A particle executing SHM has a maximum speed of `0.5ms^(-1)` and maximum acceleration of `1.0ms^(-2)`. The angular frequency of oscillation is

Options

A. `2 " rad s"^(-1)`
B. `0.5 " rad s"^(-1)`
C. `2pi " rad s"^(-1)`
D. `0.5 " rad s"^(-1)`

Answer

To find the angular frequency of a particle executing simple harmonic motion (SHM) given its maximum speed and maximum acceleration, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the formulas for maximum speed and maximum acceleration in SHM**:
- The maximum speed \( v_{\text{max}} \) is given by:
\[ v_{\text{max}} = A \omega \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.
- The maximum acceleration \( a_{\text{max}} \) is given by:
\[ a_{\text{max}} = A \omega^2 \]

2. **Given values**:
- Maximum speed \( v_{\text{max}} = 0.5 \, \text{m/s} \)
- Maximum acceleration \( a_{\text{max}} = 1.0 \, \text{m/s}^2 \)

3. **Set up the equations**:
- From the maximum speed equation:
\[ v_{\text{max}} = A \omega \quad \text{(1)} \]
- From the maximum acceleration equation:
\[ a_{\text{max}} = A \omega^2 \quad \text{(2)} \]

4. **Divide equation (1) by equation (2)**:
\[ \frac{v_{\text{max}}}{a_{\text{max}}} = \frac{A \omega}{A \omega^2} \]
This simplifies to:
\[ \frac{v_{\text{max}}}{a_{\text{max}}} = \frac{1}{\omega} \]

5. **Substitute the known values**:
\[ \frac{0.5}{1.0} = \frac{1}{\omega} \]
This simplifies to:
\[ 0.5 = \frac{1}{\omega} \]

6. **Solve for \( \omega \)**:
\[ \omega = \frac{1}{0.5} = 2 \, \text{rad/s} \]

### Final Answer:
The angular frequency of oscillation is \( \omega = 2 \, \text{rad/s} \).

---