A particle executing S.H.M. completes a distance (taking friction as negligible) in one complete time period, equal to :

Options

A. Four times the amplitude
B. Two times the amplitude
C. One times the amplitude
D. Eight times the amplitude

Answer

To solve the problem of how much distance a particle executing Simple Harmonic Motion (S.H.M.) travels in one complete time period, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding S.H.M.**:
- A particle in S.H.M. oscillates back and forth around an equilibrium position. The maximum displacement from this position is called the amplitude (A).

2. **Identifying the Motion**:
- The particle moves from the equilibrium position (0) to the maximum positive amplitude (A), then back to the equilibrium position, then to the maximum negative amplitude (-A), and finally back to the equilibrium position.

3. **Calculating Distance Traveled**:
- From the equilibrium position (0) to the maximum positive amplitude (A):
- Distance = A
- From the maximum positive amplitude (A) back to the equilibrium position (0):
- Distance = A
- From the equilibrium position (0) to the maximum negative amplitude (-A):
- Distance = A
- From the maximum negative amplitude (-A) back to the equilibrium position (0):
- Distance = A

4. **Total Distance in One Complete Time Period**:
- Total distance traveled = Distance from 0 to A + Distance from A to 0 + Distance from 0 to -A + Distance from -A to 0
- Total distance = A + A + A + A = 4A

5. **Conclusion**:
- Therefore, the total distance traveled by the particle in one complete time period is equal to **4 times the amplitude (4A)**.

### Final Answer:
The distance traveled in one complete time period is **4A**.

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