Which of the following relationships between a particle's acceleration a and its position x harmonic oscillation (a) `a= 3x^(2)`, (b) a= 5x, (c ) `a= -4x`, (d) `a = -2//x`? For the SHM. What is the angular frequency (assume the unit of rad/s)?

Explanation

To solve the problem, we need to determine which of the given relationships between a particle's acceleration \( a \) and its position \( x \) corresponds to simple harmonic motion (SHM). The key characteristic of SHM is that the acceleration \( a \) is directly proportional to the negative of the displacement \( x \).

### Step-by-Step Solution:

1. **Understanding the Condition for SHM**:
- For a particle in SHM, the acceleration \( a \) must be proportional to \( -x \). This can be expressed mathematically as:
\[ a = -k x \]
where \( k \) is a positive constant.

2. **Analyzing Each Option**:
- **Option (a)**: \( a = 3x^2 \)
- This does not satisfy the condition for SHM since it is not linear in \( x \) and does not have a negative sign.
- **Option (b)**: \( a = 5x \)
- This does not satisfy the SHM condition since it is positive and not negative.
- **Option (c)**: \( a = -4x \)
- This satisfies the SHM condition as it is linear in \( x \) and has a negative sign.
- **Option (d)**: \( a = -\frac{2}{x} \)
- This does not satisfy the SHM condition since it is not proportional to \( x \) directly.

3. **Identifying the Correct Option**:
- The only option that satisfies the condition for SHM is option (c): \( a = -4x \).

4. **Finding the Angular Frequency**:
- From the equation \( a = -k x \), we can identify that \( k = 4 \).
- The angular frequency \( \omega \) can be calculated using the relationship:
\[ \omega^2 = k \]
- Therefore, substituting \( k = 4 \):
\[ \omega^2 = 4 \implies \omega = \sqrt{4} = 2 \text{ rad/s} \]

### Final Answers:
- The correct relationship for SHM is **(c) \( a = -4x \)**.
- The angular frequency \( \omega \) is **2 rad/s**.