The potential energt of a particle of mass 0.1 kg, moving along the x-axis, is given by `U=5x(x-4)J`, where x is in meter. It can be concluded that

Options

A. The speed of the particle is maximum at x=2 m
B. The particle executes simple harmonic motion
C. The period of oscillation of the particle is `(pi)/(5)` s
D. all are correct

Explanation

To solve the problem step by step, we will analyze the potential energy function, derive the force, determine the conditions for simple harmonic motion (SHM), and calculate the period of oscillation.

### Step 1: Write down the potential energy function
The potential energy \( U \) is given by:
\[ U = 5x(x - 4) \text{ J} \]
Expanding this, we get:
\[ U = 5x^2 - 20x \text{ J} \]

### Step 2: Find the force acting on the particle
The force \( F \) acting on the particle can be found using the relation:
\[ F = -\frac{dU}{dx} \]
Calculating the derivative:
\[ \frac{dU}{dx} = \frac{d}{dx}(5x^2 - 20x) = 10x - 20 \]
Thus, the force is:
\[ F = - (10x - 20) = 20 - 10x \]

### Step 3: Determine the conditions for SHM
For the motion to be simple harmonic, the force must be proportional to the displacement and directed towards the equilibrium position. The force we derived is:
\[ F = 20 - 10x \]
This can be rewritten as:
\[ F = -10(x - 2) \]
This indicates that the force is proportional to the displacement from the equilibrium position \( x = 2 \). Therefore, the motion is indeed simple harmonic motion (SHM).

### Step 4: Find the spring constant \( k \)
From the force equation \( F = -kx \), we can identify \( k \):
\[ k = 10 \text{ N/m} \]

### Step 5: Calculate the period of oscillation
The period \( T \) of SHM is given by the formula:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
Substituting the mass \( m = 0.1 \) kg and \( k = 10 \) N/m:
\[ T = 2\pi \sqrt{\frac{0.1}{10}} = 2\pi \sqrt{0.01} = 2\pi \cdot 0.1 = \frac{2\pi}{10} = \frac{\pi}{5} \text{ seconds} \]

### Step 6: Determine where the speed is maximum
The speed of the particle is maximum at the equilibrium position where the potential energy is minimum. The equilibrium position occurs when the net force is zero:
\[ 20 - 10x = 0 \implies x = 2 \text{ m} \]
Thus, the speed is maximum at \( x = 2 \).

### Conclusion
1. The speed of the particle is maximum at \( x = 2 \) m.
2. The particle executes simple harmonic motion.
3. The period of oscillation is \( \frac{\pi}{5} \) seconds.

### Final Answer
All conclusions are correct, hence the answer is option 4.

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