A particle executes SHM with a time period of `4 s`. Find the time taken by the particle to go directly from its mean position to half of its amplitude.

Explanation

To solve the problem of finding the time taken by a particle executing Simple Harmonic Motion (SHM) to go from its mean position to half of its amplitude, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Time period \( T = 4 \) seconds.

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

3. **Set Up the Displacement Equation:**
- The displacement \( x \) in SHM is given by:
\[ x = A \sin(\omega t) \]
- We need to find the time when the particle is at half of its amplitude, which means:
\[ x = \frac{A}{2} \]

4. **Substitute the Displacement into the Equation:**
- Setting \( x = \frac{A}{2} \):
\[ \frac{A}{2} = A \sin(\omega t) \]
- Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ \frac{1}{2} = \sin(\omega t) \]

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

6. **Substitute \( \omega \) to Find \( t \):**
- Now substituting \( \omega = \frac{\pi}{2} \):
\[ \frac{\pi}{6} = \frac{\pi}{2} t \]
- Solving for \( t \):
\[ t = \frac{\frac{\pi}{6}}{\frac{\pi}{2}} = \frac{1}{3} \, \text{seconds} \]

### Final Answer:
The time taken by the particle to go directly from its mean position to half of its amplitude is \( \frac{1}{3} \) seconds.

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