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-by-Step Solution:

1. **Identify the Potential Energy Function**:
The potential energy is given by:
\[ U = U_0 - U_0 \cos(ax) \]

2. **Determine the Force**:
The force \( F \) acting on the particle can be derived from the potential energy using the relation:
\[ F = -\frac{dU}{dx} \]
We differentiate \( U \) with respect to \( x \):
\[ F = -\frac{d}{dx}(U_0 - U_0 \cos(ax)) = -\left(0 + U_0 a \sin(ax)\right) = U_0 a \sin(ax) \]

3. **Approximate for Small Displacements**:
For small displacements, we can use the approximation \( \sin(ax) \approx ax \):
\[ F \approx U_0 a (ax) = U_0 a^2 x \]

4. **Relate Force to SHM**:
In SHM, the force can also be expressed as:
\[ F = -m \omega^2 x \]
Setting the two expressions for force equal gives:
\[ -m \omega^2 x = -U_0 a^2 x \]
This simplifies to:
\[ m \omega^2 = U_0 a^2 \]

5. **Solve for Angular Frequency**:
Rearranging gives:
\[ \omega^2 = \frac{U_0 a^2}{m} \]
Taking the square root:
\[ \omega = \sqrt{\frac{U_0 a^2}{m}} \]

6. **Find the Period of Oscillation**:
The period \( T \) of oscillation is related to the angular frequency by:
\[ T = \frac{2\pi}{\omega} \]
Substituting for \( \omega \):
\[ T = \frac{2\pi}{\sqrt{\frac{U_0 a^2}{m}}} = 2\pi \sqrt{\frac{m}{U_0 a^2}} \]

### Final Result:
The period of oscillation \( T \) is given by:
\[ T = 2\pi \sqrt{\frac{m}{U_0 a^2}} \]