According to a scientists, he applied a force `F = -cx^(1//3)` on a particle and the particle is performing SHM. No other force acted on the particle. He refuses to tell whether `c` is a constant or not. Assume that he had worked only with positive `x` then :

Options

A. as `x` increases `c` also increases
B. as `x` increases `c` also decreases
C. as `x` increases `c` remain constant
D. the motion cannot be SHM

Answer

To solve the problem, we need to analyze the force applied on the particle and its relation to Simple Harmonic Motion (SHM).

### Step-by-Step Solution:

1. **Identify the Force**: The force applied on the particle is given by \( F = -c x^{1/3} \).

2. **Condition for SHM**: For a particle to be in SHM, the restoring force must be proportional to the displacement \( x \) and can be expressed as:
\[ F = -m \omega^2 x \]
where \( m \) is the mass of the particle and \( \omega \) is the angular frequency.

3. **Equate the Forces**: Since both expressions represent the force acting on the particle, we can set them equal to each other:
\[ -c x^{1/3} = -m \omega^2 x \]
By removing the negative signs, we have:
\[ c x^{1/3} = m \omega^2 x \]

4. **Rearranging the Equation**: We can rearrange this equation to express \( c \):
\[ c = m \omega^2 x^{1 - 1/3} = m \omega^2 x^{2/3} \]

5. **Analyzing the Relationship**: From the equation \( c = m \omega^2 x^{2/3} \), we can see that \( c \) is directly proportional to \( x^{2/3} \). This means that as \( x \) increases, \( c \) also increases.

6. **Conclusion**: Therefore, when \( x \) increases, \( c \) must also increase. This leads us to conclude that the correct answer to the question is that when \( x \) increases, \( c \) increases.

### Final Answer:
When \( x \) increases, \( c \) also increases.

---