A particle of mass m in a unidirectional potential field have potential energy `U (x) = a + 2b x^(2)` where a and b are positive constants. Find time period of oscillation :-

Options

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

Explanation

To find the time period of oscillation for a particle in a unidirectional potential field with the potential energy given by \( U(x) = a + 2b x^2 \), we can follow these steps:

### Step 1: Identify the Potential Energy Function
The potential energy function is given as:
\[ U(x) = a + 2b x^2 \]
where \( a \) and \( b \) are positive constants.

### Step 2: Calculate the Force
The force \( F \) acting on the particle can be found using the relation:
\[ F = -\frac{dU}{dx} \]
Calculating the derivative of \( U(x) \):
\[ \frac{dU}{dx} = \frac{d}{dx}(a + 2b x^2) = 0 + 4b x = 4b x \]
Thus, the force is:
\[ F = -4b x \]

### Step 3: Relate the Force to Hooke's Law
The force can be related to Hooke's law, which states that:
\[ F = -kx \]
From our expression for force, we can identify \( k \):
\[ -k = -4b \implies k = 4b \]

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

### Step 5: Simplify the Expression
We can simplify the expression:
\[ T = 2\pi \sqrt{\frac{m}{4b}} = 2\pi \cdot \frac{1}{2} \sqrt{\frac{m}{b}} = \pi \sqrt{\frac{m}{b}} \]

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