Chapter 2 Theory of seismic waves
2.1 The basic elastic constants
2.1a Show that when the only nonzero applied stress is
, Hooke’s law requires that the normal strains
, and that Poisson’s ratio
, defined as
, satisfies the equation
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(2.1a)
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Background
Stress is force/unit area and is denoted by
, etc., where a force in the
-direction acts upon a surface perpendicular to the
-axis. The stresses
and
are, respectively, a normal stress and a shearing stress.
Stresses produce strains (changes in size and/or shape). If the stresses cause a point
to have displacements
along the coordinate axes, the basic strains are given by derivatives of these displacements as follows:
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(2.1b)
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(2.1c)
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The vector displacement
is
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(2.1d)
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where
are unit vectors in the
-directions (see Sheriff and Geldart, 1995, problem 15.3). The dilatation
is the change in volume per unit volume, i.e.,
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(2.1e)
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A pressure
produces a decrease in volume, the proportionality constant being the bulk modulus
:
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(2.1f)
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Sometimes the compressibility,
, is used instead of
.
In addition to creating strains, stresses cause rotation of the medium, the vector rotation
being equal to
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(2.1g)
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where
,
,
For small strains and an isotropic medium (where properties are the same regard-less of the direction of measurement), Hooke’s law relates the stresses to the strains as follows:
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(2.1h)
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(2.1i)
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where
and
are Lamé’s constants (
is usually called the modulus of rigidity or the shear modulus).
Solution
Subtracting equation (2.1h) for
from the same equation for
gives
.
Dividing equation (2.1h) for
by
gives
so
2.1b Show that Young’s modulus
, defined as
, is given by the equation
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(2.1j)
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Solution
Adding the three equations (2.1h) for
,
, and
, and recalling that
, we get
Dividing both sides by
gives
Using equation (2.1a) we get
2.1c A pressure
is equivalent to stresses
. Derive the following result for the bulk modulus
:
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(2.1k)
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Solution
Since
, we add equation (2.1h) for each of the three values
, obtaining
, so from equation (2.1f),
2.2 Interrelationships among elastic constants
The entries in Table 2.2a express, for isotropic media, the quantities at the heads of the columns in terms of the pairs of elastic constants or velocities at the left ends of the rows. The first three entries in the ninth row are equations (2.1j), (2.1a), and (2.1k) and the next two entries in the same row are formulas for the P- and S- wave velocities
and
(see problem 2.5). Starting from these five relations, derive the other relations in the table.
Background
For isotropic media, any two of the elastic constants can be considered as independent and the others can be expressed in terms of these two. The P- and S-wave velocities,
and
, given by the equations
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(2.2a)
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[see Sheriff and Geldart, 1995, equations (2.28) and (2.29)] can also be expressed in terms of any two elastic constants (plus the density
).
Solution
Denoting the equations by row and column (as for matrix elements) and using a comma instead of a period, we use equations (9,1) to (9,3) to derive the equations that do not involve
and
, then we use equations (9,6) and (9,7) (see equation (2.2a)) to derive the rest. From equation (9,1),
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(3,5)
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From equation (9,2),
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(5,5)
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Solving equation (9,2) for
, we have
,
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(6,4)
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From equation (9,3),
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(7,5)
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Equating
from equations (9,2) and (9,3) [or from equations (6,4) and (7,5)] gives
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(8,4)
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thus
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(6,3)
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and
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(4,5)
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Solving equation (4,5) for
gives
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(8,2)
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Substituting equation (4,5) into equation (8,4), we get,
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(4,4)
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We use equation (7,5) to eliminate
from equation (6,3):
that is,
so
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(5,3)
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Solving for
, we get
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(7,2)
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Figure 2.2a) Relations between elastic constants and velocities for isotropic media.
We use equation (8,4) to express
in equation (9,1) in terms of
. Thus
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(8,1)
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Solving equation (8,1) for
gives
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(2,5)
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We now eliminate
from equation (9,1) using equation (7,5)
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(7,1)
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Solving equation (7,1) for
and
gives
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(3,3)
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(2,4)
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Next we use equation (6,4) to eliminate
from equation (9,1):
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(6,1)
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Using equations (9,1) and (5,5) we get
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(5,1)
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Solving equation (5,1) for
and
gives
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(3,2)
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(1,4)
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Using equations (4,4) and (4,5) to replace
in equation (9,1) by
gives
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(4,1)
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Solving equation (4,1) for
and
, we get
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(1,3)
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(2,2)
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The last equation (of this group) can be obtained by substituting equation (1,3) into equation (4,5):
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(1,5)
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Equations (10,1) to (10,3) express
,
, and
in terms of the P- and S-wave velocities,
and
(see equation 2.2a). To introduce
and
, we write equations (10,1) to (10,3) in terms of
and
. Thus,
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(10,1)
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(10,2)
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(10,3)
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Equation (10,4) is the second equation in equation (2.2a). To get equation (10,5), we write
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(10,5)
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To verify column 6, we start with equation (9,6) and express
and
in terms of the required pair of constants. Thus, using equations (1,4) and (1,5) we get
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(1,6)
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Following the same procedure, using equations (2,4) and (2,5), we get
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(2,6)
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In the same way, we get the following results:
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(3,6)
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(4,6)
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(5,6)
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(6,6)
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(7,6)
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(8,6)
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Column 7 is merely the square root of column 4 after dividing by
. Column 8 is obtained by dividing column 7 by column 6.
2.3 Magnitude of disturbance from a seismic source
2.3a Firing an air gun in water creates a pressure transient a small distance away from the air gun with peak pressure
of 5 atmospheres (
Pa). If the compressibility of water is
/Pa, what is the peak energy density?
Background
Air guns (see problem 7.7) suddenly inject a bubble of high‐pressure air into the water to generate a seismic wave.
Stresses acting upon a medium cause energy to be stored as strain energy, because the stresses are present while the medium is being displaced. Strain energy density (energy/unit volume)
is equal to [see Sheriff and Geldart, 1995, equation (2.22)]
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(2.3a)
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Solution
From problem 2.1c, we see that
Pa. Also,
for water, so
. From equation (7,5) of Table 2.2a we find that
when
. Also,
(see equation (2.1f)), so
Using equation (2.3a) we find that
[The dimensions of
are the same as those of stress, since strains are dimensionless. Thus, stress units are
.]
2.3b If the same wave is generated in rock with
Pa, what is the peak energy density? Assume a symmetrical
-wave with
for
.
Solution
We have
,
, so equation (2.3a) becomes
Equation (9,3) in Table 2.2a gives
so that
2.4 Magnitudes of elastic constants
To illustrate the relationships and magnitudes of the elastic constants, complete Table 2.4a.
[Note that these values apply to specific specimens; the elastic constants for rocks range considerably, especially as porosity and pressure change.]
Solution
We use the row-column notation to designate equations from Table 2.2a.
Water: Since water is a fluid we know that
. Equation (4,1) shows that
also. From equation (4,5) we get
Pa.
Table 2.4a. Magnitudes of elastic constants and velocities.
Constant
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Water
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Stiff mud
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Shale
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Sandstone
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Limestone
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Granite
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16
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54
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50
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2.1
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0.50
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0.43
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0.38
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0.34
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0.25
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0.20
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1.0
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1.5
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1.8
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1.9
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2.5
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2.7
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1.5
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1.6
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3.2
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Stiff mud: Because
, stiff mud is equivalent to a solid, hence
. From
and
we get
using equation (9,6), while equation (9,2) expresses
in terms of
and
, thus enabling us to find both
and
.
We have:
Solving the two equations, we get
Pa,
Pa. Using equation (6,1),
Equation (6,3) gives
Finally, to get
we note that
,
, and
are in
, and so we must express
in appropriate units of
, i.e.,
. We now have
Shale. As with stiff mud, we have been given
,
, and
, so that again
and
give us
and
gives us
so that we can solve these equations for
and
, then find
and
using equations (9,3) and (9,1). Thus,
Solving the two equations gives
Pa,
Pa.
From equation (9,3),
Pa, and equation (9,1) gives
Finally,
Sandstone: We are given the elastic constants
and
(plus
), so we get
,
,
,
,
using equations (1,3) to (1,7) in Table 2.2a. Thus
[We could also have obtained
and
by using equations (9,6) and (9,7).]
Limestone: We solve in the same way as with sandstone since we are given the same constants:
Granite: Again the solution is the same as for sandstone.
Table 2.4b summarizes the results.
Table 2.4b Magnitudes of elastic constants.
Constant
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Water
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Stiffmud
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Shale
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Sandstone
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Limestone
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Granite
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0
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1.4
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9.9
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16
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54
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50
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2.1
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3.2
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13
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17
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36
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28
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0
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0.47
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3.6
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6.0
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22
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21
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2.1
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2.9
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11
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13
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22
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14
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0.50
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0.43
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0.38
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0.34
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0.25
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0.20
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1.0
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1.5
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1.8
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1.9
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2.5
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2.7
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1.5
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1.6
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3.2
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3.6
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5.1
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4.5
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0
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0.56
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1.4
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1.8
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2.9
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2.8
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2.5 General solutions of the wave equation
2.5a Verify that
and
are solutions of the wave equation (2.5b).
Background
When unbalanced stresses act upon a medium, the strains are propagated throughout the medium according to the general wave equation
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(2.5a)
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being a disturbance such as a compression or rotation.
is propagated with velocity
(see Sheriff and Geldart, 1995, Section 2.2). The disturbance is the result of unbalanced normal stresses, shearing stresses, or a combination of both. When normal stresses create the wave, the result is a volume change and
is the dilitation [see equation (2.1e)], and we get the P-wave equation,
becoming the P-wave velocity
. Shearing stresses create rotation in the medium and
is one of the components of the rotation given by equation (2.lg) ; the result is an S-wave traveling with velocity
. Various expressions for
and
in isotropic media are given in Table 2.2a.
In one dimension the wave equation (2.5a) reduces to
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(2.5b)
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Solution
We use subscripts to denote partial derivatives and primes to denote derivatives with respect to the argument of the function. Then, writing
, we have
Substituting in equation (2.5b), we get the identity
so
is a solution. We get the same result when
. A sum of solutions is also a solution, so
is a solution.
2.5b Verify that
is a solution of equation (2.5a), where
are direction cosines.
Solution
Let
. We now must show that
is a solution of equation (2.5a). Proceeding as before, we have
In the same way we get
But
(see Sheriff and Geldart, 1995, problem 15.9a), so
.
Following the same procedure we find that
thus verifying that
is a solution of equation (2.5a).
2.5c Show that
is a solution of the wave equation in spherical coordinates (see problem 2.6b) when the wave motion is independent of
and
:
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(2.5c)
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Solution
The wave equation in spherical coordinates is given in problem 2.6b. When we drop the derivatives with respect to
and
, the equation reduces to equation (2.5c). Writing
, we proceed as in part (a). Starting with the right-hand side, we ignore
for the time being and obtain
Substitution in equation (2.5c) shows that
is a solution. In the same way we can show that
is also a solution, hence the sum is a solution.
2.6 Wave equation in cylindrical and spherical coordinates
2.6a Show that the wave equation (2.5a) can be written in cylindrical coordinates (see Figure 2.6a as
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(2.6a)
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Solution
Figure 2.6a. Cylindrical coordinates.
We shall solve by direct substitution. We have
,
,
, and
,
. The following solution is lengthy, so we use subscripts to denote partial derivatives and write
We shall require the derivatives:
To replace
and
with derivatives with respect to
and
, we write:
Then,
Thus
so
2.6b Transform the wave equation into spherical coordinates (see Figure 2.6b), showing that it becomes
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(2.6b)
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Solution
Spherical coordinates
and rectangular coordinates are related as follows (see Figure 2.6b):
We continue to use subscripts to denote derivatives and letters to denote sines and cosines:
The derivatives of
,
, and
now become:
Figure 2.6b Spherical coordinates.
Summarizing these results, we have
We now calculate the derivatives
, etc.:
Adding the three derivatives, we get
Substituting the values of
,
,
, and
, we get for the wave equation
This is often written in the more compact form
2.7 Sum of waves of different frequencies; Group velocity
2.7a A pulse composed of two frequencies,
, can be represented by factors involving the sum and difference of the two frequencies. If the two components have the same amplitudes, we can write
where
,
Show that the composite wave is given approximately by the expression
where
.
Background
When different frequency components in a pulse have different phase velocities
(the velocity with which a given frequency travels), the pulse changes shape as it moves along. The group velocity
is the velocity with which the envelope of the pulse travels.
The envelope of a pulse comprises two mirror-image curves that are tangent to the waveform at the peaks and troughs, and therefore define the general shape of the pulse.
Solution
Adding the two components and using the identity
we get for the composite wave
2.7b Why do we regard
as the amplitude? Show that the envelope of the pulse is the graph of
plus its reflection in the
-axis.
Solution
The solution in (a) can be written
. We regard
as the amplitude for two reasons: (1)
repeats every time that
increases by
or each time that
increase by
. But
,
, so
must repeat more slowly than
. (2) Each time that
attains its limiting values,
,
has the value
and therefore never exceeds
; thus the curves of
and
pass through the maxima and minima of
and therefore constitute the envelope.
2.7c Show that the envelope moves with the group velocity
where
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(2.7a)
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(see Figure 2.7a).
Solution
If we consider the quantity
as a wave superimposed on the primary wavelet
, comparison with the basic wave type
of problem 2.5a shows that
takes the place of
, i.e.,
is the velocity
with which the envelope travels. In the limit,
is given by