A particle executing a simple harmonic motion has a period of 6 s. The time taken by the particle to move from the mean position to half the amplitude, starting from the mean position is

Options

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

Answer

To solve the problem of finding the time taken by a particle executing simple harmonic motion (SHM) to move from the mean position to half the amplitude, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data**:
- The period \( T \) of the SHM is given as \( 6 \) seconds.

2. **Calculate the Angular Frequency**:
- The angular frequency \( \omega \) is calculated using the formula:
\[ \omega = \frac{2\pi}{T} \]
- Substituting the value of \( T \):
\[ \omega = \frac{2\pi}{6} = \frac{\pi}{3} \text{ rad/s} \]

3. **Write the Displacement Equation**:
- The displacement \( x \) in SHM is given by:
\[ x = a \sin(\omega t) \]
- Here, \( a \) is the amplitude.

4. **Set Up the Equation for Half Amplitude**:
- We need to find the time when the particle is at half the amplitude, which means:
\[ x = \frac{a}{2} \]
- Therefore, we have:
\[ \frac{a}{2} = a \sin(\omega t) \]
- Dividing both sides by \( a \) (assuming \( a \neq 0 \)):
\[ \frac{1}{2} = \sin(\omega t) \]

5. **Find the Angle Corresponding to \( \sin(\omega t) = \frac{1}{2} \)**:
- The angle \( \omega t \) for which \( \sin(\omega t) = \frac{1}{2} \) is:
\[ \omega t = \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \text{ radians} \]

6. **Solve for Time \( t \)**:
- Now, substituting \( \omega \) into the equation:
\[ t = \frac{\omega t}{\omega} = \frac{\frac{\pi}{6}}{\frac{\pi}{3}} = \frac{1}{2} \text{ seconds} \]

7. **Conclusion**:
- The time taken by the particle to move from the mean position to half the amplitude is:
\[ t = \frac{1}{2} \text{ seconds} \]

### Final Answer:
The time taken to move from the mean position to half the amplitude is \( \frac{1}{2} \) seconds.