A particle executing simple harmonic motion with an amplitude A. The distance travelled by the particle in one time period is

Options

A. zero
B. A
C. 2A
D. 4A

Explanation

To solve the problem of finding the distance traveled by a particle executing simple harmonic motion (SHM) with an amplitude \( A \) in one complete time period, we can break it down into the following steps:

### Step 1: Understand the motion of the particle
The particle in SHM moves back and forth between two extreme positions, which are at \( +A \) and \( -A \). The mean position is at \( x = 0 \).

**Hint:** Visualize the motion of the particle on a straight line between the two extremes.

### Step 2: Identify the distance traveled in one complete cycle
In one complete cycle of SHM, the particle moves from the maximum displacement \( +A \) to the mean position \( 0 \), then to the maximum displacement \( -A \), and finally back to the mean position \( 0 \) and back to \( +A \).

**Hint:** Break the motion into segments: \( +A \) to \( 0 \), \( 0 \) to \( -A \), \( -A \) to \( 0 \), and \( 0 \) to \( +A \).

### Step 3: Calculate the distance for each segment
1. From \( +A \) to \( 0 \): Distance = \( A \)
2. From \( 0 \) to \( -A \): Distance = \( A \)
3. From \( -A \) to \( 0 \): Distance = \( A \)
4. From \( 0 \) to \( +A \): Distance = \( A \)

**Hint:** Each segment contributes a distance of \( A \).

### Step 4: Sum the distances
Now, add up the distances from all segments:
\[ \text{Total distance} = A + A + A + A = 4A \]

**Hint:** Remember to add all the distances together to find the total distance traveled.

### Conclusion
The total distance traveled by the particle in one complete time period is \( 4A \).

**Final Answer:** The distance traveled by the particle in one time period is \( 4A \).