Analytical Solution

To find the normal acoustic modes of the tube, consideration will be given to solutions of the form:

Подпись: (1.16)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

Подпись: (1.19)U (x) = A sin —J + B cos —J. On imposing boundary conditions (1.18), it is found that

Подпись: OJL or — = nn (n = integer)- a

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.,

Подпись: (1.21). /nnx „ „ „

un (x) = sin ^~l) , n = 1, 2, 3, -.

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