Biomedical Engineering Reference
In-Depth Information
oscillations. In Fig. 2.7b, we show the temporal sequence of the air pressure
for each configuration.
As a general result, we can add two counterpropagating waves of the same
frequency to obtain
cos(
kx
−
ωt
) + cos(
kx
+
ωt
)=2cos(
kx
)cos(
ωt
)
,
(2.21)
which is a standing wave with a position-dependent amplitude 2 cos(
kx
). The
boundary condition at the open end of the tube (that is, the pressure at
x
=
L
is always zero, or atmospheric pressure) gives us
k
n
L
=(2
n
−
1)
π/
2
,
n
=1
,
2
,
3
,... .
(2.22)
This means that the frequency and wavelength of a standing wave in a tube
open at one end and closed at the other cannot take any values, but only the
allowed values
c
4
L
,
F
n
=(2
n
−
1)
(2.23)
1
2
n −
1
4
L,
λ
n
=
(2.24)
where
n
=1
,
2
,
3
,...
. In contrast, in the case of a traveling wave, the fre-
quency and wavelength have no restriction other than
ω
=
ck
.
We have to point out that establishing these stationary density configu-
rations requires a source that excites the tube of air at a precise frequency.
What happens if we excite the tube with a signal that is the sum of two
periodic signals of different frequencies? Let us analyze this by inspecting a
simple example: we take as the excitation a signal that is the sum of the first
two resonances,
P
=
A
cos(
k
1
x
−
ω
1
t
)+
A
cos(
k
1
x
+
ω
1
t
)
ω
2
t
)+
A
cos(
k
2
x
+
ω
2
t
)
=2
A
cos(
k
1
x
)cos(
ω
1
t
)+2
A
cos(
k
2
x
)cos(
ω
2
t
)
,
+
A
cos(
k
2
x
−
(2.25)
which, although it is the sum of two standing waves with different frequencies
(with the same amplitude for simplicity), is not a standing wave. Now, we
have a pattern of densities that propagates along the tube, respecting only
the condition that there are no perturbations at the open end. In Fig. 2.8,
we show a temporal succession of the density along the tube. This density
no longer constitutes a stationary structure. There are no points within the
tube with no fluctuations. On the contrary, we have a fluctuation traveling
along the tube. This can be seen, after some tedious trigonometric algebra,
by rewriting (2.25) as
P
=4
A
cos(
kx
∆
ωt
)
+4
A
cos(
kx
+
ωt
) cos(∆
kx
+∆
ωt
)
,
−
ωt
) cos(∆
kx
−
(2.26)
