If at some instant of time, the displacement of a simple harmonic oscillator is 0.02 m and its acceleration is `2ms^(-2)`, then the angular frequency of the oscillator is

Options

A. `"100 rad s"^(-1)`
B. `"10 rad s"^(-1)`
C. `"1 rad s"^(-1)`
D. `"0.1 rad s"^(-1)`

Answer

To find the angular frequency of a simple harmonic oscillator given its displacement and acceleration, we can use the relationship between acceleration, displacement, and angular frequency.

### Step-by-Step Solution:

1. **Identify the given values:**
- Displacement (x) = 0.02 m
- Acceleration (a) = 2 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, \(\omega\) is the angular frequency, and \(x\) is the displacement. The negative sign indicates that the acceleration is directed opposite to the displacement.

3. **Rearranging the formula:**
We can ignore the negative sign for the magnitude calculation:
\[ a = \omega^2 x \]
Rearranging this gives:
\[ \omega^2 = \frac{a}{x} \]

4. **Substituting the known values:**
Substitute the values of acceleration and displacement into the equation:
\[ \omega^2 = \frac{2 \, \text{m/s}^2}{0.02 \, \text{m}} \]

5. **Calculating \(\omega^2\):**
\[ \omega^2 = \frac{2}{0.02} = 100 \, \text{rad}^2/\text{s}^2 \]

6. **Finding \(\omega\):**
To find \(\omega\), take the square root of \(\omega^2\):
\[ \omega = \sqrt{100} = 10 \, \text{rad/s} \]

### Final Answer:
The angular frequency of the oscillator is \(10 \, \text{rad/s}\).

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