Problem 1:

Use the Euler-Lagrange equation to show that the shortest path between two points on a plane is a straight line.

• Concepts:
  Euler-Lagrange differential equation
• Reasoning:
  This is a variational calculus problem.
• Details of the calculation:
  The Euler-Lagrange differential equation is the fundamental equation of calculus of variations.  A typical problem consists of finding a real function y(x) of a real variable x such that a given function
  J(y) = ∫_x1^x2f(x, y(x), y'(x)) dx, where y'(x) = dy(x)/dx, has a stationary value.  The problem is to determine those functions y(x) which take given values at y₁ = y(x₁) and y₂ = y(x₂) at the end points and make J(y) an extremum.
  Then the Euler-Lagrange differential equation, ∂f/∂y - d/dx(∂f/∂y') = 0 must be satisfied.
  Assume y = f(x), x₁ ≤ x ≤ x₂, with f(x₁) = y₁ and f(x₂) = y₂.
  For any single path ds = ((dx)² + (dy)²)^½ = dx(1 + (y')²)^½.
  S = ∫_x1^x2(1 + (y')²)^½dx = ∫_x1^x2f(x, y(x), y'(x)) dx, with f(x, y(x), y'(x)) = f(x, y'(x)) = (1 + (y')²)^½.
  To find the extremum we set d/dx(∂f/∂y') = 0. 
  ∂f/∂y' = y'/(1 + (y')²)^½.  d/dx(∂f/∂y') = y''/(1 + (y')²)^½ - y'²y''/(1 + (y')²)^3/2 = 0.
  y''(1 + y'²) - y'²y''= 0.
  y'' = 0, y' = constant = m.  y = mx + b, the shortest path is a straight line.

Problem 2: 

A particle of mass m moves in one dimension under the influence of a force F(x,t) = (k/x²)exp(-t/τ), where k and τ are positive constants.  Compute the Lagrangian and the Hamiltonian, and discuss whether energy is conserved in this system.

Solution:

• Concepts:
  Lagrangian Mechanics
• Reasoning:
  We are asked to find the Lagrangian for this system one-dimensional system.
• Details of the calculation:
  F(x) = -∂U/∂x.  U(x) = (k/x)exp(-t/τ).
  L = ½m(dx/dt)² - (k/x)exp(-t/τ).  L = L(x, dx/dt, t).
  p = m(dx/dt), H = (dx/dt)p - L = p²/m - L = p²/(2m) + (k/x)exp(-t/τ).
  L does explicitly depend on time, the coordinate x does not.  So the Hamiltonian is the energy of the system, and the energy of the system is not conserved.

Aside:  See a numerical solution for the equation of motion d²x/dt² = (k/x²)exp(-t/τ).
It integrates the equation using the Numerov method O(dt⁴) and is implemented in a spreadsheet (with a short explanation).  You can implement it using any computer language and also extend it to higher order.

Problem 3:

(a)  Show that two Lagrangians L₁ and L_2, which differ only by the total derivative of a function of q and t, i.e.  L₂ = L₁ + df(q,t)/dt, describe the same motion for q.
(b)  Find the Lagrangian and Hamiltonian of a pendulum consisting of a mass m attached to a massless rigid rod AB of length l free to move in a vertical plane.  The end A of the rod is forced to move vertically, so that its displacement from the fixed point O is a given function of time γ(t).  Gravity acts vertically downward.
(c)  Show that the vertical acceleration of the point A, d²γ(t)/dt², has the same effect on the equation of motion as a time varying gravitational field.

Solution:

• Concepts:
  Lagrange's equations
• Reasoning:
  L = T - U, H(q, p, t) = ∑_i(dq_i/dt)p_i - L.  
• Details of the calculation:
  (a)  Let L₁ be the Lagrangian of the system.
  Then d/dt(∂L₁/∂(dq/dt)) -  ∂L₁/∂q = 0 yields the equations of motion.
  Let L₂ = L₁ + df(q,t)/dt = L₁ + g, with g = df(q,t)/dt.
  The equations of motion are unchanged if d/dt(∂g/∂(dq/dt)) -  ∂g/∂q = 0.
  g = df/dt = ∂f/∂t + (∂f/∂q)(dq/dt).
  Remember:  q and dq/dt are independent variables.
  
  ∂g/∂(dq/dt) = ∂f/∂q.  d/dt(∂g/∂(dq/dt)) = d/dt(∂f/∂q).
  
  ∂g/∂q = ∂/∂q(df/dt) = ∂/∂q(∂f/∂t +  (∂f/∂q)(dq/dt)) = ∂²f/∂t∂q + ∂²f/∂q² dq/dt  = d/dt(∂f/∂q).
  Therefore d/dt(∂g/∂(dq/dt)) -  ∂g/∂q = 0.
  
  (b)  L = T - U.
  z = γ - l cosθ,  x = l sinθ,  dz/dt = dγ/dt + l sinθ dθ/dt.  dx/dt = l cos dθ/dt.
  T = ½m((dz/dt)² + (dx/dt)²) = ½m((dγ/dt)² + l²(dθ/dt)²) + ml(dγ/dt)(dθ/dt) sinθ.
  U = -mg(lcosθ - γ).
  Constraint: γ = γ(t). 
  L(θ, dθ/dt) = ½m((dγ/dt)² + l²(dθ/dt)²) + ml(dγ/dt)(dθ/dt) sinθ + mg(lcosθ - γ).
  
  p_θ = ∂L/∂(dθ/dt) = ml² dθ/dt + ml(dγ/dt)sinθ.
  H = p_θ(dθ/dt) - L =  -½m(dγ/dt)² + ½ml²(dθ/dt)² + U.
  dθ/dt = (p_θ - ml(dγ/dt)sinθ)/(ml²).
  H(θ,p_θ) = (p_θ - ml(dγ/dt)sinθ)²/(2ml²) - ½m(dγ/dt)² + U.
  
  (c)  d/dt(∂L/∂(dθ/dt)) = ml²d²θ/dt² + mld²γ/dt²sinθ + ml(dγ/dt)cosθ dθ/dt/
  ∂L/∂θ = ml(dγ/dt)(dθ/dt) cosθ - mglsinθ/
  ml²d²θ/dt² + mld²γ/dt²sinθ + ml(dγ/dt)cosθ dθ/dt - ml(dγ/dt)(dθ/dt) cosθ + mglsinθ = 0.
  d²θ/dt² = -d²γ/dt²sinθ/l - gsinθ/l = -g(t)sinθ/l.  g(t) = g + d²γ/dt².
  
  Think about:  Is H conserved?  Is H the total energy of the system?

Problem 4:

Two pendulums of length L have masses m₁ and m₂ attached to their ends.  They are also connected to each other by a spring with spring constant k, as shown in the figure.  The equilibrium positions of the pendulums are vertical as shown.  Assume small displacements.
(a)  Write down the Lagrangian for the system
(b)  Find the normal mode frequencies for the system.
(c)  At t = 0, pendulum 1 is displaced by an angle θ₁, while pendulum 2
is at its vertical position. 
Find the subsequent angular position of each as a function of time.

Solution:

• Concepts:
  Coupled oscillations
• Reasoning:
  Solutions of the form θ _j = Re(A_je^iωt) can be found.  For a system with 2 degrees of freedom, 2 characteristic frequencies ω_α can be found. They may be degenerate
• Details of the calculation:
  (a)  L = T - U.
  T = ½m₁((dx₁/dt)² + (dy₁/dt)²) + ½m₂((dx₂/dt)² + (dy₂/dt)²), 
  U = -m₁gy₁ - m₂gy₂ + ½kΔ², where Δ is the change in the equilibrium length of the spring.
  To 2nd order x₁ = Lsinθ₁ = Lθ₁, x₂ = Lsinθ₂ + D = Lθ₂ + D, where D is the equilibrium length of the spring.
  y₁ = Lcosθ₁ = L(1 - θ₁²/2),  y₁ = Lcosθ₂ = L(1 - θ₁²/2).
  Define q₁ = Lθ₁,  q₂ = Lθ₂. 
  T = ½m₁(dq₁/dt)² + ½m₂(dq₂/dt)²,  U = (m₁g/L)(q₁²/2) + (m₂g/L)(q₂²/2) + ½kΔ².
  The length of the spring is [(y₂ - y₁)² + (x₂ - x₁)²]^½ = (L²[θ₁²/2 - θ₁²/2]² + [L(θ₂ - θ₁) + D]²]^½
  = L(θ₂ - θ₁) + D for small displacements (keeping only first order terms).
  Δ = L(θ₂ - θ₁),  Δ² = L²(θ₂² + θ₁² - θ₂θ₁ - θ₂θ₁).
  U = (m₁g/L)(q₁²/2) + (m₂g/L)(q₂²/2) + ½k(q₂² + q₁² - q₂q₁ - q₂q₁).
  Lagrangian = T - U = ½∑_ij[T_ij(dq_i/dt)(dq_j/dt) - k_ijq_iq_j]  with T_ij = T_ji,  k_ij = k_ji. 
  Here T₁₁ = m₁, T₂₂ = m₂,  T_ij(i≠j) = 0,  k₁₁ = m₁g/L + k,   k₂₂ = m₂g/L + k,  k₁₂ = k₂₁ = -k.
  
  Lagrangian = ½m₁(dq₁/dt)² + ½m₂(dq₂/dt)²) - (m₁g/L)(q₁²/2) - (m₂g/L)(q₂²/2) - ½k(q₂² + q₁² - q₂q₁ - q₂q₁).
  
  (b) 
  
  ω² = g/L + k/(2m₁) + k/(2m₂) ± (k/(2m₁) + k/(2m₂)).
  ω₁² = g/L,  θ₁₀ = θ₂₀, the pendulums oscillate with equal amplitudes in phase, the spring remains at its equilibrium length.
  ω₂² = g/L + k/m₁ + k/m²,   θ₁₀ = -(m₂/m₁)θ₂₀.  The pendulums oscillate with different amplitudes 180^o out of phase, the spring stretches and compresses.
  
  (c)  Most general solution: 
  θ₁(t) = Re(C₁exp(iω₁t) + C₂exp(iω₂t)),    θ₂(t) = Re(C₁exp(iω₁t) - (m₁/m₂)C₂exp(iω₂t)).
  dθ₁(t)/dt = Re(iω₁C₁exp(iω₁t) + iω₂C₂exp(iω₂t)),  dθ₂(t)/dt = Re(iω₁C₁exp(iω₁t) - iω₂(m₁/m₂)C₂expi(ω₂t)).
  Initial conditions at t = 0:
  Re(C₁ + C₂) = θ₁₀,   Re(C₁ - (m₁/m₂)C₂) = 0,  
  Re(C₁) = θ₁₀/(1 + m₂/m₁),  Re(C₂) = θ₁₀/(1 + m₁/m₂),
  Im(ω₁C₁+ ω₂C₂) = 0,  Im(ω₁C₁ - ω₂(m₁/m₂)C₂),  Im(C₁) = Im(C₂) = 0.
  θ₁(t) = θ₁₀(cos(ω₁t)/(1 + m₂/m₁) + cos(ω₂t)/(1 + m₁/m₂)).
  θ₂(t) = θ₁₀(cos(ω₁t)/(1 + m₂/m₁)  - cos(ω₂t)/(1 + m₂/m₁)).

Problem 5:

A point mass m₁ is constrained to move on a horizontal wire without friction. 
A second point mass m₂ is attached to it by a rod of negligible mass and length L.

(a)  Write down the Lagrangian of the system and find the equations of motion.
(b)  Are there any equilibrium points for the rotational motion?  Are they stable?  If yes, find the frequency of small oscillations about the equilibrium points.
(c)  Now assume the wire is not horizontal, but makes an angle φ with the horizontal direction.  Find the equations of motion.  Are there any equilibrium points for the rotational motion?  Are they stable?  If yes, find the frequency of small oscillations about the equilibrium points.

Solution:

• Concepts:
  Lagrangian Mechanics
• Reasoning:
  All forces except the forces of constraint are derivable from a potential.  The Lagrangian formalism is well suited for such a system.
• Details of the calculation:
  (a)  Divide the motion into motion of the CM and motion about the CM.
  Distance d of the CM from m₁:  m₁d = m₂(L - d), d = m₂L/(m₁ + m₂). 
  Let M = m₁ + m₂. 
  T = ½M(X'² + Y'²) + ½Iθ'².  Here I = m₁d²+ m₂(L - d)² = m₁d²+ m₂(m₁/m₂)²d² = (m₁/m₂)Md².
  The ' = d/dt denotes the derivative with respect to t.
  Constraint: Y = -d cosθ, Y' = d sinθ θ'.
  T = ½MX'² + ½(Md²sin²θ + I)θ'².   U = -Mgd cosθ.
  L = ½MX'² + ½(Md²sin²θ + I)θ'² + Mgd cosθ.
  ∂L/∂X' = MX' = constant.  (equation of motion for the X coordinate)
  The x-component of the velocity of the CM is constant.
  ∂L/∂θ' = (Md²sin²θ + I)θ', 
  d/dt(∂L/∂θ') = (Md²sin²θ + I)θ'' +  Md²2 sinθ cosθ θ'².
  ∂L/∂θ = ½Md²2 sinθ cosθ θ'² - Mgd sinθ.
  (Md²sin²θ + I)θ'' + Md²sinθ cosθ θ'² + Mgd sinθ = 0.  (equation of motion for the θ coordinate)
  (sin²θ + (m₁/m₂))θ'' + sinθ cosθ θ'² + ((m₁ + m₂)g/(m₂L)) sinθ = 0. 

  ---

  (b)  equilibrium points: θ'' = θ' = 0, --> sinθ = 0, θ = 0 or θ = π.
  The point θ = 0 is stable and the point θ = π is unstable.
  To find the frequency of small oscillations about the θ = 0 point, keep only first order terms in the small quantities in the equation of motion.
  (m₁/m₂))θ'' = -(g/d)θ,  ω² = (m₂/m₁)g/d = [(m₁ + m₂)/m₁]g/L.
  
  Different approach for part (a):
  Let x₁ denote the position of m₁ along the line and θ denote the angle the rod makes with the vertical. 
  For the mass m₂ we have x₂ = x₁ + L sinθ,  y₂ = -L cosθ.
  T = ½m₁x₁'² + ½m₂(x₂'² + y₂'²) = ½(m₁ + m₂)x₁'² + ½m₂[L²θ'² + 2Lx₁'θ' cosθ].
  U = -m₂gLcosθ.
  L = ½(m₁ + m₂)x₁'² + ½m₂[L²θ'² + 2Lx₁'θ' cosθ] + m₂gL cosθ.
  ∂L/∂x₁' = (m₁ + m₂)x₁' + m₂Lθ 'cosθ. 
  d/dt(∂L/∂x₁') = (m₁ + m₂)x₁'' + m₂Lθ'' cosθ - m₂L θ'² sinθ = 0. (equation of motion for x₁)
  ∂L/∂θ' = m₂[L²θ' + Lx₁'cosθ].  d/dt(∂L/∂θ') = m₂[L²θ'' + Lx₁''cosθ - Lx₁' sinθ θ']. 
  ∂L/∂θ = -m₂[Lx₁' sinθ θ' + gL sinθ]. 
  θ'' + (x₁''/L) cosθ + (g/L) sinθ = 0. (equation of motion for θ)
  
  The equations of motion for x₁ and θ are coupled differential equations.
  Insert x₁'' into the equation of motion for θ.
  θ'' - [m₂/(m₁ + m₂)]θ'' cos²θ + [m₂/(m₁ + m₂)θ'² sinθ cosθ + (g/L) sinθ = 0.
  θ''(m₁/(m₁ + m₂) + m₂/(m₁ + m₂)sin²θ] + [m₂/(m₁ + m₂)]θ'² sinθc osθ + (g/L) sinθ = 0.
  θ''((m₁/m₂) + sin²θ] + θ')² sinθ cosθ + ((m₁ + m₂)g/(m₂L)) sinθ = 0.
  
  We arrive at the same equation of motion for the θ coordinate.  θ is a coordinate for associate with a normal mode for small oscillations, x₁ is not.  The other normal coordinate is X, the x-coordinate of the CM.  This normal mode has ω = 0.

  ---

  (c)  If we orient our coordinate system as shown in the figure, then only the expression for the potential energy changes.

  
  U = -Mg sinφ X + Mg cosφ Y = -Mg sinφ X - Mg cosφ d cosθ.
  L = ½MX'² + ½(Md²sin²θ + I)θ'² + Mg sinφ X + Mg cosφ d cosθ.
  ∂L/∂X' = MX'.  ∂L/∂X' = Mg sinφ.  X'' = - g sinφ.
  The acceleration of the CM is constant.
  d/dt(∂L/∂θ') = (Md²sin²θ + I)θ'' +  Md²2 sinθ cosθ θ'².
  ∂L/∂θ = ½Md²2 sinθ cosθ θ'² - Mgd cosφ sinθ .
  (Md²sin²θ + I)θ'' +  Md²2 sinθ cosθ θ'² + Mgd cosφ sinθ = 0.
  (sin²θ + (m₁/m₂))θ'' + sinθ cosθ θ'² + ((m₁ + m₂)g/(m₂L)) cosφ sinθ = 0. 
  equilibrium points: θ'' = θ' = 0 --> sinθ = 0, θ = 0 or θ = π.
  The point θ = 0 is stable and the point θ = π is unstable.
  
  Small oscillations:
  θ'' = -(Mgd cosφ/I) θ = -[(m₁ + m₂)/m₁](g/L) cosφ θ.
  
  Different approach for part (c)
  T = ½m₁x₁'² + ½m₂(x₂'² + y₂'²) = ½(m₁ + m₂)x₁'² + ½m₂[L²θ'² + 2Lx₁'θ' cosθ].
  U = -(m₂ + m₂)gL sinφ x₁ - m₂g L cos(θ - φ).
  L = ½(m₁ + m₂)x₁'² + ½m₂[L²θ'² + 2Lx₁'θ' cosθ] + (m₂ + m₂)gL sinφ x₁ + m₂g L cos(θ - φ).
  ∂L/∂x₁' = (m₁ + m₂)x₁' + m₂Lθ 'cosθ. 
  d/dt(∂L/∂x₁') = (m₁ + m₂)x₁'' + m₂Lθ'' cosθ - m₂L θ'² sinθ.
  ∂L/∂x₁ = (m₂ + m₂)gL sinφ.
  (m₁ + m₂)x₁'' + m₂Lθ'' cosθ - m₂L θ'² sinθ - (m₁ + m₂)gL sinφ = 0.  (equation of motion for x₁)
  ∂L/∂θ' = m₂[L²θ' + Lx₁'cosθ].  d/dt(∂L/∂θ') = m₂[L²θ'' + Lx₁''cosθ - Lx₁' sinθ θ']. 
  ∂L/∂θ = -m₂Lx₁' sinθ θ' - m₂g L sin(θ - φ).
  θ'' + (x₁''/L) cosθ] +  (g/L) sin(θ - φ) = 0  (equation of motion for θ)
  Insert x₁'' into the equation of motion for θ.
  Mθ'' + M(x₁''/L) cosθ] +  M(g/L) sin(θ - φ) = 0.
  Mx₁''/L = -m₂θ'' cosθ + m₂θ'² sinθ + M(g/L)sinφ.
  Mθ'' - m₂θ''cos²θ + m₂θ'² sinθ cosθ + M(g/L) sinφ cosθ + M(g/L) sin(θ - φ) = 0.
  Mθ'' - m₂θ''cos²θ + m₂θ'² sinθ cosθ + M(g/L) cosφ sinθ = 0.
  (sin²θ + (m₁/m₂))θ'' + sinθ cosθ θ'² + ((m₁ + m₂)g/(m₂L)) cosφ sinθ = 0. 
  We arrive at the same equation of motion for the θ coordinate.