A student says that he had applied a force `F=-ksqrt(x)` on a particle and the particle performs in simple harmonic motion. He refuses to tell whether k is a constant or not. Assume that he has worked only with positive x and no other force acted on the particle.

Options

A. As X increases k increases
B. As X increases K decreases
C. As X increases k remains constant
D. The motion cannot be simple harmonic

Answer

To solve the problem, we need to analyze the force applied on the particle and determine how the constant \( k \) behaves as the position \( x \) changes.

### Step-by-Step Solution:

1. **Understanding the Force**: The force applied on the particle is given by:
\[ F = -k\sqrt{x} \]
where \( k \) is a parameter that we need to analyze.

2. **Condition for Simple Harmonic Motion (SHM)**: For a particle to be in simple harmonic motion, the restoring force must be proportional to the displacement from the equilibrium position and directed towards that position. This can be expressed as:
\[ F = -m\omega^2 x \]
where \( m \) is the mass of the particle and \( \omega \) is the angular frequency.

3. **Equating Forces**: We can equate the two expressions for force:
\[ -k\sqrt{x} = -m\omega^2 x \]
By removing the negative sign from both sides, we get:
\[ k\sqrt{x} = m\omega^2 x \]

4. **Rearranging the Equation**: To isolate \( k \), we can rearrange the equation:
\[ k = \frac{m\omega^2 x}{\sqrt{x}} = m\omega^2 \sqrt{x} \]

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

6. **Conclusion**: Therefore, since \( k \) is proportional to \( \sqrt{x} \), we conclude that as \( x \) increases, \( k \) increases as well.

### Final Answer:
As \( x \) increases, the value of \( k \) increases.

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