The potential energy of a particle executing SHM along the x-axis is given by `U=U_0-U_0cosax`. What is the period of oscillation?

Options

A. `2pisqrt((ma)/(U_0))`
B. `2pisqrt((U_0)/(ma))`
C. `(2pi)/(a)sqrt((m)/(U_0))`
D. `2pisqrt((m)/(aU_0))`

Explanation

To find the period of oscillation for a particle executing simple harmonic motion (SHM) with the given potential energy function \( U = U_0 - U_0 \cos(ax) \), we can follow these steps:

### Step 1: Identify the Force from Potential Energy
The force \( F \) acting on the particle can be derived from the potential energy \( U \) using the relation:
\[ F = -\frac{dU}{dx} \]
Substituting the expression for \( U \):
\[ F = -\frac{d}{dx}(U_0 - U_0 \cos(ax)) = -\left(0 + U_0 \cdot a \sin(ax)\right) = -U_0 a \sin(ax) \]

### Step 2: Approximate the Force for Small Displacements
For small displacements, we can use the small angle approximation where \( \sin(ax) \approx ax \). Therefore, the force can be approximated as:
\[ F \approx -U_0 a (ax) = -U_0 a^2 x \]

### Step 3: Relate Force to Acceleration
According to Newton's second law, \( F = ma \), we can write:
\[ ma = -U_0 a^2 x \]
This can be rearranged to express acceleration:
\[ a = -\frac{U_0 a^2}{m} x \]

### Step 4: Identify the Angular Frequency
This equation resembles the standard form of SHM, which is given by:
\[ a = -\omega^2 x \]
From our equation, we can identify:
\[ \omega^2 = \frac{U_0 a^2}{m} \]
Thus, the angular frequency \( \omega \) is:
\[ \omega = \sqrt{\frac{U_0 a^2}{m}} \]

### Step 5: Calculate the Period of Oscillation
The period \( T \) of oscillation is related to the angular frequency by the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting the expression for \( \omega \):
\[ T = \frac{2\pi}{\sqrt{\frac{U_0 a^2}{m}}} = 2\pi \sqrt{\frac{m}{U_0 a^2}} \]

### Final Answer
The period of oscillation is:
\[ T = 2\pi \sqrt{\frac{m}{U_0 a^2}} \]