A particle of mass m oscillates with a potential energy `U=U_(0)+alpha x^(2)`, where `U_(0)` and `alpha` are constants and x is the displacement of particle from equilibrium position. The time period of oscillation is

Options

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

Explanation

To find the time period of oscillation for a particle of mass \( m \) with a potential energy given by \( U = U_0 + \alpha x^2 \), we can follow these steps:

### Step 1: Understand the Potential Energy Function
The potential energy function is given as:
\[ U = U_0 + \alpha x^2 \]
where \( U_0 \) is a constant, \( \alpha \) is a constant, and \( x \) is the displacement from the equilibrium position.

### Step 2: Determine the Force Acting on the Particle
The force \( F \) acting on the particle can be derived from the potential energy using the formula:
\[ F = -\frac{dU}{dx} \]
Calculating the derivative of \( U \) with respect to \( x \):
\[ \frac{dU}{dx} = \frac{d}{dx}(U_0 + \alpha x^2) = 0 + 2\alpha x = 2\alpha x \]
Thus, the force becomes:
\[ F = -2\alpha x \]

### Step 3: Relate the Force to Hooke's Law
The force can be expressed in the form of Hooke's law:
\[ F = -kx \]
Comparing this with our expression for force, we identify:
\[ k = 2\alpha \]

### Step 4: Calculate the Time Period of Oscillation
The time period \( T \) of a simple harmonic oscillator is given by the formula:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
Substituting \( k = 2\alpha \) into the equation:
\[ T = 2\pi \sqrt{\frac{m}{2\alpha}} \]

### Final Answer
Thus, the time period of oscillation is:
\[ T = 2\pi \sqrt{\frac{m}{2\alpha}} \]

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