Problem 1:

Find the normal, longitudinal modes of vibration for three masses connected by identical springs of spring constant k.  The masses are collinear.  The end masses have mass m, while the inner mass has mass 2m.
(a)  Calculate the normal modes of the system.
(b)  Describe the relative motion of the particles for each normal mode.

Solution:

• Concepts:
  Coupled oscillations, normal modes
• Reasoning:
  We are asked to find the normal modes of coupled harmonic oscillators.
• Details of the calculation:
  (a)  The kinetic energy is T = ½[m(dx₁/dt)² +  2m(dx₂/dt)² +  m(dx₃/dt)²],
  and the potential energy is U = (k/2)[(x₂ - x₁)² + (x₃ - x₂)²].
  U = (k/2)[x₁² + 2x₂² + x₃² - x₁x₂ - x₂x₁ - x₂x₃ - x₃x₂].
  The x_i are the displacements from the equilibrium positions.  We use the x_i as our generalized coordinates q_i.
  The Lagrangian is L = T - U.  This can be put into the form
  L = 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 = T₃₃ = m, T₂₂ = 2m, T_ij(i≠j) = 0,
  k₁₁ = k₃₃ = k, k₂₂ = 2k, k₁₂ = k₂₁ = k₂₃ = k₃₂ = -k, k₁₃ = k₃₁ = 0.
  Solutions of the form x_j = Re(A_je^{iω t}) can be found.  For solutions of this form the equations of motion reduce to
  

   

  	k-ω2m	-k	0
  	-k	2k-2ω2m	-k
  	0	-k	k-ω2m

    

  	A1
  	A2
  	A3

    = 0.

  
  

  We can find the ω² from det(k_ij-ω ²T_ij) = 0.  For a system with n degrees of freedom, n characteristic frequencies ω_a can be found.  Our system has 3 degrees of freedom.

   

  	k-ω2m	-k	0
  	-k	2k-2ω2m	-k
  	0	-k	k-ω2m

    = 0.

  
  

  
  2(k-ω²m)³ - 2k²(k-ω²m) = 0.
  Solution 1: k-ω²m = 0,  ω = (k/m)^½.
  Solution 2: k-ω²m ≠ 0, then (k-ω²m)² = k², (k-ω²m) = ±k.
  For (k-ω²m) = +k, ω = 0.
  Solution 3: (k-ω²m) = -k, ω = (2k/m)^½.
  
  (b)  The displacements for each mode are determined from the equations of motion.
  Solution 1: ω = (k/m)^½.
  Equation 1 yields kA₂ = 0, equation 2 then yields kA₁ + kA₃ = 0.
  A₁ = - A₃, A₂ = 0, the central mass is stationary, m₁ and m₃ move in opposite directions with equal amplitudes.
  Solution 2: ω = 0.
  A₁ = A₂ = A₃, translation of the CM, no relative motion.
  Solution 3: ω = (2k/m)^½.
  Equation 1 yields -kA₁ - kA₂ = 0, A₁ = -A₂.
  Equation 3 yields -kA₂ - kA₃ = 0, A₃ = -A₂.
  A₁ = A₃ = -A₂, the central mass move in a direction opposite to the direction of the outer masses.  All masses oscillate with equal amplitudes.
  

Problem 2:

A triple pendulum consists of masses αm, m and m attached to a single light string at distances a, 2a, and 3a respectively from its points of suspension.  Consider only motion in a plane.
(a)  Determine the value of α such that one of the normal frequencies of the system will equal the frequency of a simple pendulum of length a/2 and mass m.  You may assume the displacements of the masses from equilibrium are small.
(b)  Find the mode corresponding to this frequency and sketch it.

Solution:

• Concepts:
  Small oscillations, coupled oscillations, normal modes
  L = ½∑_ij[T_ij(dq_i/dt)(dq_j/dt) - k_ijq_iq_j]  with T_ij = T_ji,  k_ij = k_ji. 
  Solutions of the form q_j = Re(A_je^iωt) can be found. 
  We can find the ω² from det(k_ij - ω²T_ij) = 0.
• Reasoning:
  We are asked to find the normal mode frequencies for small displacements.
• Details of the calculation:
  
  (a)  Let Φ₁, Φ₂, and Φ₃ be the generalized coordinates.  Then
  x₁ = a sinΦ₁, x₂ = a(sinΦ₁ + sinΦ₂), x₃ = a(sinΦ₁ + sinΦ₂ + sinΦ₃),
  y₁ = a cosΦ₁, y₂ = a(cosΦ₁ + cosΦ₂), x₃ = a(cosΦ₁ + cosΦ₂ + cosΦ₃).
  For small displacements we have sinΦ = Φ, cosΦ = 1 - Φ²/2.
  The kinetic energy of the system is
  T = ½αm((dx₁/dt)² + (dy₁/dt)²) + ½m((dx₂/dt)² + (dy₂/dt)²) + ½m((dx₃/dt)² + (dy₃/dt)²).
  Let us only keep terms up to second order in Φ and dΦ/dt.  Then
  (dx₁/dt)² + (dy₁/dt )² = a²(dΦ₁/dt)²,
  (dx₂/dt)² + (dy₂/dt )² = a²[(dΦ₁/dt)² + (dΦ₂/dt)² + 2(dΦ₁/dt)(dΦ₂/dt)],
  (dx₃/dt)² + (dy₃/dt )² = a²[(dΦ₁/dt)² + (dΦ₂/dt)² + (dΦ₃/dt)²
  + 2(dΦ₁/dt)(dΦ₂/dt) + 2(dΦ₂/dt)(dΦ₃/dt) + 2(dΦ₁/dt)(dΦ₃/dt)].
  
  T = ½ma²[(α + 2)(dΦ₁/dt)² + 2(dΦ₂/dt)² + (dΦ₃/dt)²
  + 4(dΦ₁/dt)(dΦ₂/dt) + 2(dΦ₂/dt)(dΦ₃/dt) + 2(dΦ₁/dt)(dΦ₃/dt)].
  
  The potential energy of the system is
  U = -αmgy₁ - mgy₂ - mgy₃ + arbitrary constant.
  U = ½mga[(α + 2)Φ₁² + 2Φ₂² + Φ₃²].
  ^.
  We have
  T = ½∑_ij[T_ij(dΦ_i/dt)(dΦ_j/dt),  U = ½∑_ijk_ijΦ_iΦ_j.
  T₁₁ = ma²(α + 2), T₂₂ = 2ma², T₃₃ = ma², T₁₂ = T₂₁ = 2ma², T₁₃ = T₃₁ = T₂₃ = T₃₂ = ma².
  k₁₁ = mga(α + 2), k₂₂ = 2mga, k₃₃ = mga, k_ij(i ≠ j) = 0.

  (b)  Let λ = ω²a/g.  Then

  mga 

  	(α + 2)(1 - λ)	-2λ	-λ
  	-2λ	2(1 - λ)	-λ
  	-λ	-λ	(1 - λ)

    = 0.

  
  
  For a simple pendulum with length a/2 we have ω = (2g/a)^½.  We are looking for a mode with ω² = 2g/a, or λ = 2.
  

   

  	α + 2	4	2
  	4	2	2
  	2	2	1

    = 0 --> α = 2.

  
  

  (b)  The mode corresponding to this frequency is found from

   

  	α + 2	4	2
  	4	2	2
  	2	2	1

    

  	A1
  	A2
  	A3

    = 0.

  
  

  A₃ = -2A₁, A₂ = 0.
  

Problem 3:

Consider an ideal gas of N particles in a cylinder with a piston so that the volume and pressure may change.  Suppose that each particle has f = 5 quadratic degrees of freedom in its energy.  Consider the three paths in a PV diagram shown.  Suppose ways 1 and 3 are straight lines on the PV diagram and way 2 is an adiabatic process.
(a)  How much work W is done on the gas for ways 1, 2, and 3?
(b)  How much heat Q is transferred to the gas for ways 1, 2, and 3?
(c)  What is the change in internal energy ΔU for ways 1, 2, 3?
(d)  What is the change in entropy ΔS for way 3?

Solution:

• Concepts:
  The ideal gas law PV = NkT,  work done by the system W = ∫PdV,  the first law of thermodynamics, entropy
• Reasoning:
  We are asked to compute ΔU, ΔQ, W, and ΔS for along different path on a PV diagram.
• Details of the calculation:
  (a)  W = -∫PdV = work done on the gas = -area under the curve on a PV diagram.
  First law of thermodynamics:  ΔU = ΔQ + W. 
  way 1:  W = -½(P₁ + P₂)(V₂ - V₁).
  way 2:  W = ΔU = (5/2)Nk(T₂ - T₁) = (5/2)(P₂V₂ - P₁V₁).
  way 3:  W = -P₁(V₂ - V₁).
  
  (b)  ΔQ = ΔU - W.
  way 1:  ΔQ =  (5/2)(P₂V₂ - P₁V₁) + ½(P₁ + P₂)(V₂ - V₁)
  = 3(P₂V₂ - P₁V₁) - ½(P₂V₁ - P₁V₂).
  way 2:  ΔQ = 0.
  way 3:  ΔQ = (5/2)(P₁V₂ - P₁V₁) + P₁(V₂ - V₁)
  = (7/2)P₁(V₂ + V₁).
  
  (c)  ΔU is a physical property of the system.  It depend only on the state of the system, not on the way the system was put into this state.
  ways 1 and 2:  ΔU = (5/2)(P₂V₂ - P₁V₁)
  way 3:  ΔU = (5/2)(P₁V₂ - P₁V₁)
  
  (d)  Change in entropy: ΔS = ∫_i^f dS = ∫_i^f dQ_r/T.
  dQ = dU - dW = (5/2)NkdT + NkdT = (7/2)NkdT.
  ΔS = (7/2)Nkln(T₂/T₁).
  way 3:  ΔS = (7/2)ln(V₂/V₁).

Problem 4:

Consider a ring of radius, r, with 4 identical point particles of mass m interconnected by identical springs with spring constant k to their nearest neighbors. The particles move without friction.  The interconnected springs can be viewed as causing harmonic oscillations.
 
(a)  Determine the number of normal modes of oscillations, and establish the Lagrangian.
(b)  Find the frequencies for small oscillations and describe the corresponding eigen-vibrations.

Solution:

• Concepts:
  Coupled oscillations
• Reasoning:
  This is a one-dimensional problem.  We have 4 masses, so there will be 4 normal modes.
• Details of the calculation:
  (a)  L = ½m∑_i=1⁴(dq_i/dt)² - ½k∑_i=1⁴(q_i+1 - q_i)².
  
  Here the q_i are the clockwise displacements from equilibrium,
  and the index I is cyclic.  We can write L as

  L = ½∑_ij[T_ij(dq_i/dt)(dq_j/dt) - k_ijq_iq_j]  with T_ij = T_ji,  k_ij = k_ji. 
  
  Here T_ii = m, T_ij(i ≠ j) = 0,   i, j = 1, 2, 3, 4,  (cyclic).
  k_ii = 2k,  k_ij = -k if j = i ± 1,  k_ij = 0 otherwise.
  
  Equations of motion:  d/dt(∂L/∂(dq_i/dt)) - ∂L/∂q_i = 0.
  Solutions of the form q_j = Re(A_je^{iω t}) can be found.  For a system with n degrees of freedom, n characteristic frequencies ω_α can be found.  Some frequencies may be degenerate.  

  For a particular frequency ω_α we solve ∑_j[k_ij - ω_α²T_ij]A_jα = 0 to find the A_jα.   
  Since it is difficult to evaluate the determinant of a 4 x 4 matrix, we find the solutions of this system of coupled equations using physical insight.
  
  (b)  Equation of motion for mass i:
  (-kq_i-1 + 2kq_i - kq_i+1) + md²q_i/dt² = 0, i,j = 1, 2, 3, 4, (cyclic).

  Normal modes:

  1.)  ω₁ = 0, A₁ = A₂ = A₃ = A₄,  ∑_jk_ij = 2k - 2*k = 0.
  All other modes can have no net angular momentum.
  
  2.)  Assume A₁ and A₃ are fixed.  A₁ = A₃ = 0, A₂ = -A₄.
  The equation of motion for i = 2 is k₂₁A₁ + k₂₂A₂ + k₂₃A₃ - ω₂²T₂₂ A₂ = 0,
  2k  - ω₂²m = 0, ω₂² = 2k/m.
  
  3.)  Assume A₂ and A₄ are fixed.  A₁ = -A₃ = 0.
  For this mode ω₃² = 2k/m.  Mode 2 and 3 are degenerate.  Any linear combination of these modes is also a normal mode.  For example, the mode with A₂ = A₃ = 0, A₁ = A₄, , A₁ = - A₂ is a linear combination of modes 2 and 3 and is not linearly independent.
  
  4.)  Assume A₁ = A₃, A₂ = A₄, A₁ = - A₂.
  The equation of motion for i = 2 is k + 2k  + k - ω₂²m = 0, ω₄² = 4k/m.
  
  For the eigen-vibrations we have:
  1.)  q_i = vt.  (uniform rotation)
  2.)  q₁ = q₃ = 0,  q₂ = C₂cos(ω₂t + φ₂),  q₄ = -C₂cos(ω₂t + φ₂).
  3.)  q₂ = q₄ = 0,  q₃ = C₃cos(ω₃t + φ₃),  q₁ = -C₃cos(ω₃t + φ₃).
  4.)  q₁ = q₃ = C₄cos(ω₄t + φ₄),  q₂ = q = -C₄cos(ω₄t + φ₄).
  The C_i and φ_i depend on the initial conditions.  The mode with the highest frequency is mode 4 (non-degenerate) and the mode with the lowest frequency is mode 1 (non-degenerate).

Problem 5:

A metal block of mass m and specific heat c with temperature T_b is placed into the ocean which has temperature T.  Assume T_b > T.  What is the total change in the entropy, ΔS_total, for the system?  Express your answer in terms of m, c, T_b, and T.

Solution:

• Concepts:
  Change in entropy: dS = dQ/T
• Reasoning:
  We are asked calculate the change in entropy along a reversible path.
• Details of the calculation:
  For the metal block ΔS_block is negative.  ΔS_{block =} ∫_Tb^T dQ/T =  ∫_Tb^T mcdT/T = mc ln(T/T_b).
  For the ocean ΔS_ocean is positive.  ΔS_ocean =  Q/T =  mc(T_b - T)/T = mc(T_b/T - 1).
  ΔS_total = ΔS_block + ΔS_ocean = mc(T_b/T - ln(T_b/T) - 1),  ΔS_total is positive.