Which of the following relationships between the force F on a particle and the particle's position x gives SHM: (a) `F= -5x`, (b) `F= -400x^(2)`, (c ) F= 10x, (d) `F= 3x^(2)` ?

Explanation

To determine which of the given relationships between force \( F \) and position \( x \) results in simple harmonic motion (SHM), we need to analyze each option based on the condition that the force must be proportional to the negative of the displacement from the equilibrium position. This means the relationship must be of the form:

\[ F = -kx \]

where \( k \) is a positive constant. Let's evaluate each option:

### Step 1: Analyze each relationship

**Option (a):** \( F = -5x \)

- This is in the form \( F = -kx \) where \( k = 5 \).
- The force is directly proportional to the displacement and acts in the opposite direction.
- **Conclusion:** This relationship gives SHM.

**Option (b):** \( F = -400x^2 \)

- This is in the form \( F = -k x^2 \) where \( k = 400 \).
- The force is not linear with respect to \( x \) (it is quadratic).
- **Conclusion:** This relationship does not give SHM.

**Option (c):** \( F = 10x \)

- This is in the form \( F = kx \) where \( k = 10 \).
- The force is proportional to the displacement but does not have a negative sign.
- **Conclusion:** This relationship does not give SHM.

**Option (d):** \( F = 3x^2 \)

- This is in the form \( F = k x^2 \) where \( k = 3 \).
- Similar to option (b), the force is quadratic and does not have a negative sign.
- **Conclusion:** This relationship does not give SHM.

### Final Answer

The only relationship that gives simple harmonic motion is:

**(a) \( F = -5x \)**