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and
v
t
= p
t
/ c (5.15)
in Equation 5.12, we have
p
r
/p
i
= ( A
1
- A
2
)/(A
1
+ A
2
) (5.16)
Thus, in the limit of equal areas when A
1
= A
2
, the magnitude of the reflected
pressure wave vanishes. If the areas are unequal, p
r
is completely determined,
since the incident pressure p
i
and both cross-sectional areas are known. Then,
the magnitude of the transmitted pressure wave can be obtained from Equation
5.10. In fact, it is easily seen that p
t
= p
i
+ p
r
= p
i
(1 + p
r
/p
i
) or
p
t
/p
i
= 1 + ( A
1
- A
2
)/(A
1
+ A
2
) = 2A
1
/(A
1
+ A
2
) (5.17)
5.1.6 Example 5-2. Classical water hammer.
Equation 5.13 is the well known formula for one-dimensional “water
hammer,” and it can be interpreted as follows. If a piston mounted at the end of
a semi-infinite pipe containing a stationary fluid with density and speed of
sound c is struck with a velocity v, a propagating pressure pulse with magnitude
p = vc is created. The fluid, being compressible, does not move as a rigid
column, which would require infinite energy. Rather, it remains undisturbed
except during the passage of the wave, when local fluid particles are
momentarily displaced a distance u with the speed v = u/ t. Pressure, fluid
velocity, and particle displacement all propagate together at the speed of sound c
(note c >> v
!
).
5.1.7 Example 5-3. Acoustic pipe resonances.
Again, recall that the displacement u(x,t) satisfies
2
u/ t
2
- c
2
2
u/ x
2
= 0 ,
where p(x,t) = - B u/ x is the acoustic pressure, and c
2
= B /
. As in Chapter 1,
we take
u(x,t) = X(x)T(t)
(5.18)
This separation of variables leads to
T"(t)/c
2
T(t) = X"(x)/X(x) = -
p
2
< 0 (5.19)
and the solution X(x) = A sin
p
x + B cos
p
x. The function X(x) is proportional
to the fluid displacement u(x,t), and it vanishes at closed ends. On the other
hand, note that
p(x,t) = - BX'(x)T(t) (5.20)
Thus, p(x,t) is proportional to the spatial derivative X'(x). Our finite system is
defined on 0 < x < L.
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