If a simple harmonic oscillator has got a displacement of 0.02 m and acceleration equal to `0.02m//s^(2)` at any time, the angular frequency of the oscillator is equal to:

Options

A. 10 rad/s
B. 0.1 rad/s
C. 100 rad/s
D. 1 rad/s

Explanation

To solve the problem, we need to find the angular frequency (ω) of a simple harmonic oscillator given its displacement (x) and acceleration (a).

### Step-by-Step Solution:

1. **Identify the given values:**
- Displacement (x) = 0.02 m
- Acceleration (a) = 0.02 m/s²

2. **Use the formula for acceleration in simple harmonic motion:**
The acceleration (a) of a simple harmonic oscillator is given by the formula:
\[ a = -\omega^2 x \]
Here, the negative sign indicates that the acceleration is in the opposite direction to the displacement, but for our calculation, we can ignore the negative sign.

3. **Rearranging the formula:**
We can rearrange the formula to solve for ω²:
\[ a = \omega^2 x \implies \omega^2 = \frac{a}{x} \]

4. **Substituting the known values:**
Now, substitute the values of a and x into the equation:
\[ \omega^2 = \frac{0.02 \, \text{m/s}^2}{0.02 \, \text{m}} = 1 \]

5. **Calculating ω:**
To find the angular frequency (ω), take the square root of ω²:
\[ \omega = \sqrt{1} = 1 \, \text{rad/s} \]

6. **Conclusion:**
The angular frequency of the oscillator is:
\[ \omega = 1 \, \text{rad/s} \]

### Final Answer:
The angular frequency of the oscillator is **1 rad/s**.