A body of mass m is situated in a potential field `U(x)=U_(0)(1-cosalphax)` when `U_(0)` and `alpha` are constant. Find the time period of small oscialltions.

Options

A. `2pisqrt((m)/(U_(0)alpha))`
B. `2pisqrt((m)/(U_(0)alpha^(2)))`
C. `2pisqrt((m)/(2U_(0)alpha))`
D. `2pisqrt((2m)/(U_(0)alpha^(2)))`

Explanation

To find the time period of small oscillations for a body of mass \( m \) situated in a potential field given by \( U(x) = U_0(1 - \cos(\alpha x)) \), we can follow these steps:

### Step 1: Write down the expression for potential energy
The potential energy is given as:
\[ U(x) = U_0(1 - \cos(\alpha x)) \]

### Step 2: Find the force acting on the body
The force \( F \) acting on the body is related to the potential energy by the relation:
\[ F = -\frac{dU}{dx} \]
Calculating the derivative:
\[ F = -\frac{d}{dx}[U_0(1 - \cos(\alpha x))] = -U_0 \cdot \frac{d}{dx}[-\cos(\alpha x)] = -U_0 \cdot \alpha \sin(\alpha x) \]
Thus, we have:
\[ F = U_0 \alpha \sin(\alpha x) \]

### Step 3: Use the small angle approximation
For small displacements, we can use the approximation \( \sin(\alpha x) \approx \alpha x \). Therefore, the force can be approximated as:
\[ F \approx U_0 \alpha^2 x \]

### Step 4: Identify the form of simple harmonic motion
The equation \( F \approx -k x \) indicates that the motion is simple harmonic, where \( k = U_0 \alpha^2 \). Thus, we can write:
\[ F = -U_0 \alpha^2 x \]

### Step 5: Write down the formula for the time period of SHM
The time period \( T \) of simple harmonic motion is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
Substituting \( k = U_0 \alpha^2 \) into the equation:
\[ T = 2\pi \sqrt{\frac{m}{U_0 \alpha^2}} \]

### Step 6: Final expression for the time period
Thus, the time period of small oscillations is:
\[ T = 2\pi \sqrt{\frac{m}{U_0 \alpha^2}} \]

### Summary
The time period of small oscillations for the given potential field is:
\[ T = 2\pi \sqrt{\frac{m}{U_0 \alpha^2}} \]