Problem 1:

A  block with mass m is hanging on a spring with spring constant k. 
The spring is held on the other side by a hand. 

(a)  If the hand moves as  y_h(t) = Asin(ωt), use the Lagrangian formalism to find the equation of motion of the block?  Assume that the spring is ideal and there is no damping.  Assume that ω is not the natural frequency of the spring.

(b)  What "initial conditions" at t = 0 are needed so that the block oscillates with a single frequency ω?  Under those conditions, what is the oscillation amplitude of the block?  What is the oscillation phase of the block relative to the hand?

Solution:

• Concepts:
  Lagrange's equation, forced oscillation
• Reasoning:
  The problem is equivalent to a forced oscillation problem with no damping. 
• Details of the calculation:
  (a) Pick a reference point.  Let the y-axis point downward.  The position of the hand with respect to the reference point is y_h = Asin(ωt).
  The position of the block is y_h + d + y_c.  Here d is the equilibrium length of the string under gravity, and y_c is the change the length of the spring from this equilibrium length.
  T = ½m[(d/dt)(y_h + y_c)]² = ½m[(dy_h/dt)² + (dy_c/dt)² + 2(dy_h/dt)(dy_c/dt)].
  U = -mg(y_h + y_c) + ½ky_c².  L = T - U.
  dL/dv_yc = mdy_c/dt + mdy_h/dt.  d/dt(dL/dv_yc) = md²y_c/dt² - mω²Asin(ωt).
  dL/dy_c = mg - ky_c.
  Equation of motion:   md²y_cdt² - mω²Asin(ωt) = mg - ky_c.
  d²y_cdt² + (k/m)y_c  - g = ω²Asin(ωt).
  Let y = y_c - mg/k,  ω₀² = k/m. 
  Then the equation of motion for the block becomes d²y/dt² + ω₀²y = ω²Asin(ωt).
  
  (b)  A particular solution of the inhomogeneous equation is y = Bsin(ωt). 
  This yields B = ω²A/(ω₀² - ω²).  To find the most general solution we add the homogeneous solution y = Csin(ω₀t + δ).
  Most general solution:  y(t) = Csin(ω₀t + δ) + [ω²A/(ω₀² - ω²)]sin(ωt).
  For the block to oscillate with a single frequency ω we need C = 0.
  Initial conditions needed:  y(0) = 0,  dy/dt|₀ = [ω³A/(ω₀² - ω²)].
  Then the oscillation amplitude is [ω²A/(ω₀² - ω²)].
  The oscillation phase of the block relative to the hand is either 0 or π, depending on if
  ω₀ > ω or if ω₀ < ω.

Problem 2:

Consider the N equally spaced, identical balls connected by springs.  All balls have mass m, all springs have spring constant β, and adjacent balls are separated by distance a at equilibrium.

(a)  Write down the equation of motion for each ball.  You can assume periodic boundary conditions.
(b)  Longitudinal wave can propagate along the springs and balls. 
Prove that the dispersion relation between the wave angular frequency ω and wave vector k is given by
ω_n = 2(β/m)^½|sin(k_n*a/2)|.
Here ω₀² = 4β/m.

Solution:

• Concepts:
  Lagrangian Mechanics, coupled oscillations, normal modes
• Reasoning:
  This is a one-dimensional problem.  We are asked to find the normal mode frequencies of coupled harmonic oscillators
• Details of the calculation:
  Let q_i (i = 0 to N - 1) denote the displacement of the ith ball (point mass) from its equilibrium position.
  (a)  L = T - U = ∑_{i = 0}^N-1[½m(dq_i/dt)² - ½β(q_i+1 - q_i)²]
  = ∑_{i = 0}^N-1[½m(dq_i/dt)² - ½β(q_i+1² + q_i² - 2q_iq_i+1].
  
  d/dt(∂L/∂(dq_j/dt)) -  ∂L/∂q_j = 0.
  The terms in the Lagrangian that depend on q_j an dq_j/dt are
  ½m(dq_j/dt)² - ½k(2q_j² - 2q_j-1q_j - 2q_jq_j+1).
  (They come from the terms in the sum i = j and i + 1 = j.)
  The equation of motion for the jth point mass therefore is
  md²q_j/dt² - β(q_j+1 - 2q_j + q_j-1) = 0.
  
  (b)  Assume q_j(t)  = |A|exp(i(k*ja - ωt).  The periodic boundary conditions imply that
  k*ja = k*(j + N)a ± n*2π,  k_nNa = n*2π,  k_n = n2π/(Na),  n = integer.
  
  Inserting solutions of the form q_j = |A|exp(i(k_n*ja - ωt)) into the equation of motion we obtain
  -mω_n² exp(ik_n*ja) - k[exp(ik_n*(j+1)a) - 2exp(ik_n*ja) + exp(ik_n*(j-1)a)] = 0,
  or
  ω_n² = -(β/m)[exp(ik_n*a) - 2 + exp(-ik_n*a)] = -¼ω₀²[2cos(k*a) - 2]
  = ½ω₀²[1 - cos(k_n*a)] = ω₀²sin²(k_n*a/2) = ω₀²sin²(nπ/N).
  Here ω₀² = 4β/m.
  ω_n = ω₀|sin(nπ/N)| = 2(β/m)^½|sin(k_n*a/2)| are the normal mode frequencies.
  This is the dispersion relation between the wave angular frequency ω and wavevector k.

Problem 3:

A particle of mass m is moving on the surface of a cone placed vertically with the vertex downwards in a uniform gravitational field.  The vertical angle of the cone is 2α .
(a)  Obtain the Lagrange equations of motion.
(b)  What quantities are conserved?  What is the energy of the system?
(c)  Determine the number of physical turning points in the orbit.  That is, does an infalling particle plunge, bounce to infinity, or oscillate back and forth in distance?
Show your work.

Solution:

• Concepts:
  Lagrange's equations of motion
• Reasoning:
  All forces except the forces of constraint are derivable from a potential. To find the constants of motion we find the Lagrangian and look for cyclic coordinates.
• Details of the calculation:
• (a)  L = T - U, T = ½m((dx/dt)² + (dy/dy)² + (dz/dt)²),  U = mgz.
  x = r cosθ , y = r sinθ , z = r cotα .
  The Cartesian velocities are
  dx/dt = dr/dt cosθ − r sinθ dθ/dt ,
  dy/dt = dr/dt sinθ + r cosθ dθ/dt ,
  dz/dt = dr/dt cotα ,
  The Lagrangian is
  L = ½m((dr/dt)² + r²(dθ/dt)² + (dr/dt)² cot²α) - mgr cotα
  = ½m((dr/dt)²/sin²α + r²(dθ/dt)²) - mgr cotα.
  
  p_θ = ∂L/∂(dθ/dt) = mr²dθ/dt.  ∂L/∂θ = 0.
  (d/dt)∂L/∂(dθ/dt) =  2mrdθ/dt + mr²d²θ/dt², 
  d²θ/dt² = -2(dθ/dt)/r.  (second order equation of motion for θ)
  p_θ = mr²dθ/dt = constant, the angular momentum L_z is constant.
  
  p_r = ∂L/∂(dr/dt) = m(dr/dt)/sin²α.  (d/dt)∂L/∂(dr/dt) = m(d²r/dt²)/sin²α. 
  ∂L/∂r = mr(dθ/dt)² - mg cotα.
  d²r/dt² = r sin²α (dθ/dt)²- g cotα sin²α.  (second order equation of motion for r)
  d²r/dt² = sin²αL_z²/(m²r³) - g cosα sinα. 
  
  (b)  p_θ = L_z is conserved since θ is a cyclic coordinate (i.e., it does not appear explicitly in L or H).  H is also conserved since it (or the Lagrangian) does not depend explicitly on time.
  Since L or H do not explicitly on time, and the coordinates and constraints do not explicitly depend on time, and the potential is velocity-independent, H is equal to the total energy H = E = T + U.
  
  (c)  E = T + U = constant = ½m(dr/dt)²/sin²α + L_z²/(2mr²) + mgr cotα.  
  At a turning point dr/dt = 0.
  If L_z = 0  there is one turning point for a particle initially moving upward (E = mgr cotα) and no turning point for a particle initially moving downward since the acceleration d²r/dt² is always negative.
  If L_z ≠ 0 there are two turning points, at E = L_z/(2mr²) + mgr cotα,  mgr³cotα  - Er² + L_z/(2m) = 0. 
  The value of r oscillates about r₀, where  d²r/dt²|_r0 = 0,  L_z²/(m²r₀³) = g cotα. 

Problem 4:

A small object with mass m moves on a smooth, friction-free horizontal surface.  It is attached to a peg at the origin by an ideal massless spring with spring constant k and equilibrium length r₀.  At time t = 0, the mass is set in motion in an arbitrary direction from point (r,φ).  Assume the spring can rotate about the origin, but not bend.

(a)  Find the Lagrangian L for the system, then
(b)  calculate the generalized momenta p_j.
(c)  Construct the Hamiltonian function, H(p_j, q_j, t),
(d)  then work out the equations of motion dp_j/dt and dq_j/dt. 
(e)  Are any of the variables cyclic, thereby giving especially simple equations of motion?  If so, integrate the equation(s) and interpret your results physically.
(f)  Consider the special case that r = constant.  Deduce the condition(s) that allow this case and discuss how this occurs physically.

Solution:

• Concepts:
  Lagrangian and Hamiltonian mechanics
• Reasoning:
  We are asked to find the Lagrangian and Hamiltonian for the system.
  This is a two-dimensional problem.
• Details of the calculation:
  (a)  L = T - U.
  T = ½m(v_x² + v_y²) = ½m((dφ/dt)²r² + (dr/dt)²),  U = ½k(r - r₀)².
  L = ½m((dφ/dt)²r² + (dr/dt)²) - ½k(r - r₀)².
  (b)  ∂L/∂(dr/dt) = p_r,   ∂L/∂(dφ/dt) = p_φ,   p_r = m(dr/dt),  p_φ  =  mr²(dφ/dt).
  (c)  H = p_r²/(2m)  + p_φ²/(2mr²) + ½k(r - r₀)², since  H(q, p, t) = ∑_i(dq_i/dt)p_i - L.
  (d)  dp_r/dt = -∂H/∂r = -k(r - r₀) + p_φ²/(mr³),  dp_φ/dt = -∂H/∂φ  = 0.
  dr/dt = ∂H/∂p_r = p_r/m,  dφ/dt = ∂H/∂p_φ  = p_φ/(mr²).
  (e)  The coordinate φ is cyclic.
  If the Lagrangian does not contain a given coordinate q_j then the coordinate is said to be cyclic and the corresponding conjugate momentum p_j is conserved.
  p_φ = constant = M, dφ/dt = M/(mr²).
  The Lagrangian and the coordinates do not contain time explicitly, therefore E = T + U is a constant of motion.
  (f)  We want dr/dt  and d²r/dt²  to be zero.
  md²r/dt²  = -k(r - r₀) + M²/(mr³) = 0,  k(r- r₀) = M²/(mr³).
  k(r - r₀)  = mr(dφ/dt) =  mv²/r.
  Rotation with uniform angular velocity, spring is stretched providing the centripetal force.