The potential energy of a particle of mass `1kg` in motion along the x- axis is given by: `U = 4(1 - cos 2x)`, where `x` in metres. The period of small oscillation (in sec) is

Options

A. `2pi`
B. `pi`
C. `pi//2`
D. `pi//4`

Explanation

To find the period of small oscillations for the given potential energy function \( U = 4(1 - \cos 2x) \), we can follow these steps:

### Step 1: Find the Force
The force \( F \) can be derived from the potential energy \( U \) using the relation:
\[ F = -\frac{dU}{dx} \]
Calculating the derivative:
\[ U = 4(1 - \cos 2x) \]
\[ \frac{dU}{dx} = 4 \cdot \frac{d}{dx}(1 - \cos 2x) = 4 \cdot 2\sin 2x = 8\sin 2x \]
Thus, the force is:
\[ F = -8\sin 2x \]

### Step 2: Relate Force to Acceleration
Using Newton's second law, \( F = ma \), where \( m = 1 \, \text{kg} \):
\[ a = \frac{F}{m} = -8\sin 2x \]
Since \( m = 1 \, \text{kg} \), we have:
\[ a = -8\sin 2x \]

### Step 3: Small Angle Approximation
For small oscillations, we can use the small angle approximation where \( \sin \theta \approx \theta \). Therefore, we can write:
\[ a \approx -8(2x) = -16x \]

### Step 4: Identify the Form of Simple Harmonic Motion
The equation \( a = -16x \) can be compared to the standard form of simple harmonic motion:
\[ a = -\omega^2 x \]
From this, we can identify:
\[ \omega^2 = 16 \implies \omega = 4 \]

### Step 5: Calculate the Period
The period \( T \) of the oscillation is related to \( \omega \) by the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{4} = \frac{\pi}{2} \, \text{seconds} \]

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

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