In simple harmonic motion, the ratio of acceleration of the particle to its displacement at any time is a measure of

Options

A. spring constant
B. angular frequency
C. `("angular frequency")^(2)`
D. restoring force

Explanation

To solve the question regarding the ratio of acceleration to displacement in simple harmonic motion (SHM), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Definitions**:
- In SHM, the acceleration \( a \) of a particle is given by the formula:
\[ a = -\omega^2 x \]
where:
- \( a \) is the acceleration,
- \( x \) is the displacement from the mean position,
- \( \omega \) is the angular frequency.

2. **Set Up the Ratio**:
- We need to find the ratio of acceleration \( a \) to displacement \( x \):
\[ \frac{a}{x} \]

3. **Substitute the Expression for Acceleration**:
- Substitute the expression for acceleration into the ratio:
\[ \frac{a}{x} = \frac{-\omega^2 x}{x} \]

4. **Simplify the Ratio**:
- Simplifying the above expression gives:
\[ \frac{a}{x} = -\omega^2 \]

5. **Consider the Magnitude**:
- Since the question asks for the ratio in terms of magnitude, we take the absolute value:
\[ \left| \frac{a}{x} \right| = |\omega^2| \]
- Since \( \omega^2 \) is always positive, we can write:
\[ \left| \frac{a}{x} \right| = \omega^2 \]

6. **Conclusion**:
- Therefore, the ratio of acceleration of the particle to its displacement at any time is a measure of the square of the angular frequency \( \omega \).

### Final Answer:
The ratio of acceleration to displacement in simple harmonic motion is \( \omega^2 \), which is a measure of the angular frequency.

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