If a particle executes SHM with angular velocity 3.5 rad/s and maximum acceleration `7.5m//s^(2)`, then the amplitude of oscillation will be

Options

A. 0.8 m
B. 0.69 m
C. 0.41 m
D. 0.61 m

Explanation

To find the amplitude of oscillation for a particle executing simple harmonic motion (SHM) given the angular velocity and maximum acceleration, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Angular velocity (ω) = 3.5 rad/s
- Maximum acceleration (A_max) = 7.5 m/s²

2. **Understand the Relation of Maximum Acceleration in SHM**:
- The maximum acceleration in SHM is given by the formula:
\[ A_{max} = \omega^2 \cdot A \]
where \( A \) is the amplitude of the oscillation.

3. **Rearranging the Formula to Solve for Amplitude**:
- To find the amplitude \( A \), we can rearrange the formula:
\[ A = \frac{A_{max}}{\omega^2} \]

4. **Substituting the Known Values**:
- Substitute the known values into the equation:
\[ A = \frac{7.5}{(3.5)^2} \]

5. **Calculating \( \omega^2 \)**:
- Calculate \( (3.5)^2 \):
\[ (3.5)^2 = 12.25 \]

6. **Calculating Amplitude**:
- Now substitute back to find \( A \):
\[ A = \frac{7.5}{12.25} \approx 0.6122 \text{ m} \]

7. **Rounding the Result**:
- The amplitude can be approximated to:
\[ A \approx 0.61 \text{ m} \]

8. **Final Answer**:
- Therefore, the amplitude of oscillation is approximately **0.61 m**.