Chapter 7: Problem 2
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.
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.
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:
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:
- \(T = \frac{1}{2}mv^2\)
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:
- \(T = \frac{1}{2}m\dot{x}^2\)
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:
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.
- \(F = -kx\)
- \(V = \frac{1}{2}kx^2\)
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:
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:
- \(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right) - \frac{\partial L}{\partial x} = 0\)
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:
- \(m\ddot{x} + kx = 0\)