A body is executing simple harmonic motion with an angular frequency 2 rad/s . The velocity of the body at 20 mm displacement, when the amplitude of motion is 60 mm is -

Options

A. 131 `mms^(-1)`
B. 118 `mms^(-1)`
C. 113 `mms^(-1)`
D. 90 `mms^(-1)`

Explanation

To solve the problem of finding the velocity of a body executing simple harmonic motion (SHM) at a specific displacement, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Angular frequency (ω) = 2 rad/s
- Amplitude (A) = 60 mm = 0.06 m (convert to meters for standard SI units)
- Displacement (x) = 20 mm = 0.02 m (convert to meters)

2. **Use the Velocity Formula for SHM**:
The formula for the velocity (v) of a particle in simple harmonic motion is given by:
\[ v = \omega \sqrt{A^2 - x^2} \]

3. **Substitute the Known Values**:
Substitute the values of ω, A, and x into the formula:
\[ v = 2 \cdot \sqrt{(0.06)^2 - (0.02)^2} \]

4. **Calculate the Squares**:
- \( A^2 = (0.06)^2 = 0.0036 \)
- \( x^2 = (0.02)^2 = 0.0004 \)

5. **Subtract the Squares**:
\[ A^2 - x^2 = 0.0036 - 0.0004 = 0.0032 \]

6. **Take the Square Root**:
\[ \sqrt{0.0032} = 0.05657 \text{ (approximately)} \]

7. **Calculate the Velocity**:
Now substitute back into the velocity formula:
\[ v = 2 \cdot 0.05657 \approx 0.11314 \text{ m/s} \]

8. **Convert to mm/s**:
Since we initially worked in mm, we convert the final answer back to mm/s:
\[ v \approx 113.14 \text{ mm/s} \]

9. **Determine the Sign of Velocity**:
The problem states the velocity is negative, indicating the direction of motion. Therefore, the final answer is:
\[ v \approx -113.14 \text{ mm/s} \]

### Final Answer:
The velocity of the body at 20 mm displacement when the amplitude of motion is 60 mm is approximately **-113.14 mm/s**.