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Write down the Lagrangian for a one-dimensional particle moving along the \(x\) axis and subject to a force \(F=-k x\) (with \(k\) positive). Find the Lagrange equation of motion and solve it.

Short Answer

Expert verified
The equation of motion is \(m\ddot{x} + kx = 0\), with solution \(x(t) = A\cos(\omega t) + B\sin(\omega t)\).

Step by step solution

01

Define Kinetic and Potential Energy

In classical mechanics, the kinetic energy (T) of a particle with mass \(m\) moving with velocity \(v\) is given by \(T = \frac{1}{2}mv^2\). The potential energy (V) of the particle is determined by the force it experiences. Given the force \(F = -kx\), the potential energy is \(V = \frac{1}{2}kx^2\).
02

Construct the Lagrangian

The Lagrangian \(L\) is defined as the difference between kinetic energy \(T\) and potential energy \(V\). Therefore, we have \(L = T - V = \frac{1}{2}mv^2 - \frac{1}{2}kx^2\). Since the velocity \(v\) is the derivative of position, \(v = \dot{x}\), the Lagrangian becomes \(L = \frac{1}{2}m\dot{x}^2 - \frac{1}{2}kx^2\).
03

Derive the Euler-Lagrange Equation

The Euler-Lagrange equation is \(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right) - \frac{\partial L}{\partial x} = 0\). We compute these partial derivatives: \(\frac{\partial L}{\partial \dot{x}} = m\dot{x}\) and \(\frac{\partial L}{\partial x} = -kx\). Differentiating \(m\dot{x}\) with respect to time gives \(m\ddot{x}\). Therefore, the equation of motion is \(m\ddot{x} + kx = 0\).
04

Solve the Equation of Motion

The equation \(m\ddot{x} + kx = 0\) is a simple harmonic oscillator equation. Its general solution is \(x(t) = A\cos(\omega t) + B\sin(\omega t)\), where \(\omega = \sqrt{\frac{k}{m}}\). \(A\) and \(B\) 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.

Kinetic Energy
Kinetic energy is a core concept in mechanics, representing the energy that a particle possesses due to its motion. For a particle with mass \(m\) moving with velocity \(v\), the kinetic energy \(T\) is given by the formula:
  • \(T = \frac{1}{2}mv^2\)
This formula tells us that kinetic energy depends on both the mass of the object and the square of its velocity. A particle's kinetic energy increases as it moves faster or gains more mass.
In the context of Lagrangian mechanics, kinetic energy plays a vital role in deriving the equations of motion. For the particle moving along the \(x\) axis, the velocity is expressed as \(v = \dot{x}\), where \(\dot{x}\) denotes the time derivative of position, or simply, the speed in the \(x\) direction. Subsequently, the kinetic energy is represented as:
  • \(T = \frac{1}{2}m\dot{x}^2\)
This concise representation allows us to easily incorporate kinetic energy into the Lagrangian framework.
Potential Energy
Potential energy represents the energy stored within a system due to its position or configuration. In our scenario, it is associated with the force the particle experiences. Specifically, we were given the force:
  • \(F = -kx\)
Here, the potential energy \(V\) can be derived as the work done against this force and is expressed by the formula:
  • \(V = \frac{1}{2}kx^2\)
This expression implies that the potential energy depends on the displacement \(x\) of the particle from the equilibrium position. The potential energy increases with the square of the displacement, just as kinetic energy does with velocity.
When building the Lagrangian, potential energy is considered because it accounts for how energy converts from stored potential energy to kinetic form. Our particle's potential energy forms half of the components necessary to construct the Lagrangian.
Euler-Lagrange Equation
The Euler-Lagrange equation is a fundamental principle in Lagrangian mechanics used to derive the equations of motion for a system. For a given Lagrangian \(L\), which is the difference between kinetic energy \(T\) and potential energy \(V\), it is expressed as:
  • \(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right) - \frac{\partial L}{\partial x} = 0\)
This equation ensures that the trajectory of the particle is consistent with the principle of least action, meaning it follows a path that minimizes the action over time.
Given the Lagrangian \(L = \frac{1}{2}m\dot{x}^2 - \frac{1}{2}kx^2\), we first calculate the partial derivatives \(\frac{\partial L}{\partial \dot{x}} = m\dot{x}\) and \(\frac{\partial L}{\partial x} = -kx\). By differentiating \(m\dot{x}\) with respect to time, we obtain \(m\ddot{x}\), leading to the equation of motion:
  • \(m\ddot{x} + kx = 0\)
This equation describes a harmonic oscillator, a common physical system, and showcases the power of Lagrangian mechanics in modeling dynamic behaviors.

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Most popular questions from this chapter

The center of a long frictionless rod is pivoted at the origin, and the rod is forced to rotate in a horizontal plane with constant angular velocity \(\omega\). Write down the Lagrangian for a bead threaded on the rod, using \(r\) as your generalized coordinate, where \(r, \phi\) are the polar coordinates of the bead. (Notice that \(\phi\) is not an independent variable since it is fixed by the rotation of the rod to be \(\phi=\omega t\).) Solve Lagrange's equation for \(r(t) .\) What happens if the bead is initially at rest at the origin? If it is released from any point \(r_{\mathrm{o}}>0,\) show that \(r(t)\) eventually grows exponentially. Explain your results in terms of the centrifugal force \(m \omega^{2} r\).

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 mass \(m_{1}\) rests on a frictionless horizontal table. Attached to it is a string which runs horizontally to the edge of the table, where it passes over a frictionless, small pulley and down to where it supports a mass \(m_{2} .\) Use as coordinates \(x\) and \(y\) the distances of \(m_{1}\) and \(m_{2}\) from the pulley. These satisfy the constraint equation \(f(x, y)=x+y=\) const. Write down the two modified Lagrange equations and solve them (together with the constraint equation) for \(\ddot{x}, \ddot{y},\) and the Lagrange multiplier \(\lambda\). Use (7.122) (and the corresponding equation in \(y\) ) to find the tension forces on the two masses. Verify your answers by solving the problem by the elementary Newtonian approach.

Using the usual angle \(\phi\) as generalized coordinate, write down the Lagrangian for a simple pendulum of length \(l\) suspended from the ceiling of an elevator that is accelerating upward with constant acceleration \(a\). (Be careful when writing \(T\); it is probably safest to write the bob's velocity in component form.) Find the Lagrange equation of motion and show that it is the same as that for a normal, nonaccelerating pendulum, except that \(g\) has been replaced by \(g+a\). In particular, the angular frequency of small oscillations is \(\sqrt{(g+a) / l}\).

(a) Write down the Lagrangian for a particle moving in three dimensions under the influence of a conservative central force with potential energy \(U(r),\) using spherical polar coordinates \((r, \theta, \phi)\). (b) Write down the three Lagrange equations and explain their significance in terms of radial acceleration, angular momentum, and so forth. (The \(\theta\) equation is the tricky one, since you will find it implies that the \(\phi\) component of \(\ell\) varies with time, which seems to contradict conservation of angular momentum. Remember, however, that \(\ell_{\phi}\) is the component of \(\ell\) in a variable direction.) (c) Suppose that initially the motion is in the equatorial plane (that is, \(\theta_{0}=\pi / 2\) and \(\dot{\theta}_{0}=0\) ). Describe the subsequent motion. (d) Suppose instead that the initial motion is along a line of longitude (that is, \(\dot{\phi}_{0}=0\) ). Describe the subsequent motion.

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