A particle of mass m moves in a one dimensional potential energy `U(x)=-ax^2+bx^4`, where a and b are positive constant. The angular frequency of small oscillation about the minima of the potential energy is equal to

Options

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

Explanation

To find the angular frequency of small oscillations about the minima of the potential energy given by \( U(x) = -ax^2 + bx^4 \), we will follow these steps:

### Step 1: Find the expression for force
The force \( F \) acting on the particle can be derived from the potential energy function using the relation:
\[ F = -\frac{dU}{dx} \]
Calculating the derivative:
\[ U(x) = -ax^2 + bx^4 \]
\[ \frac{dU}{dx} = -2ax + 4bx^3 \]
Thus, the force is:
\[ F = -\left(-2ax + 4bx^3\right) = 2ax - 4bx^3 \]

### Step 2: Find the equilibrium position
To find the equilibrium position, we set the force to zero:
\[ 2ax - 4bx^3 = 0 \]
Factoring out \( 2x \):
\[ 2x(a - 2bx^2) = 0 \]
This gives us two solutions:
1. \( x = 0 \)
2. \( a - 2bx^2 = 0 \) which leads to \( x^2 = \frac{a}{2b} \) or \( x = \pm \sqrt{\frac{a}{2b}} \)

### Step 3: Determine the nature of the equilibrium points
To determine which of these points is a minimum, we need to check the second derivative of \( U \):
\[ \frac{d^2U}{dx^2} = -2a + 12bx^2 \]
Evaluating this at \( x = 0 \):
\[ \frac{d^2U}{dx^2}\bigg|_{x=0} = -2a < 0 \quad (\text{not a minimum}) \]
Evaluating at \( x = \sqrt{\frac{a}{2b}} \):
\[ \frac{d^2U}{dx^2}\bigg|_{x=\sqrt{\frac{a}{2b}}} = -2a + 12b\left(\frac{a}{2b}\right) = -2a + 6a = 4a > 0 \quad (\text{a minimum}) \]

### Step 4: Small oscillation approximation
Now, we will consider small oscillations about the point \( x = \sqrt{\frac{a}{2b}} \). Let \( \delta x \) be the small displacement from this point:
\[ x = \sqrt{\frac{a}{2b}} + \delta x \]
Substituting this into the force equation:
\[ F = 2a\left(\sqrt{\frac{a}{2b}} + \delta x\right) - 4b\left(\sqrt{\frac{a}{2b}} + \delta x\right)^3 \]
Neglecting higher-order terms in \( \delta x \) (like \( \delta x^2 \) and \( \delta x^3 \)), we simplify to find the restoring force:
\[ F \approx -k \delta x \]
where \( k \) is the effective spring constant.

### Step 5: Finding the angular frequency
From the linear approximation of the force, we have:
\[ F = -4a \delta x \]
Thus, we can write:
\[ m \frac{d^2(\delta x)}{dt^2} = -4a \delta x \]
This gives us the equation of motion:
\[ \frac{d^2(\delta x)}{dt^2} + \frac{4a}{m} \delta x = 0 \]
Comparing this with the standard form \( \frac{d^2x}{dt^2} + \omega^2 x = 0 \), we find:
\[ \omega^2 = \frac{4a}{m} \]
Thus, the angular frequency \( \omega \) is:
\[ \omega = 2\sqrt{\frac{a}{m}} \]

### Final Answer
The angular frequency of small oscillations about the minima of the potential energy is:
\[ \omega = 2\sqrt{\frac{a}{m}} \]