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

Question: Determine the three equilibrium points for an object moving along the x-axis with a given potential energy function U(x) = a(x^4 - 2b^2x^2), and classify each as stable, unstable, or neutral. Answer: The equilibrium points are at x = 0, x = b, and x = -b. The equilibrium at x = 0 is unstable, while the equilibria at x = b and x = -b are both stable.

Step-by-step solution

Find the force equation

To find the force at a given position x, we must evaluate the derivative of the potential energy function U(x) with respect to x, and take the negative of that value: \(F(x) = -\frac{dU(x)}{dx}\) Step 2: Calculate the derivative of the potential energy function

Derive U(x)

To find the equilibrium points, we must first calculate the derivative of the potential energy function U(x) with respect to x: \(\frac{dU(x)}{dx} = \frac{d}{dx}(a(x^4-2b^2x^2))\) Step 3: Simplify the derivative

Simplify the derivative

Using basic rules of differentiation, we can simplify the derivative of U(x) as follows: \(\frac{dU(x)}{dx} = 4ax^3 - 4ab^2x\) Step 4: Determine the equilibrium points

Calculate equilibrium points

An equilibrium point occurs where the force F(x) is zero. Thus, we can set the force equation F(x) equal to zero and solve for x: \(0 = -(4ax^3 - 4ab^2x) \Rightarrow 0= x^3 - b^2x\) Step 5: Solve for x values

Solve for x values

Factor out x from the equation derived in Step 4: \(0 = x(x^2 - b^2)\) This equation has three solutions: 1. \(x = 0\) 2. \(x = b\) 3. \(x = -b\) Step 6: Analyze the second derivative of U(x)

Calculate the second derivative of U(x)

To determine the stability of each equilibrium found in Step 5, we must analyze the second derivative of the potential energy function U(x) at each equilibrium point: \(\frac{d^2U(x)}{dx^2} = \frac{d^2}{dx^2}(a(x^4-2b^2x^2))\) Step 7: Simplify the second derivative

Simplify the second derivative

Using basic rules of differentiation, we can simplify the second derivative of U(x) as follows: \(\frac{d^2U(x)}{dx^2} = 12ax^2 - 8ab^2\) Step 8: Analyze the stability of each equilibrium point

Determine stability

Evaluate the second derivative of U(x) at each equilibrium point found in Step 5: 1. \(x = 0\) \(\frac{d^2U(0)}{dx^2} = -8ab^2\), which is negative, indicating an unstable equilibrium. 2. \(x = b\) \(\frac{d^2U(b)}{dx^2} = 12ab^2 - 8ab^2 = 4ab^2\), which is positive, indicating a stable equilibrium. 3. \(x = -b\) \(\frac{d^2U(-b)}{dx^2} = 12a(-b)^2 - 8ab^2 = 4ab^2\), which is positive, indicating a stable equilibrium. The solution of the problem is as follows: There are three equilibrium points at \(x = 0\), \(x = b\), and \(x = -b\). The equilibrium at \(x = 0\) is unstable, while the equilibria at \(x = b\) and \(x = -b\) are both stable.

Key concepts

Potential Energy Function

The potential energy function, commonly denoted as U(x), describes the energy stored in a system due to its position or configuration. In physics, this concept is vital when analyzing the dynamics of a system, particularly in conservative force fields, where energy is not lost but transformed from kinetic to potential energy and vice versa.

In the context of the exercise, the potential energy function is given by a polynomial expression involving the object's position x along a one-dimensional axis. The expression includes constants that indicate the specific characteristics of the system, where 'a' represents the scaling factor of the potential energy, and 'b' represents the characteristic length scale. By understanding this function, one can predict how an object will behave under the influence of the conservative forces at play.

Knowing the potential energy helps in visualizing the system's behavior. For instance, points where this function takes a local minimum or maximum are of particular interest because they correspond to equilibrium positions of the system. Just as a ball at rest at the bottom of a valley or on top of a hill, particles will be at equilibrium where the potential energy function has stationary points.

Differentiation in Physics

Differentiation is a mathematical process of finding the rate at which a quantity changes. In physics, this concept is employed to determine how one variable changes with respect to another. One common application of differentiation is in determining the force exerted on an object by a potential energy field. The force exerted at a point can be found by taking the negative of the derivative of the potential energy function with respect to position, denoted by \( F(x) = -\frac{dU(x)}{dx} \).

In the given exercise, differentiation allows us to derive the force equation from the potential energy function. By calculating the first derivative, one finds the points where the force vanishes, which correspond to the equilibrium points. Further differentiation, by taking the second derivative, helps in understanding how strong the forces are around these points and informs about the nature of the equilibrium, be it stable, unstable, or neutral.

This process highlights the essential role differentiation plays in translating between the abstract concept of potential energy and the more tangible concept of force experienced by objects within a physical system.

Stability of Equilibrium

When discussing the stability of equilibrium points, it's about understanding whether an object at rest will remain at rest when slightly perturbed. An equilibrium point can be categorized as stable, unstable, or neutral depending on how the system responds to small displacements from that point.

An equilibrium is stable if, upon being slightly displaced, the system experiences a restoring force that brings it back to the equilibrium point. On the other hand, an equilibrium is unstable if the system, when displaced, experiences a force that further moves it away from equilibrium. Neutral equilibrium occurs when the system remains in equilibrium even after being displaced.

To mathematically assess the stability, one uses the second derivative of the potential energy function at the equilibrium points. This is evident in the exercise, where a positive second derivative suggests a concave upward curve, indicating a stable equilibrium. Conversely, a negative second derivative implies a concave downward curve, indicating an unstable equilibrium. The analysis of these second derivatives is key to predicting how the system will behave if disturbed, providing insight into potential oscillations, shifts from equilibrium, or a state of balance.