A body executing S.H.M.along a straight line has a velocity of `3 ms^(-1)` when it is at a distance of 4m from its mean position and`4ms^(-1)` when it is at a distance of 3m from its mean position.Its angular frequency and amplitude are

Options

A. `2 rad s^(-1) & 5m`
B. `1 rad s^(-1) & 10 m`
C. ` 2 rad s^(-1)& 10 m`
D. ` 1 rad s^(-1) & 5m`

Answer

To solve the problem, we need to find the angular frequency (ω) and amplitude (A) of a body executing simple harmonic motion (SHM) given its velocities at specific displacements from the mean position.

### Step-by-Step Solution:

1. **Understand the relationship between velocity, angular frequency, amplitude, and displacement:**
The velocity (v) of a body in SHM is given by the formula:
\[ v = \omega \sqrt{A^2 - x^2} \]
where:
- \( v \) = velocity
- \( \omega \) = angular frequency
- \( A \) = amplitude
- \( x \) = displacement from the mean position

2. **Set up the equations for the two given conditions:**
We have two scenarios:
- When \( x_1 = 4 \, m \), \( v_1 = 3 \, m/s \)
- When \( x_2 = 3 \, m \), \( v_2 = 4 \, m/s \)

From the formula, we can write:
\[ v_1 = \omega \sqrt{A^2 - x_1^2} \quad \text{(1)} \]
\[ v_2 = \omega \sqrt{A^2 - x_2^2} \quad \text{(2)} \]

3. **Substituting the known values into the equations:**
From equation (1):
\[ 3 = \omega \sqrt{A^2 - 4^2} \]
\[ 3 = \omega \sqrt{A^2 - 16} \quad \text{(3)} \]

From equation (2):
\[ 4 = \omega \sqrt{A^2 - 3^2} \]
\[ 4 = \omega \sqrt{A^2 - 9} \quad \text{(4)} \]

4. **Square both equations to eliminate the square root:**
Squaring equation (3):
\[ 9 = \omega^2 (A^2 - 16) \quad \text{(5)} \]

Squaring equation (4):
\[ 16 = \omega^2 (A^2 - 9) \quad \text{(6)} \]

5. **Rearranging equations (5) and (6):**
From (5):
\[ \omega^2 A^2 - 16\omega^2 = 9 \quad \Rightarrow \quad \omega^2 A^2 = 9 + 16\omega^2 \quad \text{(7)} \]

From (6):
\[ \omega^2 A^2 - 9\omega^2 = 16 \quad \Rightarrow \quad \omega^2 A^2 = 16 + 9\omega^2 \quad \text{(8)} \]

6. **Setting equations (7) and (8) equal to each other:**
\[ 9 + 16\omega^2 = 16 + 9\omega^2 \]
Rearranging gives:
\[ 16\omega^2 - 9\omega^2 = 16 - 9 \]
\[ 7\omega^2 = 7 \quad \Rightarrow \quad \omega^2 = 1 \quad \Rightarrow \quad \omega = 1 \, \text{rad/s} \]

7. **Substituting ω back to find A:**
Using equation (3):
\[ 3 = 1 \cdot \sqrt{A^2 - 16} \]
\[ 3 = \sqrt{A^2 - 16} \]
Squaring both sides:
\[ 9 = A^2 - 16 \quad \Rightarrow \quad A^2 = 25 \quad \Rightarrow \quad A = 5 \, m \]

### Final Results:
- Angular frequency \( \omega = 1 \, \text{rad/s} \)
- Amplitude \( A = 5 \, m \)