Question

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

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

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 \(V = \frac{1}{2}kx^2\).

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 \(L = \frac{1}{2}m\dot{x}^2 - \frac{1}{2}kx^2\).

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 \(m\ddot{x} + kx = 0\).

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.

Key concepts

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.