## Assignment 8, solutions

#### 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 xi are the displacements from the equilibrium positions.  We use the xi as our generalized coordinates qi.
The Lagrangian is L = T - U.  This can be put into the form
L = ½∑ij[Tij(dqi/dt)(dqj/dt) - kijqiqj]  with Tij = Tji,  kij = kji
Here T11 = T33 = m, T22 = 2m, Tij(i≠j) = 0,
k11 = k33 = k, k22 = 2k, k12 = k21 = k23 = k32 = -k, k13 = k31 = 0.
Solutions of the form xj = Re(Ajeiω 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(kij2Tij) = 0.  For a system with n degrees of freedom, n characteristic frequencies ωa 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[Tij(dqi/dt)(dqj/dt) - kijqiqj]  with Tij = Tji,  kij = kji
Solutions of the form qj = Re(Ajeiωt) can be found.
We can find the ω2 from det(kij - ω2Tij) = 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[Tij(dΦi/dt)(dΦj/dt),  U = ½∑ijkijΦiΦj.
T11 = ma2(α + 2), T22 = 2ma2, T33 = ma2, T12 = T21 = 2ma2, T13 = T31 = T23 = T32 = ma2.
k11 = mga(α + 2), k22 = 2mga, k33 = mga, kij(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 = ∫if dS = ∫if dQr/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(dqi/dt)2 - ½k∑i=14(qi+1 - qi)2.

Here the qi are the clockwise displacements from equilibrium,
and the index I is cyclic.  We can write L as

L = ½∑ij[Tij(dqi/dt)(dqj/dt) - kijqiqj]  with Tij = Tji,  kij = kji

Here Tii = m, Tij(i ≠ j) = 0,   i, j = 1, 2, 3, 4,  (cyclic).
kii = 2k,  kij = -k if j = i ± 1,  kij = 0 otherwise.

Equations of motion:  d/dt(∂L/∂(dqi/dt)) - ∂L/∂qi = 0.
Solutions of the form qj = Re(Ajeiω t) can be found.  For a system with n degrees of freedom, n characteristic frequencies ωα can be found.  Some frequencies may be degenerate.

For a particular frequency ωα we solve ∑j[kij - ωα2Tij]A = 0 to find the A.
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:
(-kqi-1 + 2kqi - kqi+1) + md2qi/dt2 = 0, i,j = 1, 2, 3, 4, (cyclic).

Normal modes:

1.)  ω1 = 0, A1 = A2 = A3 = A4,  ∑jkij = 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.)  qi = 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 Ci and φi 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 Tb is placed into the ocean which has temperature T.  Assume Tb > T.  What is the total change in the entropy, ΔStotal, for the system?  Express your answer in terms of m, c, Tb, 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 ΔSblock is negative.  ΔSblock = TbT dQ/T =  TbT mcdT/T = mc ln(T/Tb).
For the ocean ΔSocean is positive.  ΔSocean =  Q/T =  mc(Tb - T)/T = mc(Tb/T - 1).
ΔStotal = ΔSblock + ΔSocean = mc(Tb/T - ln(Tb/T) - 1),  ΔStotal is positive.