Solution Manual for Introduction to Seismology, 3rd Edition

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Chapter 1
Exercise solutions
1. The radii of the Earth, Moon, and Sun are 6,371 km, 1,738 km, and 6.951 ×
105 km, respectively. From Figures 1.1, 1.5, and 1.6, make a rough estimate
of how long it takes a P -wave to traverse the diameter of each body.
Crude estimates based on mean velocity follow:
Earth: 2 × 6371 km / 11 km/s = 1160 s = 19.3 min
Moon: 2 × 1738 km / 7.6 km/s = 460 s = 7.7 min
Sun: 2 × 7 × 105 km / 250 km/s = 5600 s = 93 min
2. The P to S velocity ratio for most common rocks is about 1.7 (∼ √3). What
solid part of the Earth has a very different P /S velocity ratio? Hint: Look at
Figure 1.1.
The inner core, where the P /S velocity ratio is about 3.
3. Assume that the S velocity perturbations plotted at 150 km depth in Figure
1.7 extend throughout the uppermost 300 km of the mantle. Estimate how
many seconds earlier a vertically upgoing S-wave will arrive at a seismic station
in the middle of Canada, compared to a station in the eastern Pacific. Ignore
any topographic or crustal thickness differences between the sites; consider
only the integrated travel time difference through the upper mantle.
S velocity is about 4.5 km/s, so wave takes 300/4.5 = 67 s to go 300 km.
Canada is about 1.4% fast; eastern Pacific is about 1.8% slow. The difference
is 3.2% or 0.032.
Thus, S-wave will arrive earlier at central Canadian station by about (0.032)67
= 2.1 s . This may be an underestimate because the plot is saturated so the
perturbations could exceed these values.
4. Assuming that the P velocity in the ocean is 1.5 km/s, estimate the minimum
and maximum water depths shown in Figure 1.8. If the crustal P velocity is 5
km/s, what is the depth to the top of the magma chamber from the sea floor?
1

2 CHAPTER 1. EXERCISE SOLUTIONS
Minimum water depth = 3.5 × 1.5 / 2 = 2.62 km
Maximum water depth = 3.93 × 1.5 / 2 = 2.95 km
Top of magma chamber = (3.95 - 3.5) × 5 / 2 = 1.1 km below sea floor
5. Earthquake moment is defined as M0 = μDA, where μ is the shear modulus,
D is the average displacement on the fault, and A is the fault area that slipped.
The moment magnitude, MW , is defined as MW = 2
3 [log10 M0 − 9.1], where
the moment M0 is in N m.
(a) The moment of the 2004 Sumatra-Andaman earthquake has been esti-
mated to be about 1.0 × 1023 N m. What moment magnitude does this
correspond to? Assuming that the fault is horizontal, crudely estimate
the slip area from the image shown in Figure 1.9. Assuming that the
shear modulus μ = 3.0 × 1010 N/m2, then compute the average displace-
ment on the fault.
From the equation given, we have
MW = 2
3
[
log10(1023) − 9.1
]
= 2
3 (23 − 9.1) = 9.27
The moment magnitude is 9.27 .
The slip area is about 200 km wide and 1200 km long, thus A = 240,000
km2 = 2.4 × 1011 m2, and we have
D = M0
μA = 1023
3.0 × 1010 × 2.4 × 1011 = 100
7.2 = 14 m
The average displacement is about 14 m .
(b) A Hollywood director wants to make a movie about a devastating magni-
tude 10 earthquake in California with 40 m of slip (displacement) along
the San Andreas Fault. Given that the crust is about 35 km thick in
California, how scientifically plausible is this scenario? Hint: you may
assume that the shear modulus μ = 3 × 1010 N/m2
We also need to know the length of the San Andreas Fault (SAF) that
ruptures, which is not given, so some research is needed. The SAF is
about 1200 km long, which is also roughly the N-S extent of California.
So 1200 km is a reasonable upper bound on the maximum SAF rupture
length. If we assume that the rupture extends throughout the crust (an
assumption implied by the question wording, but which is probably an
overestimate for typical SAF earthquakes), then the fault area is A =
1.2 × 106 × 3.5 × 104 m2 = 4.2 × 1010 m. We then have
M0 = μDA = 3 × 1010 · 40 · 4.2 × 1010 ≈ 5 × 1022 N m
The moment magnitude is
MW = 2
3
[
log10(5 × 1022) − 9.1
]
= 2
3 (22.7 − 9.1) = 9.07

3
The moment magnitude is about 9.1 , which is much less than 10. Thus
the director’s idea for a magnitude 10 earthquake is not plausible for
the SAF in California.
6. The 2004 Sumatra earthquake lasted about 10 minutes and radiated about
2 ×1017 joules of seismic energy. Compute the corresponding power output
in terrawatts (TW). Then do some research on the web and compare this
power to: (a) average rate of electricity consumption in the United States, (b)
average dissipation rate of tidal energy in the world’s oceans, and (c) total
heat flow out of the Earth. Note that the total energy release (including heat
generated on the fault, etc.) of the Sumatra earthquake may be significantly
greater than the seismically radiated energy. This is discussed in Chapter 9.
A watt is one joule per second. 10 minutes is 600 s. Thus, the radiated power
during the earthquake was 2×1017/600 = 3.33×1014 watts. A terrawatt (TW)
is 1012 watts, so the radiated power was about 330 TW . For comparison,
(a) 2017 US power consumption was about 4000 TW hours (source Wikipedia).
Thus the average power usage is 4000/(365 · 24) = 0.46 TW .
(b) Earth’s tidal energy dissipation is about 3.7 TW , most of which occurs
in the oceans (see p. 106–108 of Stacey and Davis, Physics of the Earth, 2008).
The Wikipedia article on tidal acceleration cites 3.3 TW.
(c) Earth’s integrated heat flux is about 44 TW (p. 337, Stacey and Davis,
Physics of the Earth, 2008). The Wikipedia article on Earth’s energy budget
cites 47 TW from Davies and Davies, Earth’s surface heat flux, Solid Earth,
1(1), 524, 2010.

Chapter 2
Exercise solutions
1. Assume that the horizontal components of the 2-D stress tensor are
τττ =
[ τxx τxy
τyx τyy
]
=
[ −30 −20
−20 −40
]
MPa
(a) Compute the normal and shear stresses on a fault that strikes 10◦ east
of north.
cos 10◦ = 0.9848, sin 10◦ = 0.1736, thus fault parallel vector ˆf = (0.1736,
0.9848), fault normal vector ˆn = (0.9848, -0.1736). The traction on the
fault plane is given by
t(ˆn) = τττ ˆn =
[ −30 −20
−20 −40
] [ 0.9848
−0.1736
]
=
[ −26.07
−12.75
]
MPa.
and
tN = t · ˆn = (−26.07, −12.75) · (0.9848, −0.1736) = -23.46 MPa
tS = t · ˆf = (−26.07, −12.75) · (0.1736, 0.9848) = -17.08 MPa
The fault normal compression is 23.46 MPa. The shear stress is 17.08
MPa.
(b) Compute the principal stresses, and give the azimuths (in degrees east of
north) of the maximum and minimum compressional stress axes.
λ1 = −55.61 MPa, 38◦ E of N
λ2 = −14.38 MPa, 128◦ E of N
(solution computed using Matlab)
2. The principal stress axes for a 2-D geometry are oriented at N45◦E and
N135◦E, corresponding to principal stresses of -15 and -10 MPa. What are the
4 components of the 2-D stress tensor in a (x = east, y = north) coordinate
system?
The eigenvector matrix is
N =
[ 1/√2 1/√2
1/√2 −1/√2
]
1

2 CHAPTER 2. EXERCISE SOLUTIONS
Note that N = N T = N −1. The principal (diagonalized) stress tensor is
τττ P =
[ −15 0
0 −10
]
which we can rotate to the stress tensor in E-N coordinates as
τττ = Nτττ P N T =
[ −12.5 −2.5
−2.5 −12.5
]
MPa
3. Figure 2.5 shows a vertical-component seismogram of the 1989 Loma Prieta
earthquake recorded in Finland.
(a) Estimate the dominant period, T , of the surface wave from its first ten
cycles. Then compute the frequency f = 1/T .
There are about 8 peaks in 200 s so the period, T , is about 25 s . The
frequency, f , is 1/25 = 0.04 Hz .
(b) Make an estimate of the maximum surface-wave strain recorded at this
site. Hints: 1 micron = 10−6 m, assume the Rayleigh surface wave phase
velocity at the dominant period is 3.9 km/s, remember that strain is ∂uz /∂x,
Table 3.1 may be helpful.
The amplitude A = 300 microns = 3 × 10−4 m. The wavelength Λ = cT =
(3.9 km/s)(25 s) = 97.5 km = about 100 km = 1 ×105 m. The wavenumber
k = 2π/Λ = 0.0628/km = 6.28 × 10−5/m. We can approximate displacement
as a harmonic wave as ux = A sin(kx). Strain = ∂ux/∂x = kA cos(kx) so the
maximum strain occurs when cos = 1 and is Ak = (3 × 10−4)(6.28 × 10−5) =
1.9 × 10−8 .
Note that this value is halved if we express this in terms of the strain tensor,
i.e.,
emax ≈
[ 0 10−8
10−8 0
]
4. Using Equations (2.4), (2.23), and (2.30), show that the principal stress axes
always coincide with the principal strain axes for isotropic media. In other
words, show that if x is an eigenvector of e, then it is also an eigenvector of τττ .
We can show that the principal strain axes coincide if we prove that if x is an
eigenvector of the strain tensor e then it also must be an eigenvector of the
stress tensor τ , that is if
eij xj = αxi
then
τij xj = α′xi

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3
Using the isotropic stress/strain relation τij = λekkδij + 2μeij (equation 2.29),
we have
τij xj = λekkδij xj + 2μeij xj
= λekkxi + 2μαxi
= (λekk + 2μα)xi
= α′xi (2.1)
where α′ = λekk + 2μα. This completes the proof.
5. From Equations (2.34) and (2.35) derive expressions for the Lam´e parameters
in terms of the seismic velocities and density.
μ = ρβ2
λ = ρ(α2 − 2β2)
6. Seismic observations of S velocity can be directly related to the shear modulus
μ. However, P velocity is a function of both the shear and bulk moduli. For
this reason, sometimes seismologists will compute the bulk sound speed, defined
as:
Vc =
√ κ
ρ (2.2)
which isolates the sensitivity to the bulk modulus κ.
(a) Derive an equation for Vc in terms of the P velocity, α, and the S velocity,
β.
We have
Vc =
√ κ
ρ
We need to solve for κ in terms of α and β. We have from the P and S velocity
definitions:
λ + 2μ = ρα2
μ = ρβ2
λ = ρ(α2 − 2β2)
The bulk modulus κ = λ + 2/3μ and thus
κ = ρ(α2 − 2ρβ2 + 2/3ρβ2)
= ρ(α2 − 4/3β2)
and thus
Vc =
√
α2 − 4
3 β2

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4 CHAPTER 2. EXERCISE SOLUTIONS
(b) For the specific case of a Poisson solid, express Vc as a fraction of the P
velocity.
For a Poisson solid, α = √3β, so β2 = 1
3 α2. Substituting into our result from
(a), we have
Vc =
√
α2 − 4
9 α2 = α
√ 9
9 − 4
9 =
√5
3 α = 0.745 α
7. What is the P /S velocity ratio for a rock with a Poisson’s ratio of 0.30?
Starting with equation (2.36), we have
σ = (α/β)2 − 2
2(α/β)2 − 2
2σ(α/β)2 − 2σ = (α/β)2 − 2
2 − 2σ = (α/β)2 − 2σ(α/β)2 = (α/β)2[1 − 2σ]
α
β =
√
2 − 2σ
1 − 2σ
α
β =
√
2 − 0.6
1 − 0.6 =
√ 1.4
0.4 = √3.5 = 1.87 for σ = 0.30
8. A sample of granite in the laboratory is observed to have a P velocity of 5.5
km/s and a density of 2.6 Mg/m3. Assuming it is a Poisson solid, obtain values
for the Lam´e parameters, Young’s modulus, and the bulk modulus. Express
your answers in pascals.
For a Poisson solid, β = α/√3 = 3.1754 km/s for α = 5.5 km/s. From Exercise
2.5, we have
λ = ρ(α2 − 2β2) = 2.6(5.52 − 2(3.1754)2)
λ = 26.22 Mg
m3
km2
s2
106m2
km2
103kg
Mg
λ = 26.22 × 109 kg
m s2
λ = 26.22 GPa
and
μ = ρβ2 = 2.6(3.1754)2
μ = 26.22 GPa
Note that λ = μ as should be the case for a Poisson solid. The Young’s
modulus is given by
E = (3λ + 2μ)μ
λ + μ = 2.5μ since λ = μ

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5
E = 65.55 GPa
The bulk modulus is given by
κ = λ + 2
3 μ = 5
3 μ
κ = 43.7 GPa
Summarizing, we have: λ = μ = 26.22 GPA, E = 65.55 GPa, κ = 43.7 GPa .
9. Using values from the PREM model (Appendix 1), compute values for the
bulk modulus on both sides of (a) the core–mantle boundary (CMB) and (b)
the inner-core boundary (ICB). Express your answers in pascals.
The bulk modulus is given by
κ = λ + 2
3 μ
From Exercise 2.5 we have μ = ρβ2 and λ = ρ(α2 − 2β2) and thus
κ = ρ(α2 − 4
3 β2)
From the PREM model in the Appendix, we can thus construct a table with
the κ values:
Vp Vs den kappa
CMB top 13.72 7.26 5.57 657 Gpa
CMB bot 8.06 0 9.90 643 GPa
ICB top 10.36 0 12.17 1306 GPa
ICB bot 11.03 3.5 12.76 1344 GPa
10. Figure 2.6 shows surface displacement rates as a function of distance from the
San Andreas Fault in California.
(a) Consider this as a 2-D problem with the x-axis perpendicular to the fault
and the y-axis parallel to the fault. From these data, estimate the yearly
strain (e) and rotation (ΩΩΩ) tensors for a point on the fault. Express your
answers as 2 × 2 matrices.
Each year, ∂y/∂x ≈ 43 mm / 50 km = (43 × 10−3 m) / (50 × 103 m) =
0.86×10−6 = 8.6 × 10−7/yr . This is the only non-zero partial derivative
in the J matrix. We thus have
J =
[ 0 θ
0 0
]
= e + Ω =
[ 0 θ/2
θ/2 0
]
+
[ 0 θ/2
−θ/2 0
]
where θ = ∂y/∂x = 8.6 × 10−7 and thus
e =
[ 0 4.3 × 10−7
4.3 × 10−7 0
]

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6 CHAPTER 2. EXERCISE SOLUTIONS
and
Ω =
[ 0 −4.3 × 10−7
4.3 × 10−7 0
]
(b) Assuming the crustal shear modulus is 27 GPa, compute the yearly
change in the stress tensor. Express your answer as a 2 × 2 matrix with
appropriate units.
τττ =
[ 0 2.3 × 104
2.3 × 104 0
]
Pa
(c) If the crustal shear modulus is 27 GPa, what is the shear stress across
the fault after 200 years, assuming zero initial shear stress?
4.6 × 106 Pa (4.6 MPa)
(d) If large earthquakes occur every 200 years and release all of the distributed
strain by movement along the fault, what, if anything, can be inferred
about the absolute level of shear stress?
The absolute shear stress across the fault before the earthquake must be
at least 4.6 MPa, but it could also be much bigger, depending upon how
large the stress drop is relative to the absolute stress. The involves the
frictional properties of the fault, and is discussed in Chapter 9. (Note:
this question is to get students to think or do research—the answer is not
in the chapter)
(e) What, if anything, can be learned about the fault from the observation
that most of the deformation occurs within a zone less than 50 km wide?
Vertical strike-slip faults like the San Andreas are typically modeled as
a locked zone (where the earthquakes occur) above a creeping zone at
depth. The width of the deformation zone is proportional in some sense
to the depth of the locked zone. (Note: this question is to get students
to think or do research—the answer is not in the chapter.)
11. Do some research on the observed density of the Sun. Are the high sound
velocities in the Sun (see Fig. 1.6) compared to Earth’s P velocities caused
primarily by low solar densities compared to the Earth, a higher bulk modulus
or some combination of these factors?
The average solar density is about 1.4 g/cc, compared to 4 to 5 g/cc in Earth’s
mantle. This would make solar velocities higher by about a factor of √3 ∼ 1.7.
But the average solar P velocity is about 50 times higher than in Earth, so the
more important difference is that the bulk modulus, κ, is much higher for the
Sun. Students can make this more complicated by trying to take into account
the velocity depth dependence, but I was just looking for a rough estimate
based on average properties.

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7
12. The University of California, San Diego, operates the Pi˜non Flat Observatory
(PFO) in the mountains northeast of San Diego (near Anza). Instruments
include high-quality strain meters for measuring crustal deformation.
(a) Assume, at 5 km depth beneath PFO, the seismic velocities are α = 6
km/s and β = 3.5 km/s and the density is ρ = 2.7 Mg/m3. Compute
values for the Lam´e parameters, λ and μ, from these numbers. Express
your answer in units of pascals.
From the solution to Exercise 5, we have
λ = ρ(α2 − 2β2) = 2.7(6.02 − 2(3.5)2)
λ = 31.05 Mg
m3
km2
s2
106m2
km2
103kg
Mg
λ = 31.05 × 109 kg
m s2
λ = 31.05 GPa
μ = ρβ2 = 2.7(3.5)2
μ = 33.07 GPa
Summarizing, λ = 31.05 GPa, μ = 33.07 GPa .
(b) Following the 1992 Landers earthquake (MS = 7.3), located in south-
ern California 80 km north of PFO (Fig. 2.7), the PFO strain meters
measured a large static change in strain compared to values before the
event. Horizontal components of the strain tensor changed by the follow-
ing amounts: e11 = −0.26 × 10−6, e22 = 0.92 × 10−6, e12 = −0.69 × 10−6.
In this notation 1 is east, 2 is north, and extension is positive. You may
assume that this strain change occurred instantaneously at the time of
the event. Assuming these strain values are also accurate at depth, use
the result you obtained in part (a) to determine the change in stress
due to the Landers earthquake at 5 km, that is, compute the change in
τ11, τ22, and τ12. Treat this as a two-dimensional problem by assuming
there is no strain in the vertical direction and no depth dependence of
the strain.
From (2.30), we have
τττ =
[ λ tr[e] + 2μe11 2μe12
2μe21 λ tr[e] + 2μe22
]
Given the Landers strain changes of e11 = −0.26×10−6, e22 = 0.92×10−6,
e12 = −0.69 × 10−6, then tr[e] = 0.66 × 10−6 and
τττ =
[ 3300 −45600
−45600 81300
]
Pa
or τ11 = 3.3 × 103 Pa, τ22 = 8.13 × 104 Pa, and τ12 = −4.56 × 104 Pa
(c) Compute the orientations of the principal strain axes (horizontal) for the
response at PFO to the Landers event. Express your answers as azimuths
(degrees east of north).

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8 CHAPTER 2. EXERCISE SOLUTIONS
The strain tensor is
[ −0.26 −0.69
−0.69 0.92
]
× 10−6
The solution to the eigenvalue problem for this matrix gives orientations
of principal strains as N65◦E and N155◦E (or N25◦W) .
(d) A steady long-term change in strain at PFO has been observed to occur in
which the changes in one year are: e11 = 0.101×10−6, e22 = −0.02×10−6,
e12 = 0.005 × 10−6. Note that the long-term strain change is close to sim-
ple E–W extension. Assuming that this strain rate has occurred steadily
for the last 1,000 years, from an initial state of zero stress, compute the
components of the stress tensor at 5 km depth. (Note: This is proba-
bly not a very realistic assumption!) Don’t include the large hydrostatic
component of stress at 5 km depth.
From (2.30), we have
τττ =
[ λ tr[e] + 2μe11 2μe12
2μe21 λ tr[e] + 2μe22
]
Given the 1000 year strain changes of e11 = 0.101 × 10−3, e22 = −0.02 ×
10−3, e12 = 0.005 × 10−3, then tr[e] = 0.081 × 10−3 and using λ = 31.05
GPa and μ = 33.07 GPa from 2.6a we have
τττ =
[ 9.20 0.33
0.33 1.19
]
MPa
(e) Farmer Bob owns a 1 km2 plot of land near PFO that he has fenced and
surveyed with great precision. How much land does Farmer Bob gain or
lose each year? How much did he gain or lose as a result of the Landers
earthquake? Express your answers in m2.
Use tr[e] as a measure of area change. Each year, tr[e] = 0.081 × 10−6 so
Farmer Bob gains 0.081 m2 of land each year .
For Landers tr[e] = 0.66 × 10−6 so
Farmer Bob gained 0.66 m2 of land following the Landers earthquake .
(f) (COMPUTER) Write a computer program that computes the stress across
vertical faults at azimuths between 0 and 170 degrees (east from north,
at 10 degree increments). For the stress tensors that you calculated in (b)
and (d), make a table that lists the fault azimuth and the correspond-
ing shear stress and normal stress across the fault (for Landers these
are the stress changes, not absolute stresses). At what azimuths are the
maximum shear stresses for each case?

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9
(g) (COMPUTER) Several studies (e.g., Stein et al., 1992, 1994; Harris and
Simpson, 1992; Harris et al., 1995; Stein, 1999; Harris, 2002) have mod-
eled the spatial distribution of events following large earthquakes by as-
suming that the likelihood of earthquake rupture along a fault is related
to the Coulomb failure function (CFF). Ignoring the effect of pore fluid
pressure, the change in CFF may be expressed as:
∆CFF = ∆|τs| + μs∆τn,
where τs is the shear stress (traction), τn is the normal stress, and μs is
the coefficient of static friction (don’t confuse this with the shear mod-
ulus!). Note that CFF increases as the shear stress increases, and as
the compressional stress on the fault is reduced (recall in our sign con-
vention that extensional stresses are positive and compressional stresses
are negative). Assume that μs = 0.2 and modify your computer pro-
gram to compute ∆CFF for each fault orientation. Make a table of the
yearly change in ∆CFF due to the long-term strain change at each fault
azimuth.
(h) (COMPUTER) Now assume that the faults will fail when their long-
term CFF reaches some critical threshold value. The change in time to
the next earthquake may be expressed as
∆t = CFF1000+L − CFF1000
CFFa
,
where CFFa is the annual change in CFF, CFF1000 is the thousand year
change in CFF, and CFF1000+L is the thousand year + Landers change
in CFF (note that CFF1000+L 6 = CFF1000 + CFFL). Compute the effect
of the Landers earthquake in terms of advancing or retarding the time
until the next earthquake for each fault orientation. Express your answer
in years, using the sign convention of positive time for advancement of
the next earthquake and negative time for retardation.

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10 CHAPTER 2. EXERCISE SOLUTIONS
Results for parts f –h are given in the following table:
----Landers---- Annual (*1000)
azi tau-s tau-n tau-s tau-n CFF-L CFF-a dt L/a
0 -45643.5 3294.0 330.7 9196.2 46302.3 2170.0 -20.7 21.3
10 -56239.4 21258.7 1679.6 8841.7 60491.1 3447.9 -15.1 17.5
20 -60052.0 41764.0 2825.9 8047.3 68404.8 4435.3 -11.7 15.4
30 -56621.4 62336.7 3631.3 6908.7 69088.8 5013.0 -8.8 13.8
40 -46361.5 80495.3 3998.7 5563.4 62460.5 5111.4 -5.9 12.2
50 -30509.7 94049.8 3883.8 4173.4 49319.6 4718.5 -2.5 10.5
60 -10977.9 101365.2 3300.5 2906.6 31251.0 3881.9 2.4 8.1
70 9877.9 101559.1 2319.1 1915.8 30189.8 2702.3 11.2 11.2
80 29542.3 94608.3 1058.0 1320.3 48464.0 1322.0 36.7 36.7
90 45643.5 81351.0 -330.8 1192.0 61913.7 569.2 -51.6 108.8
100 56239.4 63386.3 -1679.6 1546.5 68916.6 1988.9 -21.9 34.7
110 60052.0 42881.0 -2825.9 2341.0 68628.2 3294.0 -15.6 20.8
120 56621.4 22308.3 -3631.3 3479.5 61083.1 4327.2 -12.1 14.1
130 46361.5 4149.6 -3998.7 4824.9 47191.4 4963.7 -9.2 9.5
140 30509.6 -9404.8 -3883.8 6214.8 28628.7 5126.8 -6.3 5.6
150 10977.9 -16720.2 -3300.5 7481.6 7633.9 4796.8 -3.0 1.6
160 -9877.9 -16914.1 -2319.1 8472.5 6495.1 4013.6 1.6 1.6
170 -29542.3 -9963.3 -1058.0 9068.0 27549.7 2871.6 9.6 9.6
(i) No increase in seismicity (small earthquake activity) has been observed
near PFO following the Landers event. Does this say anything about the
validity of the threshold CFF model?
Maybe! (another food for thought question to get students thinking, with
no simple answer)

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Chapter 3
Exercise solutions
1. Period T is to angular frequency ω as wavelength Λ is to: (a) wavenumber k,
(b) velocity c, (c) frequency f , (d) time t, (e) none of the above.
(a) wave number
2. Figure 3.7 plots a harmonic plane wave at t = 0, traveling in the x direction
at 5 km/s. (a) Write down an equation for this wave using the sine function
that describes displacement, u, as a function of x and t.
We have wavelength Λ = 8000 m, velocity c = 5000 m/s, thus f = c/Λ = 5/8
Hz. We then have ω = 2πf = 5π/4 and k = 2π/Λ = π/4000. Thus we have
for a wave traveling in the +x direction:
u = 0.04 sin[(π/4000)x − (5π/4)t]
= 0.04 sin[π( x
4000 − 5
4 t)]
= 0.04 sin[0.000785 x − 3.927 t] (t in seconds, x in meters)
The correct signs of the x and t terms in this equation can be verified by
inspecting the figure and considering the cases t = 0 and x = 0. Alternatively,
we could reverse the signs for t and x by adding a phase term π, i.e.,
u = 0.04 sin
[ 5π
4 t − π
4000 x + π
]
(b) Write the corresponding equation using the cosine function.
u = 0.04 cos[0.000785 x − 3.927 t − π/2] (t in seconds, x in meters)
(c) What is the displacement at x = 6 km and t = 30 s?
Zero, since 6000/4000 - 5(30)/4 = 1.5 - 37.5 = -36, sin −36π = 0.
(d) What is the maximum strain for this wave?
1

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2 CHAPTER 3. EXERCISE SOLUTIONS
The strain is ∂u/∂x = (0.000785)(0.04) cos[0.000785 x − 3.927 t] so the maxi-
mum is (0.000785)(0.04) = 0.0000314 = 3.14 × 10−5 .
If this is a shear wave, the 2-D strain tensor would have half this value as the
off-diagonal terms.
3. Consider two types of monochromatic plane waves propagating in the x di-
rection in a uniform medium: (a) P -wave in which ux = A sin(ωt − kx), (b)
S-wave with displacements in the y direction, i.e., uy = A sin(ωt − kx). For
each case, derive expressions for the nonzero components of the stress tensor.
Refer to 2.17 to get the components of the strain tensor; then use 2.30 to
obtain the stress components. Hint: Look at Example 3.4.1.
(a)
From (2.17) the only nonzero component of the strain tensor is
e11 = ∂ux
∂x = −kA cos(ωt − kx)
Then from (2.29) we have
τ11 = −(λ + 2μ)kA cos(ωt − kx)
τ22 = τ33 = −λkA cos(ωt − kx)
(b)
We are given uy = A sin(ωt − kx) and are asked to solve for the nonzero
components of the stress tensor. From (2.17) the only nonzero components of
the strain tensor are
e12 = e21 = 1
2
∂uy
∂x = − 1
2 kA cos(ωt − kx)
Then from (2.30) we have
τ12 = τ21 = −μkA cos(ωt − kx)
4. Assume harmonic P -waves are traveling through a solid with α = 10 km/s. If
the maximum strain is 10−8, what is the maximum particle displacement for
waves with periods of: (a) 1 s, (b) 10 s, (c) 100 s?
For harmonic P -wave traveling in x direction we have
ux = A sin(ωt − kx)
The only nonzero strain component is
e11 = ∂ux
∂x = −kA cos(ωt − kx)

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