Geology Reference
In-Depth Information
p
1
(x,t) = p
i
e
i(
t-kx)
+ p
r
e
i(
t+kx)
(5.6)
v
1
(x,t) = v
i
e
i(
t-kx)
+ v
r
e
i(
t+kx)
(5.7)
whereas in Section 2, the only family of waves that exists is the right-going one
given by
p
2
(x,t) = p
t
e
i(
t-kx)
(5.8)
v
2
(x,t) = v
t
e
i(
t-kx)
(5.9)
---------------+
Section 1 |
+---------------
i ---> x=0 t --->
<--- r Section 2
+---------------
|
---------------+
Figure 5.1.
Wave transition at area change.
In these equations, the subscripts i, r, and t refer to
incident
,
reflected
and
transmitted
, respectively. At the junction x = 0, the acoustic pressure and the
volume velocity must be continuous. If we substitute Equations 5.6 to 5.9 in our
matching conditions, set x = 0 and cancel the e
i t
' s, we obtain
p
i
+ p
r
= p
t
(5.10)
A
1
(v
i
+ v
r
) = A
2
v
t
(5.11)
Now, let us divide the first equation by the second, so that
(p
i
+ p
r
)/(v
i
+ v
r
) = ( A
1
/A
2
)( p
t
/v
t
) (5.12)
For a right-going wave, the displacement function takes the form u(x,t) = f(x-ct),
where f may be arbitrary. Since p = -B u
x
= -Bf ' and v = u
t
= -c f ', where
primes denote differentiation with respect to the argument, it follows upon
division that p/v = B/c, or p/v = B/c = c since c
2
= B / . This leads to
p = vc (5.13)
Now consider the analogous derivation for left-going waves. We replace -c
above by +c, for which p = -Bf ' and v = cf '. This leads to p/v = - B/c, so that p
=
vc. If we substitute
v
i
= p
i
/ c
(5.14a)
v
r
= - p
r
/ c
(5.14b)
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