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 can follow these steps:

### Step 1: Find the Force
The force \( F \) acting on the particle is given by the negative gradient of the potential energy:
\[ F = -\frac{dU}{dx} \]
Calculating the derivative of \( U(x) \):
\[ U(x) = -ax^2 + bx^4 \]
\[ \frac{dU}{dx} = -2ax + 4bx^3 \]
Thus, the force is:
\[ F = -\left(-2ax + 4bx^3\right) = 2ax - 4bx^3 \]

### Step 2: Find the Equilibrium Position
At the equilibrium position, the net force is 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
To determine if these points are minima, we need to check the second derivative of \( U \):
\[ \frac{d^2U}{dx^2} = -2a + 12bx^2 \]
At \( x = 0 \):
\[ \frac{d^2U}{dx^2}\bigg|_{x=0} = -2a \quad (\text{which is negative, indicating a maximum}) \]
At \( x = \sqrt{\frac{a}{2b}} \):
\[ \frac{d^2U}{dx^2}\bigg|_{x=\sqrt{\frac{a}{2b}}} = -2a + 12b\left(\frac{a}{2b}\right) = -2a + 6a = 4a \quad (\text{which is positive, indicating a minimum}) \]

### Step 4: Calculate the Angular Frequency
The angular frequency \( \omega \) of small oscillations about the equilibrium position can be found using:
\[ \omega^2 = \frac{1}{m} \frac{d^2U}{dx^2}\bigg|_{x_0} \]
Substituting \( x_0 = \sqrt{\frac{a}{2b}} \):
\[ \omega^2 = \frac{1}{m} \cdot 4a \]
Thus,
\[ \omega = \sqrt{\frac{4a}{m}} = 2\sqrt{\frac{a}{m}} \]

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

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