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8.1 2D Traveling Wave Solutions

199

Based on the formulas shown above, topics such as cutoff frequency, echoes, and sound distortion in waveguides will be discussed in this chapter.

8.12D Traveling Wave Solutions

8.1.1Definition of Wavenumber Vectors

We have derived the relationship between the combined wavenumber k and the component wavenumbers kx and ky in Chap. 7:

k2 ¼ k2x þ k2y

Substituting the relationship between the wavenumber k and the wavelength λ:

2π k ¼ λ

into the rst equation gives the relationship between the combined wavelength λ and the component wavelengths λx and λy as:

1 ¼ 1 þ 1

λ2 λ2x λ2y

The relationship above can also be proved geometrically with trigonometric relationships:

cos θ ¼

λ

; sin θ ¼

 

λ

 

 

 

 

 

 

λx

λy

 

 

 

 

 

 

 

and the trigonometric identity:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

cos 2θ þ sin 2θ ¼ 1 ! 1

¼

λ2

þ

λ2

!

1

¼

1

þ

1

λ2

λ2

λ2

 

λ2

λ2

 

 

 

x

 

y

 

 

 

 

 

x

 

y

200

8 Acoustic Waveguides

= +

1

1 = 1 + 1

Where

= ; =

̂

̂

The combined wavenumber is a vector that presents the propagation direction and the magnitude of a 2D propagating plane wave as:

!

k¼ kxbex þ kybey

¼k nxbex þ nybey

where:

nx ¼ kkx ; ny ¼ kky

where bex and bey are the base vectors of the Cartesian coordinate systems and kx and ky

!

are vector components of k .

!

Note that nx and ny indicate the direction of the wavenumber vector k and have the following properties:

nx ¼ cos θ; ny ¼ sin θ

! n2x þ n2y ¼ 1

8.1 2D Traveling Wave Solutions

201

8.1.2Wavenumber Vectors in 2D Traveling Wave Solutions

The sound pressure and ow velocity of 1D plane waves were derived in Chap. 4 and are listed below for reference:

p ðx, tÞ ¼ A cos ðωt kx þ θ Þ

1

u ðx, tÞ ¼ ρoc p ðx, tÞ

The formulas above will be transferred to 2D plane waves using the wavenumber vector we learned in the previous section:

1D Plan Wave

±

±

±

 

 

± = ±

1

 

 

 

±

 

A single variable

ρ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

A single variable

 

 

 

 

 

 

 

 

 

Note that the formulas of pressure and velocity on the left-hand side of the gure are for the plane wave traveling along the x-axis. The right-hand side of the gure shows the same plane wave rotated by an angle. The rotated plane wave becomes 2D and can be formulated with a new variable ξ. The formulas of pressure and velocity of the 2D plane wave are the same as the 1D plane wave except using the different variable names x and ξ as:

p ðξ, tÞ ¼ A cos ðωt kξ þ θ Þ

1

u ðξ, tÞ ¼ ρoc p ðξ, tÞ along traveling direction

The new variable ξ is the result of the dot product of the wavenumber vector

! !

k kx, ky and the position vector r ðx, yÞ as:

202

 

 

 

 

 

 

 

 

 

 

 

 

 

8

 

Acoustic Waveguides

 

!

!

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ξ ¼

 

k

r

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

!k

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ð

 

Þ

 

 

þ

 

 

 

 

k

 

 

 

 

kx

, ky

 

x, y

 

kxx

kyy

 

kxx

þ

kyy

 

 

 

 

 

 

 

 

 

 

q

 

 

 

 

 

 

¼

 

 

 

!

 

 

 

 

¼

¼

 

 

 

 

 

 

 

 

 

k

 

 

 

 

 

kx þ ky

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

¼

nxx

þ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

nyy

 

 

 

 

 

 

 

 

 

 

 

 

!

And the velocity of the plane wave above is along the traveling direction k :

̂= ̂+ ̂

̂

̂̂

The value of is the same on a plane that is perpendicular to the wavenumber vector

Based on the formulas above, the pressure and velocity of a 2D plane wave can be formulated in the Cartesian coordinate as:

p ðx, y, tÞ ¼ A cos ðωt kξ þ θ Þ

 

¼ A cos

ωt k nxx þ nyy þ θ

¼ A cos

ωt kxx þ kyy

þ θ

¼ A cos hωt !k !r þ θ i

where:

!k

 

k ¼

 

kx

 

ky

 

nx ¼

 

; ny ¼

 

 

 

k

k

 

!

The 2D ow velocity u of the 2D plane wave can be derived using Euler s force equation as the following:

8.1 2D Traveling Wave Solutions

 

 

Z

 

 

 

 

 

 

 

 

 

203

 

ð

 

Þ ¼ ρo

 

½

 

ð

 

 

 

 

Þ&

 

 

!u

 

x, y, t

1

b

 

p

b

x, y, t

 

dt

Þ

 

 

 

¼

 

ρ0c

 

 

ð

 

 

 

 

 

 

 

nxex

þ nyey

p

 

x, y, t

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

!

Note that the sound pressure p is a scalar but the ow velocity u is a vector because of the gradient operator .

!

The detailed derivation of the backward ow velocity u 2 is shown in Example

!

8.1. The derivation of the forward ow velocity u þ is one of the homework problems.

Example 8.1: (Derivation of 2D Plane Traveling Waves: Backward)

!

Derive the 2D backward ow velocity vector u ðx, y, tÞ of a 2D plane backward traveling wave from the 2D acoustic pressure p (x, y, t) using Eulers force equation:

p ðx, y, tÞ ¼ A cos

ωt þ

kxx þ kyy þ θ

 

ð

 

Þ ¼

 

b 0

b

ð

 

Þ

!u

 

x, y, t

 

nxex þ nyey

p

x, y, t

 

ρ c

 

where:

nx

Eulers force equation:

Example 8.1: Solution

 

kx

, and ny ¼

ky

¼

q

q

kx2 þ ky2

kx2 þ ky2

pðx, y, tÞ ¼ ρo

!

ðx, y, tÞ

t

u

!

Backward ow velocity u ðx, y, tÞ is a vector and can be calculated from acoustic pressure based on Eulers force equation:

pðx,

!

! uðx, y,

y, tÞ ¼ ρo

 

!

ðx, y, tÞ

 

u

t

tÞ ¼ ρo

Z

½ pðx, y, tÞ&dt

1

 

 

 

 

 

Eulers force equation for the 2D wave equation in Cartesian coordinate is formatted as:

!uðx, y, tÞ ¼ ρo

Z ex x

þ ey y pðx, y, tÞ dt

1

b

b

 

 

 

Remarks

204

8 Acoustic Waveguides

Eulers force equation is valid for both forward and backward waves.

!

Pressure p(x, y, t) is a scalar function and velocity uðx, y, tÞ is a vector function. Therefore, the backward velocity can be calculated from pressure as:

!u 2

ðx, y, tÞ ¼ ρo

Z

ex x

þ ey y p ðx, y, tÞ dt

 

 

 

 

 

 

 

1

 

 

 

b

 

b

 

 

 

 

 

 

 

 

 

 

 

 

 

 

¼ ρo Z

 

 

 

 

 

 

e

 

Z

y p ðx, y, tÞ dt

 

 

x p ðx, y, tÞ dt

ρo

 

 

 

 

ex

 

 

 

 

 

 

 

 

 

y

 

 

 

 

 

 

 

 

 

 

b

 

 

 

 

 

 

 

 

 

 

 

b

 

 

 

 

 

 

 

 

 

Separate the velocity vector

 

into two

vector

components,

bex

u

x ð

x, y, t

Þ

and eyuy ðx, y, tÞ:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

b

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

where:

ux ðx, y,

uy ðx, y,

x

!

u ðx, y, tÞ ¼ bexux ðx, y, tÞ þ beyuy ðx, y, tÞ

tÞ ¼ ρo

Z x p ðx, y, tÞ dt ¼ ρoω p ðx, y, tÞ

1

 

 

kx

tÞ ¼ ρo

Z y p ðx, y, tÞ dt ¼ ρoω p ðx, y, tÞ

1

 

 

ky

p ðx, y, tÞ ¼ A cos ωt þ

kxx þ kyy þ θ

p ðx, y, tÞ ¼ A x cos ωt þ kxx þ kyy

þ θ

 

 

¼ A kx sin ωt þ kxx þ kyy þ θ

8.1 2D Traveling Wave Solutions

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

205

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ux ðx, y, tÞ ¼ Z

 

p ðx, y, tÞ dt

 

 

 

 

 

 

 

 

 

 

x

 

 

 

 

 

 

 

 

 

 

¼ A kx Z sin ωt þ kxx þ kyy þ θ dt

 

 

 

 

 

 

 

 

 

¼ A ω cos ωt þ kxx þ kyy þ θ

 

 

 

 

 

 

 

 

 

 

 

 

kx

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

¼

ω p ðx, y, tÞ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

kx

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Similarly:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

uy ðx, y, tÞ ¼ Z y p ðx, y, tÞ dt ¼ ω

 

 

 

 

 

 

 

 

 

 

 

 

p ðx, y, tÞ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ky

 

 

 

 

 

 

 

 

 

Combining two vector components back to one velocity vector yields

 

 

!

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

u ðx, y, tÞ ¼ exux ðx, y, tÞ þ eyuy ðx, y, tÞ

ω p x,

 

 

t

 

 

 

 

 

b e x

 

ω p b

t ey

 

 

 

 

 

 

 

¼ b 1

kx

 

 

ð

x, y,

Þ b

 

ky

 

 

ð

 

y,

 

Þ

 

 

 

 

ρo

 

 

 

 

 

ρo

 

 

 

 

 

 

 

 

 

 

¼ ρoω

kxex

þ kyey p ðx, y, tÞ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

b

 

 

 

 

b

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

7

 

 

 

 

ρo

 

 

q6

kx

 

ky

 

 

 

kx

 

 

ky

ð

Þ

 

 

1

 

 

2

 

 

 

2

2

 

 

kx

 

 

 

 

 

 

ky

 

 

 

 

3

 

x, y, t

 

¼ ω

 

kx þ ky

4q b

þ

q b

5

 

 

 

 

 

2

þ

2 ex

 

 

2

þ

 

2 ey

p

 

¼ ρo

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ω nxex þ nyey p ðx, y, tÞ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

k

 

 

b

 

 

 

 

b

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

where:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

¼ q

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

!

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

kx

k

 

 

 

 

 

 

 

 

ky

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

k

 

 

 

 

 

kx þ ky

 

 

 

 

 

 

 

 

 

 

 

 

 

 

nx ¼

q

, and ny ¼

q

 

 

 

 

 

 

 

 

 

 

 

 

kx2 þ ky2

 

kx2 þ ky2

 

 

 

 

 

 

 

 

 

Replacing ω and k with c yields the nal form: