A `20g` particle is executing `SHM` between the limits `(5,0,0)cm` and `(15, 0, 0)cm`. The total distance convered during one oscillation is

Options

A. `10 cm`
B. `15cm`
C. `20 cm`
D. `25 cm`

Explanation

To find the total distance covered by a particle executing simple harmonic motion (SHM) between the limits of \(5.0 \, \text{cm}\) and \(15.0 \, \text{cm}\), we can follow these steps:

### Step 1: Identify the limits of oscillation
The particle oscillates between two points:
- Minimum position (left limit): \(x_1 = 5.0 \, \text{cm}\)
- Maximum position (right limit): \(x_2 = 15.0 \, \text{cm}\)

### Step 2: Determine the mean position
The mean position (equilibrium position) in SHM is the midpoint between the two limits:
\[ \text{Mean position} = \frac{x_1 + x_2}{2} = \frac{5.0 \, \text{cm} + 15.0 \, \text{cm}}{2} = 10.0 \, \text{cm} \]

### Step 3: Calculate the distance from the mean position to the limits
The distance from the mean position to either limit is:
- Distance from mean position to minimum position:
\[ d_1 = 10.0 \, \text{cm} - 5.0 \, \text{cm} = 5.0 \, \text{cm} \]
- Distance from mean position to maximum position:
\[ d_2 = 15.0 \, \text{cm} - 10.0 \, \text{cm} = 5.0 \, \text{cm} \]

### Step 4: Calculate the total distance covered in one complete oscillation
In one complete oscillation, the particle travels from the minimum position to the maximum position and back to the minimum position:
- Distance from minimum to maximum: \(5.0 \, \text{cm} + 5.0 \, \text{cm} = 10.0 \, \text{cm}\)
- Distance from maximum back to minimum: \(5.0 \, \text{cm} + 5.0 \, \text{cm} = 10.0 \, \text{cm}\)

Thus, the total distance covered in one oscillation is:
\[ \text{Total distance} = 10.0 \, \text{cm} + 10.0 \, \text{cm} = 20.0 \, \text{cm} \]

### Final Answer
The total distance covered during one oscillation is \(20.0 \, \text{cm}\).

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