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Consider a particle of mass \(m\) moving in two dimensions, subject to a force \(\mathbf{F}=-k x \hat{\mathbf{x}}+K \hat{\mathbf{y}}\) where \(k\) and \(K\) are positive constants. Write down the Hamiltonian and Hamilton's equations, using \(x\) and \(y\) as generalized coordinates. Solve the latter and describe the motion.

Short Answer

Expert verified
The particle oscillates harmonically in \(x\) and moves linearly in \(y\).

Step by step solution

01

Define the Lagrangian

To find the Hamiltonian, we first need the Lagrangian. The Lagrangian, \(L\), is given by the kinetic energy minus the potential energy. The kinetic energy \(T\) is \(T = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2)\). The potential energy \(V\) can be deduced from the force \(\mathbf{F}\). Since \(\mathbf{F} = -abla V\), and \(\mathbf{F} = -kx \hat{\mathbf{x}} + K \hat{\mathbf{y}}\), the potential energy is \(V = \frac{1}{2}kx^2 - Ky\). Thus, \(L = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2) - \left(\frac{1}{2}kx^2 - Ky\right)\).
02

Derive the Hamiltonian

The Hamiltonian \(H\) is found via the Legendre transformation: \(H = \sum p_i \dot{q_i} - L\). The conjugate momenta are \(p_x = \frac{\partial L}{\partial \dot{x}} = m\dot{x}\) and \(p_y = \frac{\partial L}{\partial \dot{y}} = m\dot{y}\). Substituting these into the Hamiltonian expression, we have \(H = \frac{p_x^2}{2m} + \frac{p_y^2}{2m} + \frac{1}{2}kx^2 - Ky\).
03

Write Hamilton's Equations

Hamilton's equations are given by \(\dot{q_i} = \frac{\partial H}{\partial p_i}\) and \(\dot{p_i} = -\frac{\partial H}{\partial q_i}\). For \(x\) and \(y\), Hamilton's equations are: \(\dot{x} = \frac{p_x}{m}\), \(\dot{y} = \frac{p_y}{m}\), \(\dot{p_x} = -kx\), and \(\dot{p_y} = 0\).
04

Solve the Differential Equations

Solving \(\dot{p_x} = -kx\), we get a simple harmonic oscillator equation: \(m\ddot{x} = -kx\), whose solution is \(x(t) = A\cos(\omega t + \phi)\) with \(\omega = \sqrt{\frac{k}{m}}\). The equation \(\dot{p_y} = 0\) implies \(p_y = \text{constant}\). Thus, \(y(t) = \frac{p_y}{m} t + C\), where \(C\) is a constant.
05

Describe the Motion

The motion of the particle is given by: along the \(x\)-axis, it executes simple harmonic motion with an angular frequency \(\omega = \sqrt{\frac{k}{m}}\); along the \(y\)-axis, it moves with constant velocity \(\dot{y} = \frac{p_y}{m}\). The trajectory in the \(xy\)-plane is a straight line superimposed with oscillations along the \(x\)-axis.

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Key Concepts

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

Lagrangian mechanics
Lagrangian mechanics is a reformulation of classical mechanics, which provides a powerful method for analyzing systems. It is based on the Lagrangian function, denoted as \( L \), which is defined as the difference between the kinetic and potential energy of the system. For a particle of mass \( m \) moving in two dimensions, the kinetic energy \( T \) is expressed as \( T = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2) \), where \( \dot{x} \) and \( \dot{y} \) represent the time derivatives of the particle's position, or its velocities in the \( x \) and \( y \) directions.

The potential energy, \( V \), hinges on the forces acting on the particle. If the force \( \mathbf{F} \) is given by \( \mathbf{F} = -kx \hat{\mathbf{x}} + K \hat{\mathbf{y}} \), the potential energy can be found through the relationship \( \mathbf{F} = - abla V \). This yields \( V = \frac{1}{2}kx^2 - Ky \). Thus, the Lagrangian \( L \) becomes \( L = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2) - \left(\frac{1}{2}kx^2 - Ky \right) \).
  • Lagrangian mechanics simplifies the process of finding the equations of motion, especially for complex systems.
  • It does this by using generalized coordinates, which can be customized to suit the specific problem.
  • This approach is particularly useful for systems with constraints.
Hamilton's equations
Hamilton's equations are a set of first-order differential equations that provide an efficient way to describe the evolution of a system over time. These equations are derived from a transformation of the Lagrangian, known as the Legendre transformation, which allows us to express the dynamics in terms of conjugate variables, namely the generalized coordinates \( q_i \) and their conjugate momenta \( p_i \).

For our problem with the mass moving under the forces \( -kx \) and \( K \), the conjugate momenta are derived as \( p_x = m\dot{x} \) and \( p_y = m\dot{y} \). The Hamiltonian \( H \) can be calculated as:
\[H = \frac{p_x^2}{2m} + \frac{p_y^2}{2m} + \frac{1}{2}kx^2 - Ky\]
Hamilton's equations then provide:
  • \( \dot{x} = \frac{p_x}{m} \)
  • \( \dot{y} = \frac{p_y}{m} \)
  • \( \dot{p_x} = -kx \)
  • \( \dot{p_y} = 0 \)

These equations map out the system's behaviors in terms of the positions and velocities, offering a complete picture of its dynamics. They effectively replace the traditional Newton's second law for many applications. Instead of directly finding accelerations, you solve for velocities and momenta, which can be advantageous in certain analyses.
Simple Harmonic Motion
Simple harmonic motion (SHM) is a fundamental type of periodic motion where the restoring force is directly proportional to the displacement. This principle is exemplified in our exercise by the particle's movement in the \( x \)-direction under the influence of the force \( -kx \).

Solving Hamilton's equation for \( \dot{p_x} = -kx \) yields the differential equation for a harmonic oscillator:
\[ m\ddot{x} = -kx \]
This indicates that the system's response is sinusoidal and can be described by:
\[ x(t) = A\cos(\omega t + \phi) \]
where \( A \) is the amplitude, \( \omega \) is the angular frequency \( \omega = \sqrt{\frac{k}{m}} \), and \( \phi \) is the phase constant. This solution describes oscillations with a specific frequency dictated by the system's properties.
  • SHM is characterized by a regular, back-and-forth motion around an equilibrium position.
  • This type of motion is prevalent in physics, seen in systems such as pendulums and springs.
  • In our problem, the particle on the \( x \)-axis oscillates, while on the \( y \)-axis, it moves with a constant velocity.
SHM is a core concept that helps in understanding more complex oscillatory motions encountered in different physical contexts.

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Most popular questions from this chapter

Consider a mass \(m\) constrained to move in a vertical line under the influence of gravity. Using the coordinate \(x\) measured vertically down from a convenient origin \(O\), write down the Lagrangian \(\mathcal{L}\) and find the generalized momentum \(p=\partial \mathcal{L} / \partial \dot{x}\). Find the Hamiltonian \(\mathcal{H}\) as a function of \(x\) and \(p\) and write down Hamilton's equations of motion. (It is too much to hope with a system this simple that you would learn anything new by using the Hamiltonian approach, but do check that the equations of motion make sense.)

Here is a simple example of a canonical transformation that illustrates how the Hamiltonian formalism lets one mix up the \(q^{\prime}\) s and the \(p\) 's. Consider a system with one degree of freedom and Hamiltonian \(\mathcal{H}=\mathcal{H}(q, p)\). The equations of motion are, of course, the usual Hamiltonian equations \(\dot{q}=\partial \mathcal{H} / \partial p\) and \(\dot{p}=-\partial \mathcal{H} / \partial q .\) Now consider new coordinates in phase space defined as \(Q=p\) and \(P=-q .\) Show that the equations of motion for the new coordinates \(Q\) and \(P\) are \(\dot{Q}=\partial \mathcal{H} / \partial P\) and \(\dot{P}=-\partial \mathscr{H} / \partial Q ;\) that is, the Hamiltonian formalism applies equally to the new choice of coordinates where we have exchanged the roles of position and momentum.

A bead of mass \(m\) is threaded on a frictionless wire that is bent into a helix with cylindrical polar coordinates \((\rho, \phi, z)\) satisfying \(z=c \phi\) and \(\rho=R,\) with \(c\) and \(R\) constants. The \(z\) axis points vertically up and gravity vertically down. Using \(\phi\) as your generalized coordinate, write down the kinetic and potential energies, and hence the Hamiltonian \(\mathcal{H}\) as a function of \(\phi\) and its conjugate momentum \(p\) Write down Hamilton's equations and solve for \(\ddot{\phi}\) and hence \(\ddot{z}\). Explain your result in terms of Newtonian mechanics and discuss the special case that \(R=0\).

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.

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).

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