The potential energy of a particle of mass 1 kg in motin along the x-axis is given by
U = 4(1-cos2x)J
Here, x is in meter. The period of small osciallationis (in second) is

Options

A. `2pi)`
B. `pi`
C. `pi/2`
D. `sqrt(2pi)`

Answer

To solve the problem, we need to find the period of small oscillations for a particle of mass 1 kg whose potential energy is given by \( U = 4(1 - \cos(2x)) \) Joules. Let's go through the steps systematically.

### Step 1: Understanding the Potential Energy Function
The potential energy function is given as:
\[ U = 4(1 - \cos(2x)) \]
This function describes the potential energy of the particle as it moves along the x-axis.

### Step 2: Finding the Force
The force acting on the particle can be found using the relation:
\[ F = -\frac{dU}{dx} \]
We need to differentiate \( U \) with respect to \( x \).

### Step 3: Differentiate the Potential Energy
Differentiating \( U \):
\[ \frac{dU}{dx} = \frac{d}{dx}[4(1 - \cos(2x))] = 4 \cdot 0 + 4 \cdot \sin(2x) \cdot 2 = 8 \sin(2x) \]
Thus, the force becomes:
\[ F = -\frac{dU}{dx} = -8 \sin(2x) \]

### Step 4: Small Angle Approximation
For small oscillations, we can use the small angle approximation, where \( \sin(2x) \approx 2x \) (in radians). Therefore, the force can be approximated as:
\[ F \approx -8(2x) = -16x \]

### Step 5: Identifying the System as Simple Harmonic Motion
The force \( F = -16x \) indicates that the motion is simple harmonic motion (SHM), where the restoring force is proportional to the displacement \( x \) and acts in the opposite direction.

### Step 6: Relating Force to the Spring Constant
In SHM, the force can be expressed as:
\[ F = -kx \]
where \( k \) is the spring constant. From our expression, we see that:
\[ k = 16 \]

### Step 7: Finding the Angular Frequency
The angular frequency \( \omega \) is related to the spring constant and mass by the equation:
\[ k = m\omega^2 \]
Given that the mass \( m = 1 \) kg, we have:
\[ 16 = 1 \cdot \omega^2 \implies \omega^2 = 16 \implies \omega = 4 \, \text{rad/s} \]

### Step 8: Calculating the Period
The period \( T \) of the oscillation is given by:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{4} = \frac{\pi}{2} \, \text{seconds} \]

### Final Answer
The period of small oscillation is:
\[ \boxed{\frac{\pi}{2}} \, \text{seconds} \]