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Analytical Solution
To find the normal acoustic modes of the tube, consideration will be given to solutions of the form:
и (x, t) = Re[U(x)e ш],
where Re[] is the real part of [ ]. Substitution of Eq. (1.16) into Eqs. (1.15) and (1.14) yields the following eigenvalue problem:
|
d2U щ2 „ |
(1.17) |
|
+ -2 U = 0 dx2 a2 |
|
|
U (0) = U (L) = 0- |
(1.18) |
The two linearly independent solutions of Eq. (1.17) are
U (x) = A sin —J + B cos —J. On imposing boundary conditions (1.18), it is found that
 |
 |
B = 0, and A sin ^-“j = 0- For a nontrivial solution A cannot be zero, this leads to,
Therefore, (1.20)
is the eigenvalue or eigenfrequency. The eigenfunction or mode shape is obtained from Eq. (1.19); i. e.,
. /nnx „ „ „
un (x) = sin ^~l) , n = 1, 2, 3, -.
