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Chapter 7: Problem 3

Consider a mass \(m\) moving in two dimensions with potential energy \(U(x, y)=\frac{1}{2} k r^{2},\) where \(r^{2}=x^{2}+y^{2} .\) Write down the Lagrangian, using coordinates \(x\) and \(y,\) and find the two Lagrange equations of motion. Describe their solutions. [This is the potential energy of an ion in an "ion trap," which can be used to study the properties of individual atomic ions.]

Short Answer

Expert verified
The Lagrangian is \( \mathcal{L} = \frac{1}{2} m (\dot{x}^2 + \dot{y}^2) - \frac{1}{2} k (x^2 + y^2) \). Equations of motion are \( m\ddot{x} = -kx \) and \( m\ddot{y} = -ky \), which are simple harmonic oscillators.

Step by step solution

01

Understanding the Potential Energy

The potential energy given is \( U(x, y) = \frac{1}{2} k r^2 \) where \( r^2 = x^2 + y^2 \). This represents the energy associated with the position of the mass in two-dimensional space.
02

Writing the Kinetic Energy

In two dimensions, the kinetic energy \( T \) for a mass \( m \) is given by \( T = \frac{1}{2} m (\dot{x}^2 + \dot{y}^2) \), where \( \dot{x} \) and \( \dot{y} \) are the velocities in the \( x \) and \( y \) directions.
03

Formulating the Lagrangian

The Lagrangian \( \mathcal{L} \) is the difference between the kinetic and potential energy: \( \mathcal{L} = T - U \). Substituting the expressions from previous steps, \( \mathcal{L} = \frac{1}{2} m (\dot{x}^2 + \dot{y}^2) - \frac{1}{2} k (x^2 + y^2) \).
04

Finding the Lagrange Equations of Motion

Using the Euler-Lagrange equation \( \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{q}} \right) - \frac{\partial \mathcal{L}}{\partial q} = 0 \), apply it for both coordinates. For \( x \), this results in \( m \ddot{x} = -kx \), and for \( y \), it results in \( m \ddot{y} = -ky \).
05

Solving the Equations of Motion

The equations \( m \ddot{x} = -kx \) and \( m \ddot{y} = -ky \) describe harmonic oscillators in both \( x \) and \( y \) directions. The general solutions are \( x(t) = A \cos(\omega t + \phi_x) \) and \( y(t) = B \cos(\omega t + \phi_y) \) where \( \omega = \sqrt{\frac{k}{m}} \), and \( A, B, \phi_x, \phi_y \) are constants determined by initial conditions.

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Key Concepts

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

Two-Dimensional Motion
When a mass moves in two dimensions, it means it has freedom to travel both in the "x" direction and the "y" direction. Imagine tracking the path of an object on a flat surface: it can go left, right, forward, or backward. This is the essence of two-dimensional motion.

In our exercise, we deal with a particle that moves within this plane. Its position at any time can be represented by coordinates \(x\) and \(y\). Understanding how these coordinates change over time can help us describe the particle's overall motion.

The potential energy \(U(x, y)\) is a function of these coordinates, meaning it gives us a measure of the energy based on the object's position. The kinetic energy, however, depends on the changes in these coordinates 鈥� specifically, the velocities which are \(\dot{x}\) and \(\dot{y}\). By combining both kinetic and potential energies, we create a comprehensive picture of the system using Lagrangian mechanics.
Harmonic Oscillator
A harmonic oscillator is a system that experiences a restoring force proportional to the displacement from an equilibrium position. In simple terms, if you pull it out of its rest position, it will try to go back, like a bouncy spring.

In our problem, the equations of motion \( m \ddot{x} = -kx \) and \( m \ddot{y} = -ky \) clearly show harmonic behavior. These equations mean the force trying to bring the particle back to center increases with the distance it moves from the center. It鈥檚 like the stronger you stretch a spring, the harder it pulls back.

These equations describe the behavior in both the \(x\) and \(y\) directions, independently. Each direction can oscillate with its own frequency, but, as they have the same constants, they share a frequency \(\omega = \sqrt{\frac{k}{m}}\). It tells us how fast these oscillations occur and is determined by the mass \(m\) and spring constant \(k\). This results in periodic motion represented as a cosine function.
Euler-Lagrange Equation
The Euler-Lagrange equation is a fundamental equation in Lagrangian mechanics. It allows us to derive the equations of motion for a system from a given Lagrangian.

The Lagrangian \(\mathcal{L}\) is given by the difference between kinetic energy \(T\) and potential energy \(U\). For our system, \(\mathcal{L} = \frac{1}{2} m (\dot{x}^2 + \dot{y}^2) - \frac{1}{2} k (x^2 + y^2)\). This captures the dynamics of our problem in a single formula.

The Euler-Lagrange equation itself is \(\frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{q}} \right) - \frac{\partial \mathcal{L}}{\partial q} = 0\). You apply this to each coordinate: \(q\) becomes \(x\) and \(y\). For \(x\), this leads to \(m \ddot{x} = -kx\), and similarly for \(y\).

What's fascinating about this is how it connects the change in motion (acceleration and force) to the properties of the physical system. It gives a beautiful way to look at the physics of complex systems.

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Most popular questions from this chapter

Noether's theorem asserts a connection between invariance principles and conservation laws. In Section 7.8 we saw that translational invariance of the Lagrangian implies conservation of total linear momentum. Here you will prove that rotational invariance of \(\mathcal{L}\) implies conservation of total angular momentum. Suppose that the Lagrangian of an \(N\) -particle system is unchanged by rotations about a certain symmetry axis. (a) Without loss of generality, take this axis to be the \(z\) axis, and show that the Lagrangian is unchanged when all of the particles are simultaneously moved from \(\left(r_{\alpha}, \theta_{\alpha}, \phi_{\alpha}\right)\) to \(\left(r_{\alpha}, \theta_{\alpha}, \phi_{\alpha}+\epsilon\right)\) (same \(\epsilon\) for all particles). Hence show that $$\sum_{\alpha=1}^{N} \frac{\partial \mathcal{L}}{\partial \phi_{\alpha}}=0.$$ (b) Use Lagrange's equations to show that this implies that the total angular momentum \(L_{z}\) about the symmetry axis is constant. In particular, if the Lagrangian is invariant under rotations about all axes, then all components of \(\mathbf{L}\) are conserved.

Lagrange's equations in the form discussed in this chapter hold only if the forces (at least the nonconstraint forces) are derivable from a potential energy. To get an idea how they can be modified to include forces like friction, consider the following: A single particle in one dimension is subject to various conservative forces (net conservative force \(=F=-\partial U / \partial x)\) and a nonconservative force (let's call it \(F_{\text {fric }}\) ). Define the Lagrangian as \(\mathcal{L}=T-U\) and show that the appropriate modification is $$\frac{\partial \mathcal{L}}{\partial x}+F_{\mathrm{fric}}=\frac{d}{d t} \frac{\partial \mathcal{L}}{\partial \dot{x}}.$$

A particle is confined to move on the surface of a circular cone with its axis on the \(z\) axis, vertex at the origin (pointing down), and half-angle \(\alpha\). The particle's position can be specified by two generalized coordinates, which you can choose to be the coordinates \((\rho, \phi)\) of cylindrical polar coordinates. Write down the equations that give the three Cartesian coordinates of the particle in terms of the generalized coordinates ( \(\rho, \phi\) ) and vice versa.

Consider the well-known problem of a cart of mass \(m\) moving along the \(x\) axis attached to a spring (force constant \(k\) ), whose other end is held fixed (Figure 5.2 ). If we ignore the mass of the spring (as we almost always do) then we know that the cart executes simple harmonic motion with angular frequency \(\omega=\sqrt{k / m} .\) Using the Lagrangian approach, you can find the effect of the spring's mass \(M,\) as follows: (a) Assuming that the spring is uniform and stretches uniformly, show that its kinetic energy is \(\frac{1}{6} M \dot{x}^{2} .\) (As usual \(x\) is the extension of the spring from its equilibrium length.) Write down the Lagrangian for the system of cart plus spring. (Note: The potential energy is still \(\frac{1}{2} k x^{2}\).) (b) Write down the Lagrange equation and show that the cart still executes SHM but with angular frequency \(\omega=\sqrt{k /(m+M / 3)} ;\) that is, the effect of the spring's mass \(M\) is just to add \(M / 3\) to the mass of the cart.

The "spherical pendulum" is just a simple pendulum that is free to move in any sideways direction. (By contrast a "simple pendulum"- unqualified - is confined to a single vertical plane.) The bob of a spherical pendulum moves on a sphere, centered on the point of support with radius \(r=R\) the length of the pendulum. A convenient choice of coordinates is spherical polars, \(r, \theta, \phi,\) with the origin at the point of support and the polar axis pointing straight down. The two variables \(\theta\) and \(\phi\) make a good choice of generalized coordinates. (a) Find the Lagrangian and the two Lagrange equations. (b) Explain what the \(\phi\) equation tells us about the \(z\) component of angular momentum \(\ell_{z^{*}}\) (c) For the special case that \(\phi=\) const, describe what the \(\theta\) equation tells us. (d) Use the \(\phi\) equation to replace \(\dot{\phi}\) by \(\ell_{z}\) in the \(\theta\) equation and discuss the existence of an angle \(\theta_{\mathrm{o}}\) at which \(\theta\) can remain constant. Why is this motion called a conical pendulum? (e) Show that if \(\theta=\theta_{0}+\epsilon,\) with \(\epsilon\) small, then \(\theta\) oscillates about \(\theta_{\mathrm{o}}\) in harmonic motion. Describe the motion of the pendulum's bob.
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