Question

(a) Write down the Lagrangian \(\mathcal{L}\left(x_{1}, x_{2}, \dot{x}_{1}, \dot{x}_{2}\right)\) for two particles of equal masses, \(m_{1}=m_{2}=m,\) confined to the \(x\) axis and connected by a spring with potential energy \(U=\frac{1}{2} k x^{2} .\) [Here \(x\) is the extension of the spring, \(x=\left(x_{1}-x_{2}-l\right),\) where \(l\) is the spring's unstretched length, and I assume that mass 1 remains to the right of mass 2 at all times.] (b) Rewrite \(\mathcal{L}\) in terms of the new variables \(X=\frac{1}{2}\left(x_{1}+x_{2}\right)\) (the CM position) and \(x\) (the extension), and write down the two Lagrange equations for \(X\) and \(x\). (c) Solve for \(X(t)\) and \(x(t)\) and describe the motion.

Short answer

The Lagrangian is \( \mathcal{L}(X, x, \dot{X}, \dot{x}) = 2m \dot{X}^2 + \frac{m}{2} \dot{x}^2 - \frac{1}{2} k x^2 \). Motion: CM moves at constant velocity, extension oscillates harmonically.

Step-by-step solution

Write the Kinetic Energy Expression

The kinetic energy of the system consists of the kinetic energies of both particles. For two particles of equal mass, the kinetic energy is given by:\[ T = \frac{1}{2} m \dot{x}_1^2 + \frac{1}{2} m \dot{x}_2^2 \]

Write the Potential Energy Expression

The potential energy is due to the spring and is given by:\[ U = \frac{1}{2} k (x_1 - x_2 - l)^2 \] where \(x = x_1 - x_2 - l\) is the extension of the spring.

Construct the Lagrangian

The Lagrangian \( \mathcal{L} \) is the difference between the kinetic energy \( T \) and the potential energy \( U \):\[ \mathcal{L}(x_1, x_2, \dot{x}_1, \dot{x}_2) = \frac{1}{2} m \dot{x}_1^2 + \frac{1}{2} m \dot{x}_2^2 - \frac{1}{2} k (x_1 - x_2 - l)^2 \]

Change Variables to Center of Mass and Extension

Introduce new variables \(X = \frac{1}{2}(x_1 + x_2)\) for the center of mass position and \(x = x_1 - x_2 - l\) for the extension. Re-express the Lagrangian in terms of these variables. Begin by finding expressions for \(x_1\) and \(x_2\): \[ x_1 = X + \frac{x}{2} + \frac{l}{2}, \quad x_2 = X - \frac{x}{2} - \frac{l}{2} \] Then substitute these into \( \mathcal{L} \).

Simplify the Lagrangian

After substitution, the Lagrangian is expressed as a function of \(X\) and \(x\):\[ \mathcal{L}(X, x, \dot{X}, \dot{x}) = 2m \dot{X}^2 + \frac{m}{2} \dot{x}^2 - \frac{1}{2} k x^2 \] This expression arises because the terms involving \(l\) cancel out or are constant.

Derive Lagrange Equations for New Variables

Using the derived Lagrangian in terms of \(X\) and \(x\), apply the Euler-Lagrange equation \( \frac{d}{dt}\left( \frac{\partial \mathcal{L}}{\partial \dot{q}} \right) - \frac{\partial \mathcal{L}}{\partial q} = 0 \) for each variable \(X\) and \(x\). For \(X\): \[ 4m \ddot{X} = 0 \]For \(x\): \[ m \ddot{x} + k x = 0 \]

Solve the Lagrange Equations

Solve the differential equations derived:1. For \(X(t)\): Since \(\ddot{X} = 0\), the solution is \(X(t) = C_1 + C_2 t\), where \(C_1\) and \(C_2\) are constants.2. For \(x(t)\): The equation \(\ddot{x} + \frac{k}{m} x = 0\) is a simple harmonic oscillator with solution \(x(t) = A \cos(\omega t) + B \sin(\omega t)\), where \(\omega = \sqrt{\frac{k}{m}}\) and \(A\) and \(B\) are constants.

Key concepts

Kinetic Energy

Kinetic energy is a fundamental concept in physics. It represents the energy an object possesses due to its motion. In the context of two particles of equal mass moving along the x-axis, we calculate the kinetic energy for each particle using the formula: \[ T = \frac{1}{2} m \dot{x}_1^2 + \frac{1}{2} m \dot{x}_2^2 \]This equation shows the sum of the kinetic energies of both particles. The terms \(m\) and \(\dot{x}_i\) refer to the mass of each particle and their velocities, respectively. The key takeaway here is that kinetic energy is dependent on both the mass of the objects and the square of their velocity. As velocity increases, kinetic energy grows rapidly.

• Kinetic energy increases with velocity.
• It is a crucial component of the Lagrangian formulation.

Potential Energy

Potential energy arises from the system's configuration and is associated with forces that don't depend on velocity but on position. In our problem, the potential energy is related to a spring connecting two masses. The potential energy stored in the spring system when it is displaced by an extension \(x\) is expressed as:\[ U = \frac{1}{2} k (x_1 - x_2 - l)^2 \]This equation highlights that potential energy is determined by the spring constant \(k\) and the square of the displacement. A stiffer spring (larger \(k\)) or greater extension results in higher potential energy. Potential energy is crucial because it describes the stored energy that can potentially be converted into kinetic energy.

• Extension \(x\) alters the potential energy.
• High spring constant \(k\) means more potential energy.

Euler-Lagrange Equation

The Euler-Lagrange equation is an essential part of Lagrangian mechanics, providing a powerful method for deriving the equations of motion for a system. It is formulated as:\[ \frac{d}{dt}\left( \frac{\partial \mathcal{L}}{\partial \dot{q}} \right) - \frac{\partial \mathcal{L}}{\partial q} = 0 \]where \(q\) and \(\dot{q}\) are generalized coordinates and velocities, respectively. Applying this equation to each system's coordinate allows us to find the motions' equations, representing the dynamics of the system. For the system in the exercise, the Euler-Lagrange equations govern the movement of both the center of mass \(X\) and the spring extension \(x\). This method provides insight into how complex interacting systems behave.

• Derived from the principle of least action.
• Helps find the true path of motion.

Harmonic Oscillator

The harmonic oscillator is a central model in physics describing systems where the force can be approximated as proportional to the displacement. In this exercise, the dynamics of the spring connecting the particles can be described as a harmonic oscillator. The differential equation that emerges for \(x(t)\) is:\[ m \ddot{x} + k x = 0 \]This equation is characteristic of simple harmonic motion, where the solution \(x(t)\) is a sinusoidal function, given by:\[ x(t) = A \cos(\omega t) + B \sin(\omega t) \]with \(\omega = \sqrt{\frac{k}{m}}\). It reveals the system undergoes oscillations with an angular frequency \(\omega\), which depends on the mass and spring constant. Harmonic oscillators are fundamental in understanding not just mechanical systems but also electrical and quantum systems.

• Equation describes oscillatory motion.
• Frequency \(\omega\) depends on \(m\) and \(k\).