A system has potential energy U(x)=(10 J)[1−sin⁡((3.14 rad/m)x)]U(x) = (10 \, \text{J}) [1 - \sin((3.14 \, \text{rad/m}) x)] as a particle moves over the range 0 m ≤x≤3 m0\text{ m }\le x\le3\text{ m}. For each, is it a point of stable or unstable equilibrium?

Ch 10: Interactions and Potential Energy

Knight Calc - Physics for Scientists and Engineers 5th Edition

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Knight Calc 5th Edition

Ch 10: Interactions and Potential Energy

Problem 59b

Chapter 10, Problem 59b

A system has potential energy $U (x) = (10 J) [1 - \sin ((3.14 rad/m) x)]$ as a particle moves over the range $0 m is less than or equal to x is less than or equal to 3 m$ . For each, is it a point of stable or unstable equilibrium?

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Step 1: To determine the points of equilibrium, recall that equilibrium occurs where the force acting on the particle is zero. The force is related to the potential energy by the equation F(x) = -dU/dx. Begin by differentiating the given potential energy function U(x) = (10 J)[1 - sin((3.14 rad/m)x)] with respect to x.

Step 2: Compute the derivative of U(x). Using the chain rule, dU/dx = (10 J)(-cos((3.14 rad/m)x))(3.14 rad/m). Simplify this expression to find the force F(x) = -dU/dx.

Step 3: Set F(x) = 0 to find the points of equilibrium. This means solving the equation cos((3.14 rad/m)x) = 0. Identify the values of x within the given range 0 m ≤ x ≤ 3 m that satisfy this condition.

Step 4: To determine the stability of each equilibrium point, examine the second derivative of the potential energy, d²U/dx². If d²U/dx² > 0 at a point, it is a point of stable equilibrium. If d²U/dx² < 0, it is a point of unstable equilibrium. Compute d²U/dx² = (10 J)(-sin((3.14 rad/m)x))(3.14 rad/m)².

Step 5: Evaluate d²U/dx² at each equilibrium point found in Step 3. Use the sign of d²U/dx² to classify each point as stable or unstable equilibrium. Summarize the results for all equilibrium points within the range.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Potential Energy

Potential energy is the energy stored in a system due to its position or configuration. In this context, the potential energy function U(x) describes how the energy varies with the position x of a particle. Understanding this function is crucial for analyzing the forces acting on the particle and determining its equilibrium points.

Equilibrium Points

Equilibrium points occur where the net force acting on a particle is zero, meaning the particle is in a state of rest or constant motion. These points can be classified as stable or unstable based on the behavior of the potential energy function around them. A stable equilibrium is characterized by a local minimum in potential energy, while an unstable equilibrium corresponds to a local maximum.

Stability Analysis

Stability analysis involves examining how small perturbations affect the equilibrium state of a system. For potential energy functions, this is often done by evaluating the second derivative of the potential energy at equilibrium points. If the second derivative is positive, the equilibrium is stable; if negative, it is unstable. This analysis helps predict the behavior of the particle when it is slightly displaced from its equilibrium position.

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A system has potential energy U(x)=(10J)[1−sin((3.14rad/m)x)]U(x) = (10 \, \(\text{J}\)) [1 - \(\sin\)((3.14 \, \(\text{rad/m}\)) x)] as a particle moves over the range 0 m ≤x≤3 m0\(\text{ m }\]\le\) x\(\le\)3\(\text{ m}\). For each, is it a point of stable or unstable equilibrium?

Verified video answer for a similar problem:

Key Concepts

Potential Energy

Equilibrium Points

Stability Analysis

A system has potential energy $U (x) = (10 J) [1 - \sin ((3.14 rad/m) x)]$ as a particle moves over the range $0 m is less than or equal to x is less than or equal to 3 m$ . For each, is it a point of stable or unstable equilibrium?