A particle executes simple harmonic motion. Its amplitude is 8 cm and time period is 6s.The time it will take to travel from its position of maximum displacement to the point correspoding to half of its amplitude, is _________s.

Explanation

To solve the problem, we need to analyze the motion of a particle undergoing simple harmonic motion (SHM).

### Step-by-Step Solution:

1. **Understanding SHM Parameters**:
- The amplitude \( A \) of the motion is given as \( 8 \) cm.
- The time period \( T \) is given as \( 6 \) seconds.

2. **Finding the Position at Maximum Displacement**:
- The maximum displacement corresponds to the amplitude, which is \( x = A = 8 \) cm.

3. **Finding the Position at Half Amplitude**:
- Half of the amplitude is \( \frac{A}{2} = \frac{8 \, \text{cm}}{2} = 4 \, \text{cm} \).

4. **Using the Equation of Motion**:
- The position \( x \) in SHM can be described by the equation:
\[ x(t) = A \cos\left(\frac{2\pi}{T} t\right) \]
- We need to find the time \( t \) when the particle moves from \( x = 8 \) cm to \( x = 4 \) cm.

5. **Setting Up the Equation**:
- At \( x = 4 \) cm:
\[ 4 = 8 \cos\left(\frac{2\pi}{6} t\right) \]
- Simplifying this gives:
\[ \cos\left(\frac{2\pi}{6} t\right) = \frac{1}{2} \]

6. **Solving for Time**:
- The cosine function equals \( \frac{1}{2} \) at angles \( \frac{\pi}{3} \) and \( \frac{5\pi}{3} \).
- Thus, we have:
\[ \frac{2\pi}{6} t = \frac{\pi}{3} \quad \text{or} \quad \frac{2\pi}{6} t = \frac{5\pi}{3} \]
- Solving for \( t \):
- For \( \frac{\pi}{3} \):
\[ t = \frac{6}{2} \cdot \frac{1}{3} = 1 \, \text{s} \]
- For \( \frac{5\pi}{3} \):
\[ t = \frac{6}{2} \cdot \frac{5}{3} = 5 \, \text{s} \]

7. **Finding the Time Interval**:
- Since the particle starts at maximum displacement (8 cm) and moves to half amplitude (4 cm), we take the first solution:
- The time taken to go from maximum displacement to half amplitude is \( 1 \, \text{s} \).

### Final Answer:
The time it will take to travel from its position of maximum displacement to the point corresponding to half of its amplitude is **1 second**.