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 step by step, we will analyze the motion of a particle executing Simple Harmonic Motion (SHM) and determine the time taken to move from the mean position to half of its amplitude.

### Step-by-Step Solution:

1. **Understanding SHM**:
- In SHM, the position of the particle can be described by the equation:
\[ x(t) = A \sin(\omega t) \]
where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( t \) is the time.

2. **Given Data**:
- The time period \( T \) is given as \( 4 \, \text{s} \).
- We need to find the time taken to move from the mean position (where \( x = 0 \)) to half of the amplitude (where \( x = \frac{A}{2} \)).

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

4. **Setting Up the Equation**:
- At half of the amplitude, 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. **Solving for Time**:
- The sine function gives us:
\[ \sin(\omega t) = \frac{1}{2} \]
- The angle whose sine is \( \frac{1}{2} \) is \( \frac{\pi}{6} \) radians. Therefore:
\[ \omega t = \frac{\pi}{6} \]
- Substituting \( \omega = \frac{\pi}{2} \):
\[ \frac{\pi}{2} t = \frac{\pi}{6} \]

6. **Isolating \( t \)**:
- Canceling \( \pi \) from both sides:
\[ \frac{1}{2} t = \frac{1}{6} \]
- Multiplying both sides by \( 2 \):
\[ t = \frac{1}{3} \, \text{s} \]

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

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