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

Options

A. 40 mm /s
B. 60mm/s
C. 113 mm/s
D. 120 mm/s

Answer

To solve the problem, we need to find the velocity of a body executing simple harmonic motion (SHM) at a displacement of 20 mm, given that the amplitude is 60 mm and the angular frequency is 2 rad/s.

### Step-by-Step Solution:

1. **Identify the given values:**
- Angular frequency (ω) = 2 rad/s
- Displacement (x) = 20 mm = 0.02 m
- Amplitude (A) = 60 mm = 0.06 m

2. **Use the formula for velocity in SHM:**
The velocity (v) of a body in simple harmonic motion can be calculated using the formula:
\[ v = \omega \sqrt{A^2 - x^2} \]

3. **Substitute the values into the formula:**
\[ v = 2 \sqrt{(0.06)^2 - (0.02)^2} \]

4. **Calculate \(A^2\) and \(x^2\):**
\[ A^2 = (0.06)^2 = 0.0036 \, \text{m}^2 \]
\[ x^2 = (0.02)^2 = 0.0004 \, \text{m}^2 \]

5. **Subtract \(x^2\) from \(A^2\):**
\[ A^2 - x^2 = 0.0036 - 0.0004 = 0.0032 \, \text{m}^2 \]

6. **Take the square root:**
\[ \sqrt{0.0032} = \sqrt{32 \times 10^{-4}} = \sqrt{32} \times 10^{-2} = 4\sqrt{2} \times 0.01 = 0.04\sqrt{2} \, \text{m} \]

7. **Calculate the velocity:**
\[ v = 2 \times 0.04\sqrt{2} = 0.08\sqrt{2} \, \text{m/s} \]

8. **Convert to mm/s:**
Since 1 m = 1000 mm, we have:
\[ v = 0.08\sqrt{2} \times 1000 \, \text{mm/s} = 80\sqrt{2} \, \text{mm/s} \]

9. **Approximate the value:**
\[ \sqrt{2} \approx 1.414 \Rightarrow 80 \times 1.414 \approx 113.12 \, \text{mm/s} \]

### Final Answer:
The velocity of the body at a displacement of 20 mm is approximately **113 mm/s**.