The potential energy of a particle of mass 1 kg `U = 10 + (x-2)^(2)`. Herer, U is in joule and x in met. On the positive x=axis particle travels up to x=+6cm. Choose the wrong statement.

Options

A. on negative x-axis particle travels up to x = -2m
B. The maximum kinetic energy of the particle is 16 J
C. The period of oscillation of the particle is `sqrt(2pi)` seconds.
D. None of the above.

Answer

To solve the problem step by step, we will analyze the potential energy function, derive the force, determine the equilibrium position, calculate the maximum kinetic energy, and finally find the time period of oscillation. We will then identify the wrong statement among the given options.

### Step 1: Analyze the Potential Energy Function
The potential energy \( U \) is given by:
\[ U = 10 + (x - 2)^2 \]
where \( U \) is in joules and \( x \) is in meters.

### Step 2: Find the Force
The force \( F \) acting on the particle can be derived from the potential energy using the relation:
\[ F = -\frac{dU}{dx} \]
Calculating the derivative:
\[ \frac{dU}{dx} = \frac{d}{dx}(10 + (x - 2)^2) = 2(x - 2) \]
Thus, the force is:
\[ F = -2(x - 2) = -2x + 4 \]

### Step 3: Determine the Equilibrium Position
The equilibrium position occurs when the force is zero:
\[ -2x + 4 = 0 \implies x = 2 \text{ meters} \]

### Step 4: Calculate the Maximum Kinetic Energy
The particle travels up to \( x = 6 \) cm (or \( 0.06 \) m). We need to calculate the potential energy at the equilibrium position and at the maximum displacement.

1. **At the equilibrium position \( x = 2 \) m**:
\[ U(2) = 10 + (2 - 2)^2 = 10 \text{ joules} \]

2. **At maximum displacement \( x = 0.06 \) m**:
\[ U(0.06) = 10 + (0.06 - 2)^2 = 10 + (1.94)^2 = 10 + 3.7616 \approx 13.7616 \text{ joules} \]

The change in potential energy \( \Delta U \) when moving from the equilibrium position to the maximum displacement is:
\[ \Delta U = U(0.06) - U(2) = 13.7616 - 10 \approx 3.7616 \text{ joules} \]

The maximum kinetic energy \( K_{\text{max}} \) can be found using the conservation of mechanical energy:
\[ K_{\text{max}} = E_{\text{total}} - U_{\text{min}} = \Delta U = 3.7616 \text{ joules} \]

### Step 5: Find the Time Period of Oscillation
The time period \( T \) of oscillation is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
where \( k \) is the spring constant. From the force equation \( F = -kx \), we can identify \( k \) from the force expression:
\[ F = -2x \implies k = 2 \text{ N/m} \]
Thus, substituting the values:
\[ T = 2\pi \sqrt{\frac{1 \text{ kg}}{2 \text{ N/m}}} = 2\pi \sqrt{\frac{1}{2}} = \pi \sqrt{2} \text{ seconds} \]

### Step 6: Identify the Wrong Statement
Now, we need to evaluate the statements provided in the question:

1. The particle travels up to \( x = -2 \) m on the negative x-axis (True).
2. The maximum kinetic energy of the particle is \( 16 \) joules (False, it is approximately \( 3.7616 \) joules).
3. The time period of oscillation is \( \pi \sqrt{2} \) seconds (True).
4. The equilibrium position is at \( x = 2 \) m (True).

Thus, the wrong statement is:
**The maximum kinetic energy of the particle is \( 16 \) joules.**