Question

An object is restricted to movement in one dimension. Its position is specified along the \(x\) -axis. The potential energy of the object as a function of its position is given by \(U(x)=a\left(x^{4}-2 b^{2} x^{2}\right),\) where \(a\) and \(b\) represent positive numbers. Determine the location(s) of any equilibrium point(s), and classify the equilibrium at each point as stable, unstable, or neutral.

Short answer

Answer: The equilibrium points are \(x_1 = 0\) (unstable), \(x_2 = b\) (stable), and \(x_3 = -b\) (stable).

Step-by-step solution

Find the gradient of potential energy function

To find the gradient of the potential energy function, \(U(x)\), we will take the first derivative of \(U(x)\) with respect to \(x\). This will give us the force acting on the object, \(F(x)\). \(F(x) = -\frac{dU(x)}{dx} = -\frac{d}{dx}(a(x^4 - 2b^2x^2))\)

Calculate the first derivative of U(x)

Now we will find the first derivative of the potential energy function, \(U(x)\): \(\frac{dU(x)}{dx} = 4ax^3 - 4ab^2x\)

Calculate the force acting on the object

We can now calculate the force acting on the object using the expression we derived for the gradient of potential energy function: \(F(x) = -\frac{dU(x)}{dx} = -4ax^{3}+4ab^2x\)

Find the equilibrium points

The equilibrium points are the points where force acting on the object is zero. Hence, we will set \(F(x)\) to zero and solve for \(x\): \(0 = - 4ax^3 + 4ab^2x \Longrightarrow 0 = x(- 4ax^2 + 4ab^2)\) From this equation, we can see that there are three equilibrium points: \(x_1 = 0\), \(x_2 = b\), and \(x_3 = -b\).

Classify the equilibrium points

To classify the equilibrium points as stable, unstable, or neutral, we will examine the second derivative of the potential energy function, \(\frac{d^2U(x)}{dx^2}\): \(\frac{d^2U(x)}{dx^2} = 12ax^2 - 8ab^2\) Now we will evaluate the second derivative at each equilibrium point: \(\frac{d^{2} U}{d x^{2}}(x_1)=12 a(0)^2 - 8 a b^2 = -8ab^2\) Since the second derivative at \(x_1\) is negative, this equilibrium point is unstable. \(\frac{d^{2} U}{d x^{2}}(x_2)=12 a(b)^2 - 8 a b^2 = 4ab^2\) Since the second derivative at \(x_2\) is positive, this equilibrium point is stable. \(\frac{d^{2} U}{d x^{2}}(x_3)=12 a(-b)^2 - 8 a b^2 = 4ab^2\) Since the second derivative at \(x_3\) is positive, this equilibrium point is stable. In conclusion, there are three equilibrium points: \(x_1 = 0\) (unstable), \(x_2 = b\) (stable), and \(x_3 = -b\) (stable).