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2.6. Impedance of an open tube using the inhomogeneous wave equation

Solution of the inhomogeneous wave equation for an open tube

If the tube of Fig. 2.6 is open at x = 0 instead of closed, let the solution be in the form of an eigenfunction expansion

\tilde{p} (x) = \sum_{n = 0}^{\infty} {\tilde{A}}_{n} \sin ((n + 1 / 2) π x / l) .

$\tilde{p} (x) = \sum_{n = 0}^{\infty} {\tilde{A}}_{n} \sin ((n + 1 / 2) π x / l) .$

(2.101)

When the piston is stationary, this satisfies the boundary conditions

{\tilde{p} (x) |}_{x = 0} = 0, \frac{\partial}{\partial x} {\tilde{p} (x) |}_{x = l} = 0.

${\tilde{p} (x) |}_{x = 0} = 0, \frac{\partial}{\partial x} {\tilde{p} (x) |}_{x = l} = 0.$

(2.102)

Inserting Eq. (2.101) in Eq. (2.81) and multiplying both sides by sin((m+1/2)πx/l) while integrating over the length of the tube gives

\begin{matrix} \sum_{n = 0}^{\infty} (k^{2} - \frac{{(n + 1 / 2)}^{2} π^{2}}{l^{2}}) {\tilde{A}}_{n} \int_{0}^{l} \sin ((m + 1 / 2) π x / l) \sin ((n + 1 / 2) π x / l) d x \\ = j k ρ_{0} c {\tilde{u}}_{0} \int_{0}^{l} \sin ((m + 1 / 2) π x / l) δ (x - l) d x . \end{matrix}

$\begin{matrix} \sum_{n = 0}^{\infty} (k^{2} - \frac{{(n + 1 / 2)}^{2} π^{2}}{l^{2}}) {\tilde{A}}_{n} \int_{0}^{l} \sin ((m + 1 / 2) π x / l) \sin ((n + 1 / 2) π x / l) d x \\ = j k ρ_{0} c {\tilde{u}}_{0} \int_{0}^{l} \sin ((m + 1 / 2) π x / l) δ (x - l) d x . \end{matrix}$

(2.103)

The two integrals have the following identities

\int_{0}^{l} \sin ((m + 1 / 2) π x / l) \sin ((n + 1 / 2) π x / l) d x = {\begin{matrix} 0, & m \neq n \\ l / 2, & m = n \end{matrix}

$\int_{0}^{l} \sin ((m + 1 / 2) π x / l) \sin ((n + 1 / 2) π x / l) d x = {\begin{matrix} 0, & m \neq n \\ l / 2, & m = n \end{matrix}$

(2.104)

\int_{0}^{l} \sin ((m + 1 / 2) π x / l) δ (x - l) d x = \sin ((m + 1 / 2) π) = {(- 1)}^{m} .

$\int_{0}^{l} \sin ((m + 1 / 2) π x / l) δ (x - l) d x = \sin ((m + 1 / 2) π) = {(- 1)}^{m} .$

(2.105)

The first is the property of orthogonality and the second is a property of the Dirac delta function

\int_{- \infty}^{\infty} F (x) δ (x - l) d x = F (l) .

$\int_{- \infty}^{\infty} F (x) δ (x - l) d x = F (l) .$

(2.106)

Hence, sin((m+1/2)πx/l) in this case is the orthogonal (or normalizing) function and we can eliminate the summation to yield

{\tilde{A}}_{n} = 2 j k l ρ_{0} c {\tilde{u}}_{0} \frac{{(- 1)}^{n}}{k^{2} l^{2} - {(n + 1 / 2)}^{2} π^{2}},

${\tilde{A}}_{n} = 2 j k l ρ_{0} c {\tilde{u}}_{0} \frac{{(- 1)}^{n}}{k^{2} l^{2} - {(n + 1 / 2)}^{2} π^{2}},$

(2.107)

so that the solution of Eqs. (2.80) and (2.81), given by Eq. (2.101), becomes

\tilde{p} (x) = 2 j k l ρ_{0} c {\tilde{u}}_{0} \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} \sin ((n + 1 / 2) π x / l)}{k^{2} l^{2} - {(n + 1 / 2)}^{2} π^{2}} .

$\tilde{p} (x) = 2 j k l ρ_{0} c {\tilde{u}}_{0} \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} \sin ((n + 1 / 2) π x / l)}{k^{2} l^{2} - {(n + 1 / 2)}^{2} π^{2}} .$

(2.108)

Eq. (2.108) is equivalent to the solution of Eq. (2.79), which would be obtained by applying the boundary conditions to the solution to the homogeneous wave equation, given by

\tilde{p} (x) = - j ρ_{0} c {\tilde{u}}_{0} \frac{\sin k x}{\cos k l} .

$\tilde{p} (x) = - j ρ_{0} c {\tilde{u}}_{0} \frac{\sin k x}{\cos k l} .$

(2.109)

Impedance of the open tube

We saw previously that the impedance of an open tube is given by

Z_{s} = \frac{\tilde{p} (l)}{- {\tilde{u}}_{0}} = j ρ_{0} c \tan k l,

$Z_{s} = \frac{\tilde{p} (l)}{- {\tilde{u}}_{0}} = j ρ_{0} c \tan k l,$

(2.110)

which is equivalent to the following expression obtained using Eq. (2.108)

Z_{s} = \frac{\tilde{p} (l)}{- {\tilde{u}}_{0}} = j ρ_{0} c \sum_{n = 0}^{\infty} \frac{2 k l}{{(n + 1 / 2)}^{2} π^{2} - k^{2} l^{2}} .

$Z_{s} = \frac{\tilde{p} (l)}{- {\tilde{u}}_{0}} = j ρ_{0} c \sum_{n = 0}^{\infty} \frac{2 k l}{{(n + 1 / 2)}^{2} π^{2} - k^{2} l^{2}} .$

(2.111)

Because modes occur when kl = (n + ½)π, the eigenfrequencies are given by

f = \frac{(2 n + 1) c}{4 l} or l = \frac{2 n + 1}{4} λ, n = 0,1,2,3 \dots

$f = \frac{(2 n + 1) c}{4 l} or l = \frac{2 n + 1}{4} λ, n = 0,1,2,3 \dots$

(2.112)

Expansion for tan

We have thus obtained a useful expansion for tan kl, which converges when truncated to a finite number of terms and retains the singularities at kl = (n + 1/2) π, unlike the more conventional formula tan kl = (sin kl)/(cos kl), where the sine and cosine expansions are given by Eqs. (2.95) and (2.96) respectively. Hence

\tan x = \sum_{n = 0}^{\infty} \frac{2 x}{{(n + 1 / 2)}^{2} π^{2} - x^{2}} .

$\tan x = \sum_{n = 0}^{\infty} \frac{2 x}{{(n + 1 / 2)}^{2} π^{2} - x^{2}} .$

(2.113)

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Table of Contents for Acoustics

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2.6. Impedance of an open tube using the inhomogeneous wave equation

Solution of the inhomogeneous wave equation for an open tube

Impedance of the open tube

Expansion for tan

Table of Contents for
Acoustics