## Assignment 5, solutions

#### Problem 1:

A train has a whistle, which emits a 400 Hz sound.  You are stationary and you hear the whistle, but the pitch is 440 Hz.  How fast is train moving towards or away from you?

Solution:

• Concepts:
The Doppler shift
• Reasoning:
The frequency of the train's whistle is shifted, because the train is moving with respect to the observer.
• Details of the calculation:
The pitch is higher, so the train is moving towards you.  The train is a moving source.
Its speed relative to you is found from f = f0v/(v-vs).  We have
(v-vs) = f0v/f = (400/s)(340 m/s)/(440/s) = 309 m/s.
Therefore vs = 340 m/s - 309 m/s = 31 m/s = 69 mph.

#### Problem 2:

Two masses, 1 kg and 2 kg, are fixed horizontally to fixed side supports with springs as shown below. The masses are constrained to move along the horizontal line.  From their equilibrium position m1 is given a displacement L to the right, while m2 is held fixed.  At t = 0 they are released from rest.  Give the equation for the positions of m1 and m2 as a function of time.

Solution:

• Concepts:
Coupled oscillations, normal modes
• Reasoning:
Small oscillations:
L = ½∑ij[Tij(dqi/dt)(dqj/dt) - kijqiqj]  with Tij = Tji,  kij = kji
Solutions of the form qj = Re(Ajeiωt) can be found.
For a particular frequency ωα we solve
j[kij - ωαTij]A = 0
to find the A.  The most general solution for each coordinate qj is α sum of simple harmonic oscillations in all of the frequencies ωα.
qj = Re∑α(CαAexp(iωαt)).
• Details of the calculation:
The kinetic energy is T = ½[m1(dx1/dt)2 + m2(dx2/dt)2],
and the potential energy is U = ½)[k2(x2 - x1)2 + k1x12 + k3x22].
k1 = 2 N/m, k2 = 2 N/m,  k3 = 6 N/m.
U = ½[(k1 + k2)x12 + (k3 + k2)x22 - k2( x1x2 + x2x1)].
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 = m1 = 1, T 22 = m2 = 2,  Tij(i≠j) = 0,
k11= 5,  k22 = 8,  k12 = k21 = -2.
All numerical quantities are in SI units.
|kij - ω2Tij| = 0,  (k11 - ω2T11)(k22 - ω2T22) - k122 = 0,  (5 - ω2)(8 - 2ω2) - 4 = 0.
ω4 - 9ω2 + 18 = 0,  ω2= 9/2 ± 3/2.  ωa = 6, ωb = 3.
ωa:  -A1a - 2A2a = 0,   A1a = -2A2a.  Choose A1a = 2, A2a = -1.
ωa:  2A1b - 2A2b = 0,   A1b = A2b.  Choose A1b = A2b = 1.
Most general solution:
x1 = 2Cacos(ωat + φa) + Cbcos(ωbt + φb),  x2 = -Cacos(ωat + φa) + Cbcos(ωbt + φb).
dx1/dt = -2ωaCasin(ωat + φa) - ωbCbsins(ωbt + φb),
dx2/dt = ωaCasin(ωat + φa) - ωbCbsin(ωbt + φb).
Initial conditions at t = 0:
L = 2Cacos(φa) + Cbcos(φb),  0 = -Cacos(φa) + Cbcos(φb).  L = 3Cacos(φa)
0 = -2ωaCasin(φa) - ωbCbsins(φb),  0 = ωaCasin(φa) - ωbCbsin(φb).  φa = φb = 0.
Ca = Cb = L/3.
Positions of m1 and m2 as a function of time:
x1 = (2L/3)cos(√6 t) + (L/3)cos(√3 t),  x2 = -(L/3)cos(√6 t) + (L/3)cos(√3 t).

#### Problem 3:

Consider a "molecule" made up of three equal masses m connected by three equal springs with spring constant k.  The equilibrium position is an equilateral triangle.  Consider only motion in the plane of this triangle.  Find all the normal modes for motion in this plane.

Solution:

• Concepts:
Coupled oscillations, normal modes, symmetry
• Reasoning:
We are asked to find the normal modes of a system of 3 particles connected by springs.  The particles move in a plane.
m1 = m2 = m3 = m.
We have three masses moving in two dimensions, so there are six normal modes.
Three of of these modes will have ω = 0.  Two of these correspond to uniform translation of the CM along two perpendicular axes and all masses stay at their equilibrium position.  The other ω = 0 mode corresponds to uniform rotation about the CM.  The springs are stretched by an amount 2qcos(30o), so that 4kqcos2(30o) = 3kq = mrΩ2.  Ω is the angular speed of the uniform rotation.

If the equilibrium length of the springs is L then r = L/(2cos30o) + q = (2/3)Lcos30o + q.
The remaining modes keep the CM fixed and have no angular momentum about the CM.
One of the remaining modes is the "breathing mode".
Each mass moves along the line joining the CM and its equilibrium position.  The masses move in and out in unison.
T = (3/2)m(dq/dt)2,  U = (3/2)k(2qcos(30o))2 = (9/2)kq2, L = T - U.
Equation of motion.  3md2q/dt2 + 9kq = 0, ω2 = 3k/m.
• Details of the calculation:
The remaining two modes are degenerate.
Let us look at one of these.  When m1 and m2 both move up a distance χ, m3 moves down a distance 2χ.  At the same time m1 moves a distance -η and m2 moves a distance η along the x-direction.

T = 2*½m[(dη/dt)2 + (dχ/dt)2)] + ½m(d2χ/dt)2 = 3m(dχ/dt)2 + m(dη/dt)2.
U = ½k(2η)2 + 2*½kΔ2.  Δ = L' - L.
L'2 = (Lcos30o - 3χ)2 + (L/2 + η)2  = (L + Δ)2.
Assuming small displacements and keeping only terms to first order in the small quantities we have
-3χcos30o + η/2 = Δ.  -(3√3/2)χ + η/2 = Δ.
U = 2kη2 + k(27χ2/4 + η2/4 - (3√3/2)χη) = ½[(9/2)kη2 + (27/2)kχ2 - (3√3/2)k(χη + ηχ)].
T = ½∑ijTij(dqi/dt)(dqj/dt),  U = ½∑ijkijqiqj, with q1 = η and q2 = χ.
T11 = 2m,  T22 = 6m, Tij(i≠j) = 0,  k11 = 9k/2,  k22 = 27k/2,  k12 = k21 = -(3√3/2)k.
det(kij2Tij) = 0.
(9k/2 - 2mω2)(27k/2 - 6mω2) - 27k2/4 = 0.  ω4 - 4.5(k/m)ω2 + 4.5(k/m)2 = 0.
ω2 = 3 k/m or  ω2 = 1.5 k/m.
We rediscover the "breathing mode", with ω2 = 3 k/m,  χ = -η/√3.
For ω2 = 1.5 k/m we have (9/2 - 3)η - (3√3/2)χ = 0,  χ = η/√3.
As masses m1 and m2 move up a distance χ, there horizontal separation increases by 2η = 2√3x.

Another of the degenerate modes has masses m2 an m3 moving towards the CM distance χ, while m1 moves a distance 2χ towards the CM from the opposite direction.  The mode where m3 and m1 move in unison is a linear combination of the previous two independent degenerate modes.

#### Problem 4:

A space scientist proposes to measure the Gravitational constant G by locating a solid gold sphere of mass 8 x 104 kg and radius 1 m in a spaceship.  A hole is to be drilled through the diameter of the sphere and a small gold ball of mass 80 gram is released from rest at its surface so that it oscillates back and forth within the tube passing through the large sphere.
(a)  Calculate the period of the oscillatory motion of the particle, assuming that there is no friction present.
(b)  If air friction is present so that the particle returns to a radial distance of 99 cm rather than to 100 cm at the end of one full oscillation, obtain an estimate of the frictional force (assumed to be of form Ff = -bv), and the change in period caused by the frictional force.

Solution

• Concepts:
Newton's law of gravity, Newton's 2nd law, the principle of superposition, the underdamped harmonic oscillator
• Reasoning:
In part (a) the acceleration of the of the objects is due to the gravitational force.  The gravitational force on a object in the hole, a distance r from the center is in the -r direction and its magnitude is found using "Gauss' law".  [(∇∙Fg/m) = -4πGr, for a spherical mass distribution.  4πr2(Fg/m) = 4πGMinside, (Fg/m) = GMinside/r2.]
In part (b) the oscillator is underdamped, since it crosses the equilibrium position many times.
• Details of the calculation:
(a)  For an object of mass m in the hole F = (4/3)Gmπρr , with ρ = 3M/(4πR3 ).  M and R are the mass and radius of the large sphere, respectively.
For an object moving in the hole we therefore have F = -kr,  k = (4/3)Gmπρ.
The force on the object obeys Hooke's law.  The object will oscillate with angular frequency
ω = (k/m)½ = ((4/3)Gπρ)½ = 2.31*10-3/s.
Its period is T = 2π/ω = (3π/Gρ)½ = 2π(R3/GM)½ = 2720 s.
(b)  The equation of motion for the damped harmonic oscillator is
d2x/dt2 + 2bdx/dt + ω02x = 0.
The solution for underdamped motion is x(t) = A exp(-bt) cos(ω1t - δ), with ω12 = ω02 - b2.  Here ω0 is the angular frequency of an undamped oscillator with the same spring constant.  [We define T1 as the time between adjacent zero crossings, 2T1 as the "period", and ω1 = 2π/(2T1) as the "angular frequency" of the underdamped oscillator.]
After one period 2T1 the amplitude decreases from 1 m to 0.99 m.
exp(-b2T1) = 0.99,  -b2T1 = ln(0.99),  b2T1 = 0.01,
ω12 = ω02 - b2,  (2π)2/(2T1)2 = ω02 - b2,  (2π)2b2/(ω02 - b2) = 10-4.
b = 1.59*10-3 ω0 = 3.68*10-6/s.
(2T1 - T0)/T0 = (0.01/(1.59*10-3*2π) - 1) = 9.7*10-4.
ω12 ≈ ω02, 2T1 ≈ T0.
The period of the underdamped motion is practically equal to the period of the undamped motion since b << ω0.

#### Problem 5:

A set of coupled masses is constrained to move on a circular path of fixed radius.  The chain consists of four light masses m alternating with four heavy masses M, joined by identical springs with force constant k.
(a)  Find the (coupled) equations of motion for the ith light mass m and the ith heavy mass M.
(b)  Solve these equations of motion for the normal modes of the system.
(c)  Sketch the motion associated with each of the allowed frequencies

Solution:

• Concepts:
Coupled oscillations
• Reasoning:
Let qi denote the displacement from equilibrium.  Let a clockwise displacement be a positive displacement.
(a)  L = ½∑ij[Tij(dqi/dt)(dqj/dt) - kijqiqj]  with Tij = Tji,  kij = kji
Here Tii = mi, Tij(i ≠ j) = 0,   i, j = 1, 2, 3, 4, 5, 6, 7, 8, (cyclic).
mi = M if i = odd,  mi = m if i = even.
kii = 2k,  kij = -k if j = i ± 1,  kij = 0 otherwise.
The equations of motions are ∑j(Tij 2qj/∂t2 + kijqj) = 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 8 x 8 matrix, we find the solutions of this system of coupled equations using physical insight.  We need to find 8 normal modes.
• Details of the calculation:

(b)  1.)  ω1 = 0, A1 = A2 = A3 = A4 = A5 = A6, = A7 = A8,  ∑jkij = 2k - 2k = 0.
All other modes must have zero total or net angular momentum.

2.)  Assume all masses M are fixed.  A1 = A3 = A5 = A7 = 0.
The net force on each of these masses must be zero.
For the light masses we then have A2 = A6,  A4 = A8,  A2 = -A4.
To solve for ω22 we can use the equation of motion for mass 2.
k22A2 - ω22m A2 = 0,  2k - ω22m  = 0.
ω22 = 2k/m.

3.)  Assume all masses m are fixed.  A2 = A4 = A6 = A8 = 0.
The net force on each of these masses must be zero.
For the heavy masses we then have A1 = A5,  A3 = A7,  A1 = - A3.
To solve for ω32 we use the equation of motion for mass 3.
k33A3 - ω32M A3 = 0,
ω32 = 2k/M.

4.)  Assume only m1 and m5 are fixed.  A1 = A5 = 0.
How can this happen?  The net force acting on m1 and m5 must be zero.
A2 = A4,  A3 = aA2,  A6 = A8 = -A2,  A7 = -A3.
We can use the equations of motion for masses 1 and 3 to solve for ω42 and a.
mass 2:  k22A2 + k23A3 - mω2A2 = 0,  or  2kA2 - kaA2 - mω2A2 = 0,  2k - ka - mω2 = 0.
mass 3:  k32A2 + k33A3 + k34A4 - mω2A2 = 0,  or  -2kA2 + 2kaA2 - Mω2aA2 = 0,  -2k2 + 2kaA - Mω2aA = 0.
Eliminate ω2 and solve for a.  a4 = m/M ± ((M2 + m2)/M2)½.
Choose a4 = 1 - m/M + ((M2 + m2)/M2)½.  a4 > 0.
Insert to solve for ω42
ω42  = k/m + k/M - k(M2 + m2)½/(mM).

5.)  Assume m1 and m5 are fixed.  A1 = A5 = 0.
A2 = A4,  A3 = aA2,  A6 = A8 = -A2,  A7 = -A3.
Choose a5 = m/M - ((M2 + m2)/M2)½.  a5 <  0.
ω52 = k/m + k/M + k(M2 + m2)½/(mM).

6.)  Assume m2 and m6 are fixed.  A2 = A6= 0.
A3 = A5,  A4 = aA3,  A7 = A1 = -A3,  A8 = -A4.
(We are just relabeling i --> i + 1 and interchanging m an M in case 4.
a6 = 1 - M/m + ((M2 + m2)/m2)½.  a6 > 0.
ω62 = k/m + k/M - k(M2 + m2)½/(mM).
Note:  mode 4 and 6 are degenerate.  Any linear combination of those modes is also a normal mode.

7.)  Assume m2 and m6 are fixed.  A2 = A6= 0.
A3 = A5,  A4 = aA3,  A7 = A1 = -A3,  A8 = -A4.
Choose 7 = 1 - M/m - ((M2 + m2)/m2)½.  a7 < 0.
ω72 = k/m + k/M + k(M2 + m2)½/(mM).
Note:  mode 5 and 7 are degenerate.  Any linear combination of those modes is also a normal mode.

8.)   Assume none of the masses are fixed.
A1 = A3 = A5 = A7.   A2 = A4, = A6 = A8.  A2 = aA1.
All heavy masses oscillate against all light masses.
mass 1:  k11A1 + k12aA1 + k18aA1 - Mω2A1 = 0,  2k -  2ka - Mω2 = 0.
mass 2:  k21A1 + k22aA1 + k23A1 - mω2aA1 = 0,  -2k +  2ka - maω2 = 0.
Solve for a.  a8 = -M/m.
Solve for ω82.
ω82 = 2k/m + 2k/M

(c)  See diagrams.