The distance travelled by the particle is

Options

A. 8 cm
B. 2 cm
C. 4 cm
D. zero

Explanation

To solve the problem of finding the distance traveled by a particle executing simple harmonic motion (SHM) with a given amplitude, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Amplitude**:
The amplitude (A) is the maximum distance from the mean position to either extreme position in SHM. In this case, the amplitude is given as 2 cm.

2. **Identify the Motion of the Particle**:
In one complete oscillation (one time period), the particle moves from the mean position to the maximum positive displacement (A), then back to the mean position, then to the maximum negative displacement (-A), and finally returns to the mean position.

3. **Calculate the Distance Traveled in One Complete Oscillation**:
- From the mean position to the maximum positive displacement: Distance = A = 2 cm
- From the maximum positive displacement back to the mean position: Distance = A = 2 cm
- From the mean position to the maximum negative displacement: Distance = A = 2 cm
- From the maximum negative displacement back to the mean position: Distance = A = 2 cm

Therefore, the total distance traveled in one complete oscillation is:
\[ \text{Total Distance} = A + A + A + A = 4A \]

4. **Substitute the Value of Amplitude**:
Now, substituting the value of amplitude (A = 2 cm):
\[ \text{Total Distance} = 4 \times 2 \text{ cm} = 8 \text{ cm} \]

5. **Conclusion**:
The distance traveled by the particle in one time period is 8 cm.

### Final Answer:
The distance traveled by the particle in one time period is **8 cm**.

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