A particle executing of a simple harmonic motion covers a distance equal to half its amplitude in on e second . Then the time period of the simple harmonic motion is

Options

A. 4 s
B. 6 s
C. 8 s
D. 12 s

Explanation

To solve the problem of finding the time period of a particle executing simple harmonic motion (SHM) that covers a distance equal to half its amplitude in one second, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the SHM Equation**:
The general equation for displacement in SHM is given by:
\[ y = A \sin(\omega t) \]
where \( y \) is the displacement, \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( t \) is the time.

2. **Identify Given Information**:
The problem states that the particle covers a distance equal to half its amplitude in one second. Thus, we have:
\[ y = \frac{A}{2} \quad \text{and} \quad t = 1 \text{ second} \]

3. **Set Up the Equation**:
Substitute the known values into the SHM equation:
\[ \frac{A}{2} = A \sin(\omega \cdot 1) \]
Simplifying this gives:
\[ \frac{1}{2} = \sin(\omega) \]

4. **Find the Value of \(\omega\)**:
We know that \(\sin(\frac{\pi}{6}) = \frac{1}{2}\). Therefore:
\[ \omega = \frac{\pi}{6} \text{ radians/second} \]

5. **Calculate the Time Period**:
The time period \( T \) of SHM is related to the angular frequency by the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of \(\omega\):
\[ T = \frac{2\pi}{\frac{\pi}{6}} = 2 \cdot 6 = 12 \text{ seconds} \]

6. **Conclusion**:
The time period of the simple harmonic motion is:
\[ T = 12 \text{ seconds} \]

### Final Answer:
The time period of the simple harmonic motion is **12 seconds**.