- Substitute \( x = \frac{A}{2} \) into the SHM equation:
\[ \frac{A}{2} = A \sin(\omega t) \]
- Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ \frac{1}{2} = \sin(\omega t) \]

6. **Solve for \( \omega t \):**
- The sine function equals \( \frac{1}{2} \) at:
\[ \omega t = \frac{\pi}{6} \]
- Therefore:
\[ t = \frac{\pi}{6\omega} \]

7. **Substitute \( \omega \):**
- Now substitute \( \omega = \frac{\pi}{6} \):
\[ t = \frac{\pi}{6 \cdot \frac{\pi}{6}} = \frac{\pi}{\frac{\pi}{6}} = 1 \text{ second} \]

### Final Answer:
The time it takes for the particle to traverse a distance equal to half its amplitude from its equilibrium position is **1 second**.