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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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