If a simple harmonic oscillator has got a displacement of `0.02m` and acceleration equal to `2.0ms^-2` at any time, the angular frequency of the oscillator is equal to

Options

A. `10rads^-1`
B. `0.1rads^-1`
C. `100rads^-1`
D. `1 rad s^-1`

Explanation

To find the angular frequency of a simple harmonic oscillator given its displacement and acceleration, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values**:
- Displacement (x) = 0.02 m
- Acceleration (a) = 2.0 m/s²

2. **Use the formula for acceleration in simple harmonic motion**:
The acceleration (a) of a simple harmonic oscillator can be expressed as:
\[ a = -\omega^2 x \]
where:
- \( \omega \) is the angular frequency,
- \( x \) is the displacement.

3. **Consider the magnitude of acceleration**:
Since we are interested in the magnitude, we can write:
\[ |a| = \omega^2 |x| \]
Thus, we can rewrite the equation as:
\[ a = \omega^2 x \]

4. **Substitute the known values into the equation**:
Substitute \( a = 2.0 \, \text{m/s}^2 \) and \( x = 0.02 \, \text{m} \):
\[ 2.0 = \omega^2 \times 0.02 \]

5. **Solve for \( \omega^2 \)**:
Rearranging the equation gives:
\[ \omega^2 = \frac{2.0}{0.02} \]
\[ \omega^2 = 100 \, \text{s}^{-2} \]

6. **Calculate \( \omega \)**:
Taking the square root of both sides:
\[ \omega = \sqrt{100} = 10 \, \text{radians/second} \]

### Final Answer:
The angular frequency of the oscillator is \( \omega = 10 \, \text{radians/second} \).

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