Lagrangian Analysis of Two-Body Systems and Spring Dynamics

Posted on Aug 2, 2025 in Physical Engineering

Problem 1: Two Masses in a Uniform Gravitational Field

Two masses m₁ and m₂ move in a uniform gravitational field g and interact via a potential energy U(r).

1. Show that the Lagrangian can be decomposed as L = L_CM + L_rel.
2. Write down the Lagrange equations for the CM coordinates X, Y, Z and describe the motion of the CM. Then write the Lagrange equations for the relative coordinates and show that the motion of r is the same as that of a particle of reduced mass μ, position r, and potential energy U(r).

Solution for Two Masses

Definitions

• r₁, r₂: Positions of particles 1 and 2
• m₁, m₂: Masses
• g: Uniform gravitational field
• R = (m₁r₁ + m₂r₂) / (m₁ + m₂): Center of mass position
• r = r₁ – r₂: Relative position
• M = m₁ + m₂: Total mass
• μ = (m₁m₂) / M: Reduced mass

(a) Lagrangian Decomposition

Kinetic Energy

T = ½ m₁ |ṙ₁|² + ½ m₂ |ṙ₂|²

Where ṙ₁ = Ṙ + (m₂/M) ṙ, and ṙ₂ = Ṙ – (m₁/M) ṙ

Therefore, T = ½ M |Ṙ|² + ½ μ |ṙ|²

Potential Energy

V = –m₁ g · r₁ – m₂ g · r₂ + U(r)

This simplifies to V = –M g · R + U(r)

Final Lagrangian

L = T – V = ½ M |Ṙ|² + ½ μ |ṙ|² + M g · R – U(r)

Thus:

• L_CM = ½ M |Ṙ|² + M g · R
• L_rel = ½ μ |ṙ|² – U(r)

(b) Lagrange Equations and Motion Analysis

Lagrange Equations for CM Coordinates R

From L_CM = ½ M |Ṙ|² + M g · R

The Lagrange equations are:

d/dt (∂L/∂Ẋ) = ∂L/∂X   ⇒   M Ẍ = M g_x   ⇒   Ẍ = g_x

Similarly for Y and Z coordinates:

Ÿ = g_y

Z̈ = g_z

So, M R̈ = M g   ⇒   R̈ = g

The motion of the Center of Mass is that of a free particle under uniform gravitational acceleration.

Lagrange Equations for Relative Coordinates r

From L_rel = ½ μ |ṙ|² – U(r)

The Lagrange equations are:

μ r̈ = -∇U(r)

Thus, the relative motion is the same as that of a particle of reduced mass μ in the potential U(r).

Problem 2: Spring-Mass System Dynamics

Two particles of masses m₁ and m₂ are joined by a massless spring of natural length L and force constant k.

Initially, m₂ is resting on a table and m₁ is held vertically above m₂ at a height L. At time t = 0, m₁ is projected vertically upward with initial velocity v₀. Assume v₀ is small enough that the masses never collide.

Find the positions of the two masses at any subsequent time t (before either returns to the table) and describe the motion.

Solution for Spring-Mass System

Definitions

• y₁(t), y₂(t): Vertical positions of m₁ and m₂
• m₁, m₂: Masses
• g: Acceleration due to gravity
• R = (m₁y₁ + m₂y₂) / (m₁ + m₂): Center of mass position
• r = y₁ – y₂: Relative position
• M = m₁ + m₂: Total mass
• μ = (m₁ * m₂) / M: Reduced mass

Energy Components

Spring Potential Energy

U(r) = ½ k (r – L)²

Gravitational Potential Energy

V_g = –m₁ g y₁ – m₂ g y₂ = –M g R

Total Potential Energy

V = –M g R + ½ k (r – L)²

Kinetic Energy

T = ½ M Ṙ² + ½ μ ṙ²

Lagrangian

L = T – V

L = ½ M Ṙ² + ½ μ ṙ² + M g R – ½ k (r – L)²

Lagrange Equations

For R (Center of Mass)

d/dt (∂L/∂Ṙ) = ∂L/∂R

⇒ M R̈ = M g

⇒ R̈ = g

So, the CM moves like a free particle under gravity.

For r (Relative Motion)

d/dt (∂L/∂ṙ) = ∂L/∂r

⇒ μ r̈ = –k (r – L)

⇒ r̈ + (k/μ) (r – L) = 0

This is a simple harmonic oscillator equation centered at r = L.

General Solution

Let ω = √(k / μ)

Relative Motion

r(t) = L + A cos(ωt) + B sin(ωt)

Initial conditions:

• At t = 0: r(0) = L ⇒ A = 0
• ṙ(0) = v₀ ⇒ B = v₀ / ω

So:

r(t) = L + (v₀ / ω) sin(ωt)

CM Motion

Initial conditions:

• R(0) = (m₁ * L + m₂ * 0) / M = (m₁ * L) / M
• Ṙ(0) = (m₁ * v₀ + m₂ * 0) / M = (m₁ * v₀) / M

So:

R(t) = (m₁ * L) / M + (m₁ * v₀) / M * t + ½ g t²

Final Positions

To find y₁(t) and y₂(t), we use the definitions of R and r:

• y₁(t) = R(t) + (m₂ / M) * r(t)
• y₂(t) = R(t) – (m₁ / M) * r(t)