Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Consider a mass \(m\) confined to the \(x\) axis and subject to a force \(F_{x}=k x\) where \(k>0 .\) (a) Write down and sketch the potential energy \(U(x)\) and describe the possible motions of the mass. (Distinguish between the cases that \(E>0 \text { and } E<0 .)\) (b) Write down the Hamiltonian \(\mathcal{H}(x, p)\), and describe the possible phase-space orbits for the two cases \(E>0\) and \(E<0\). (Remember that the function \(\mathcal{H}(x, p)\) must equal the constant energy \(E .\) ) Explain your answers to part (b) in terms of those to part (a).

Short Answer

Expert verified
The potential energy is \( U(x) = \frac{k}{2}x^2 \) and motion is oscillatory for \( E > 0 \). The Hamiltonian yields elliptical phase-space orbits for \( E > 0 \); no orbits exist for \( E < 0 \).

Step by step solution

01

Determine the Potential Energy Function

To determine the potential energy function, we integrate the force with respect to position, using the relation that the force is the negative gradient of the potential energy: \( F_x = -\frac{dU}{dx} \). Given \( F_x = kx \), we have \( -kx = -\frac{dU}{dx} \). Integrating both sides gives \( U(x) = \frac{k}{2}x^2 \). This represents a parabolic potential centered at the origin.
02

Sketch the Potential Energy and Describe Motion

Sketch the potential energy \( U(x) = \frac{k}{2}x^2 \): It is a parabola opening upwards with its vertex at the origin. If the energy \( E > 0 \), the mass can move in bounded oscillations around the origin since the energy is insufficient to escape the potential well. If \( E < 0 \), which is physically unrealistic in classical mechanics for this scenario, the particle would not have enough energy to exist at any \( x \).
03

Formulate the Hamiltonian

The Hamiltonian \( \mathcal{H}(x, p) \) represents the total energy of the system and is given by \( \mathcal{H}(x, p) = \frac{p^2}{2m} + U(x) \). Substituting \( U(x) = \frac{k}{2}x^2 \), we obtain \( \mathcal{H}(x, p) = \frac{p^2}{2m} + \frac{k}{2}x^2 \). This Hamiltonian describes simple harmonic motion.
04

Analyze Phase-Space Orbits

Phase-space is a plot of \( x \) versus \( p \). For \( E > 0 \), the equation \( \mathcal{H}(x, p) = E \) describes an ellipse centered at the origin due to the form \( \frac{p^2}{2m} + \frac{k}{2}x^2 = E \). The orbits are closed and elliptical, indicating periodic motion. For \( E < 0 \), there are no possible phase-space orbits, as negative energy is not viable for real solutions.
05

Explain Phase-Space in Terms of Potential Energy

The elliptical phase-space orbits for \( E > 0 \) correspond to oscillations in potential energy described by \( U(x) = \frac{k}{2}x^2 \). These oscillations are confined within the potential well, representing motion that reflects back and forth in the potential. This matches the parabolic potential's confining nature. No phase-space motion exists for \( E < 0 \) since it's energetically impossible.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Potential Energy
Potential energy is a key concept in understanding harmonic oscillators. In this scenario, you have a force acting on a mass, which is described by the equation:
  • \( F_x = kx \)
To find the potential energy \( U(x) \), you integrate the force with respect to position. This is because the force is the negative gradient of the potential energy. Using the given force, the integration appears as:
  • \( -kx = -\frac{dU}{dx} \)
Upon integrating, the potential energy function becomes:
  • \( U(x) = \frac{k}{2}x^2 \)
This function forms a parabolic shape, indicating that the potential energy increases as you move away from the origin. This parabola represents a potential well, where a mass can oscillate back and forth around the center.
For energy \( E > 0 \), the mass has enough energy for oscillations around the origin without escaping. Conversely, a scenario with \( E < 0 \) would be impossible in classical terms, as the energy would not allow the mass to exist within this potential.
Hamiltonian Mechanics
Hamiltonian mechanics is a formalism in physics that allows you to describe the total energy of a system. The Hamiltonian \( \mathcal{H}(x, p) \) for a simple harmonic oscillator combines both kinetic and potential energies as:
  • \( \mathcal{H}(x, p) = \frac{p^2}{2m} + \frac{k}{2}x^2 \)
This formulation of the Hamiltonian establishes that the total energy is conserved within the system. The kinetic energy \( \frac{p^2}{2m} \) relates to the momentum \( p \), while the potential energy part \( \frac{k}{2}x^2 \) is derived from the force acting on the mass.
When you analyze this, for energies \( E > 0 \), it characterizes a system in simple harmonic motion. Such a system oscillates around its equilibrium point, and the Hamiltonian equation reflects this stable and periodic nature. It's important in this context because it allows the dynamics of the system to be understood as movements through a defined energy landscape.
Phase Space Analysis
Phase space gives you a powerful way to visualize the motion of a system. It uses position \( x \) and momentum \( p \) as axes to plot the state of the system. For a harmonic oscillator, the Hamiltonian gives:
  • \( \mathcal{H}(x, p) = E \)
Where \( E \) is the total energy. For a system with \( E > 0 \), the phase-space plot reveals elliptical orbits. These ellipses represent closed, periodic paths, indicating stable oscillations around an equilibrium point.
Such visualizations help you understand how the kinetic and potential energies exchange roles as the system evolves. The elliptical trajectories demonstrate how the motion regularly confers momentum into position and vice versa, looping around the plot predictably.
However, when \( E < 0 \), there are no phase-space orbits because such a system state is not physically possible. The idea that negative total energy could result in any real-world motion is conceptually incorrect within this framework.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

13.13 \star\star Consider a particle of mass \(m\) constrained to move on a frictionless cylinder of radius \(R\), given by the equation \(\rho=R\) in cylindrical polar coordinates \((\rho, \phi, z) .\) The mass is subject to just one external force, \(\mathbf{F}=-k r \hat{\mathbf{r}},\) where \(k\) is a positive constant, \(r\) is its distance from the origin, and \(\hat{\mathbf{r}}\) is the unit vector pointing away from the origin, as usual. Using \(z\) and \(\phi\) as generalized coordinates, find the Hamiltonian \(\mathcal{H}\). Write down and solve Hamilton's equations and describe the motion.

The simple form \(\mathcal{H}=T+U\) is true only if your generalized coordinates are "natural" (relation betweeen generalized and underlying Cartesian coordinates is independent of time). If the generalized coordinates are not "natural," you must use the definition \(\mathcal{H}=\sum p_{i} \dot{q}_{i}-\mathcal{L} .\) To illustrate this point, consider the following: Two children are playing catch inside a railroad car that is moving with varying speed \(V\) along a straight horizontal track. For generalized coordinates you can use the position \((x, y, z)\) of the ball relative to a point fixed in the car, but in setting up the Hamiltonian you must use coordinates in an inertial frame \(-\) a frame fixed to the ground. Find the Hamiltonian for the ball and show that it is not equal to \(T+U\) (neither as measured in the car, nor as measured in the ground-based frame).

Set up the Hamiltonian and Hamilton's equations for a projectile of mass \(m\), moving in a vertical plane and subject to gravity but no air resistance. Use as your coordinates \(x\) measured horizontally and \(y\) measured vertically up. Comment on each of the four equations of motion.

A roller coaster of mass \(m\) moves along a frictionless track that lies in the \(x y\) plane \((x\) horizontal and \(y\) vertically up). The height of the track above the ground is given by \(y=h(x) .\) (a) Using \(x\) as your generalized coordinate, write down the Lagrangian, the generalized momentum \(p,\) and the Hamiltonian \(\mathcal{H}=p \dot{x}-\mathcal{L}\) (as a function of \(x\) and \(p\) ). (b) Find Hamilton's equations and show that they agree with what you would get from the Newtonian approach. [Hint: You know from Section 4.7 that Newton's second law takes the form \(F_{\text {tang }}=m \ddot{s},\) where \(s\) is the distance measured along the track. Rewrite this as an equation for \(\ddot{x}\) and show that you get the same result from Hamilton's equations.]

The general proof of the divergence theorem $$\int_{S} \mathbf{n} \cdot \mathbf{v} d A=\int_{V} \nabla \cdot \mathbf{v} d V$$ is fairly complicated and not especially illuminating. However, there are a few special cases where it is reasonably simple and quite instructive. Here is one: Consider a rectangular region bounded by the six planes \(x=X\) and \(X+A, y=Y\) and \(Y+B,\) and \(z=Z\) and \(Z+C,\) with total volume \(V=A B C .\) The surface \(S\) of this region is made up of six rectangles that we can call \(S_{1}\) (in the plane \(x=X\) ), \(S_{2}\) (in the plane \(x=X+A\) ), and so on. The surface integral on the left of (13.63) is then the sum of six integrals, one over each of the rectangles \(S_{1}, S_{2},\) and so forth. (a) Consider the first two of these integrals and show that $$ \int_{S_{1}} \mathbf{n} \cdot \mathbf{v} d A+\int_{S_{2}} \mathbf{n} \cdot \mathbf{v} d A=\int_{Y}^{Y+B} d y \int_{Z}^{Z+C} d z\left[v_{x}(X+A, y, z)-v_{x}(X, y, z)\right] $$ (b) Show that the integrand on the right can be rewritten as an integral of \(\partial v_{x} / \partial x\) over \(x\) running from \(x=X\) to \(x=X+A .(\text { c) Substitute the result of part }(b)\) into part (a), and write down the corresponding results for the two remaining pairs of faces. Add these results to prove the divergence theorem (13.63).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free