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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 × 10 5 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 × 10 5 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

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    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 M 0 = 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 , M W , is defined as M W = 2 3 [log 10 M 0 − 9 . 1], where the moment M 0 is in N m. (a) The moment of the 2004 Sumatra-Andaman earthquake has been esti- mated to be about 1 . 0 × 10 23 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 × 10 10 N/m 2 , then compute the average displace- ment on the fault. From the equation given, we have M W = 2 3 [ log 10 (10 23 ) − 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 km 2 = 2 . 4 × 10 11 m 2 , and we have D = M 0 μA = 10 23 3 . 0 × 10 10 × 2 . 4 × 10 11 = 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 × 10 10 N/m 2 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 × 10 6 × 3 . 5 × 10 4 m 2 = 4 . 2 × 10 10 m. We then have M 0 = μDA = 3 × 10 10 · 40 · 4 . 2 × 10 1 0 ≈ 5 × 10 22 N m The moment magnitude is M W = 2 3 [ log 10 (5 × 10 22 ) − 9 . 1 ] = 2 3 (22 . 7 − 9 . 1) = 9 . 07

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    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 × 10 17 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 × 10 17 / 600 = 3 . 33 × 10 14 watts. A terrawatt (TW) is 10 12 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.

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    Chapter 2 Exercise solutions 1. Assume that the horizontal components of the 2-D stress tensor are τ τ τ = [ τ xx τ xy τ yx τ yy ] = [ − 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 ˆ f = (0.1736, 0.9848), fault normal vector ˆ n = (0.9848, -0.1736). The traction on the fault plane is given by t ( ˆ n ) = τ τ τ ˆ n = [ − 30 − 20 − 20 − 40 ] [ 0 . 9848 − 0 . 1736 ] = [ − 26 . 07 − 12 . 75 ] MPa . and t N = t · ˆ n = ( − 26 . 07 , − 12 . 75) · (0 . 9848 , − 0 . 1736) = -23.46 MPa t S = t · ˆ f = ( − 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. λ 1 = − 55 . 61 MPa, 38 ◦ E of N λ 2 = − 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

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    2 CHAPTER 2. EXERCISE SOLUTIONS Note that N = N T = N − 1 . The principal (diagonalized) stress tensor is τ τ τ P = [ − 15 0 0 − 10 ] which we can rotate to the stress tensor in E-N coordinates as τ τ τ = Nτ τ τ 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 ∂u z /∂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 × 10 5 m. The wavenumber k = 2 π/ Λ = 0 . 0628/km = 6 . 28 × 10 − 5 /m. We can approximate displacement as a harmonic wave as u x = A sin( kx ). Strain = ∂u x /∂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., e max ≈ [ 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 τ τ τ . 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 e ij x j = αx i then τ ij x j = α ′ x i

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    3 Using the isotropic stress/strain relation τ ij = λe kk δ ij + 2 μe ij (equation 2.29), we have τ ij x j = λe kk δ ij x j + 2 μe ij x j = λe kk x i + 2 μαx i = ( λe kk + 2 μα ) x i = α ′ x i (2.1) where α ′ = λe kk + 2 μα . 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 = ρβ 2 λ = ρ ( α 2 − 2 β 2 ) 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: V c = √ κ ρ (2.2) which isolates the sensitivity to the bulk modulus κ . (a) Derive an equation for V c in terms of the P velocity, α , and the S velocity, β . We have V c = √ κ ρ We need to solve for κ in terms of α and β . We have from the P and S velocity definitions: λ + 2 μ = ρα 2 μ = ρβ 2 λ = ρ ( α 2 − 2 β 2 ) The bulk modulus κ = λ + 2 / 3 μ and thus κ = ρ ( α 2 − 2 ρβ 2 + 2 / 3 ρβ 2 ) = ρ ( α 2 − 4 / 3 β 2 ) and thus V c = √ α 2 − 4 3 β 2

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    4 CHAPTER 2. EXERCISE SOLUTIONS (b) For the specific case of a Poisson solid, express V c as a fraction of the P velocity. For a Poisson solid, α = √ 3 β , so β 2 = 1 3 α 2 . Substituting into our result from (a), we have V c = √ α 2 − 4 9 α 2 = 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 2 σ ( α/β ) 2 − 2 σ = ( α/β ) 2 − 2 2 − 2 σ = ( α/β ) 2 − 2 σ ( α/β ) 2 = ( α/β ) 2 [1 − 2 σ ] α β = √ 2 − 2 σ 1 − 2 σ 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/m 3 . 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 ( α 2 − 2 β 2 ) = 2 . 6(5 . 5 2 − 2(3 . 1754) 2 ) λ = 26 . 22 Mg m 3 km 2 s 2 10 6 m 2 km 2 10 3 kg Mg λ = 26 . 22 × 10 9 kg m s 2 λ = 26 . 22 GPa and μ = ρβ 2 = 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 = (3 λ + 2 μ ) μ λ + μ = 2 . 5 μ 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 = ρβ 2 and λ = ρ ( α 2 − 2 β 2 ) and thus κ = ρ ( α 2 − 4 3 β 2 ) 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 (Ω Ω Ω) 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 × 10 3 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. τ τ τ = [ 0 2 . 3 × 10 4 2 . 3 × 10 4 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 × 10 6 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/m 3 . 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 ( α 2 − 2 β 2 ) = 2 . 7(6 . 0 2 − 2(3 . 5) 2 ) λ = 31 . 05 Mg m 3 km 2 s 2 10 6 m 2 km 2 10 3 kg Mg λ = 31 . 05 × 10 9 kg m s 2 λ = 31 . 05 GPa μ = ρβ 2 = 2 . 7(3 . 5) 2 μ = 33 . 07 GPa Summarizing, λ = 31.05 GPa, μ = 33.07 GPa . (b) Following the 1992 Landers earthquake ( M S = 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: e 11 = − 0 . 26 × 10 − 6 , e 22 = 0 . 92 × 10 − 6 , e 12 = − 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 τ 11 , τ 22 , and τ 12 . 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 τ τ τ = [ λ tr[ e ] + 2 μe 11 2 μe 12 2 μe 21 λ tr[ e ] + 2 μe 22 ] Given the Landers strain changes of e 11 = − 0 . 26 × 10 − 6 , e 22 = 0 . 92 × 10 − 6 , e 12 = − 0 . 69 × 10 − 6 , then tr[ e ] = 0 . 66 × 10 − 6 and τ τ τ = [ 3300 − 45600 − 45600 81300 ] Pa or τ 11 = 3 . 3 × 10 3 Pa, τ 22 = 8 . 13 × 10 4 Pa, and τ 12 = − 4 . 56 × 10 4 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: e 11 = 0 . 101 × 10 − 6 , e 22 = − 0 . 02 × 10 − 6 , e 12 = 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 τ τ τ = [ λ tr[ e ] + 2 μe 11 2 μe 12 2 μe 21 λ tr[ e ] + 2 μe 22 ] Given the 1000 year strain changes of e 11 = 0 . 101 × 10 − 3 , e 22 = − 0 . 02 × 10 − 3 , e 12 = 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 τ τ τ = [ 9 . 20 0 . 33 0 . 33 1 . 19 ] MPa (e) Farmer Bob owns a 1 km 2 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 m 2 . Use tr[ e ] as a measure of area change. Each year, tr[ e ] = 0 . 081 × 10 − 6 so Farmer Bob gains 0.081 m 2 of land each year . For Landers tr[ e ] = 0 . 66 × 10 − 6 so Farmer Bob gained 0.66 m 2 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 = ∆ | τ s | + μ s ∆ τ n , where τ s is the shear stress (traction), τ n is the normal stress, and μ s 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 μ s = 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 = CFF 1000+ L − CFF 1000 CFF a , where CFF a is the annual change in CFF, CFF 1000 is the thousand year change in CFF, and CFF 1000+ L is the thousand year + Landers change in CFF (note that CFF 1000+ L 6 = CFF 1000 + CFF L ). 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 = 2 πf = 5 π/ 4 and k = 2 π/ Λ = π/ 4000. Thus we have for a wave traveling in the + x direction: u = 0 . 04 sin[( π/ 4000) x − (5 π/ 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 [ 5 π 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 − 36 π = 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 u x = A sin( ωt − kx ), (b) S -wave with displacements in the y direction, i.e., u y = 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 e 11 = ∂u x ∂x = − kA cos( ωt − kx ) Then from (2.29) we have τ 11 = − ( λ + 2 μ ) kA cos( ωt − kx ) τ 22 = τ 33 = − λkA cos( ωt − kx ) (b) We are given u y = 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 e 12 = e 21 = 1 2 ∂u y ∂x = − 1 2 kA cos( ωt − kx ) Then from (2.30) we have τ 12 = τ 21 = − 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 u x = A sin( ωt − kx ) The only nonzero strain component is e 11 = ∂u x ∂x = − kA cos( ωt − kx )

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