68 |
3 Solutions of Acoustic Wave Equation |
3.6Wavenumber, Angular Frequency, and Wave Speed
The backward wave p (x, t) and the forward wave p+(x, t) can be expressed in Form 2 (Chap. 1) as:
p (x, t) ¼ A cos (ωt kx + ϕ ) (Form 2 : REP)
Because the harmonic wave repeats after every 2π, the above cosine function will be the same when the increase of angle, θ, is an integer of 2π as:
cos ðωt kx þ ϕ Þ ¼ cos ðωt kx þ ϕ þ θÞ
¼ cos ðωt kx þ ϕ þ 2π iÞ
where:
θ¼ 2π ∙ i.
i is an integer.
Based on this property, we can define wavelength λ and period T from the following relations:
θ ¼ ωT ¼ 2π ! ω ¼ 2Tπ
2π
θ ¼ kλ ¼ 2π ! k ¼ λ
Based on the two equations above, we can conclude the relationship between k and ω as:
kλ ¼ ωT ! ω ¼ Tλ k ¼ ck
where c is the speed of the wave and is equal to the wavelength (λ) divided by the period (T) as:
c Tλ
In real applications, angular frequency ω is a given property. Also, the speed of sound in the air is a known value, which just slightly varies with meteorological conditions. Hence, based on the known speed of sound and given a specific angular frequency, we can calculate the wavenumber and the wavelength of the acoustic wave as:
3.6 Wavenumber, Angular Frequency, and Wave Speed |
69 |
k ¼ ωc
and
2π |
¼ 2π |
c |
¼ 2π |
c |
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c |
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λ ¼ |
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¼ |
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k |
ω |
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2πf |
f |
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where the circular frequency f (unit in Hz) is related to the angular frequency ω (unit in rad/s) as:
ω ¼ 2πf
or
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rad |
1 |
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ω |
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¼ 2π½rad& f h |
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i |
time |
time |
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By replacing ω with ck, the backward wave p (x, t) and the forward wave p+(x, t) can be expressed in Form 2 as:
p (x, t) ¼A cos (ωt kx + ϕ ) |
(Form 2 : REP) |
¼A cos [k(ct x) + ϕ ] |
(Form 2 : REP) |
or: |
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p+(x, t) ¼ A+ cos [k(ct x) + ϕ+] |
(Form 2 : REP) |
p (x, t) ¼ A cos [k(ct + x) + ϕ ] |
(Form 2 : REP) |
Forward wave: 
, 
= A
cos 
− 
+ 
Backward wave: 
cos
t=
t=
= −c 
if 
> 
then 
< 
= c 
if 
> 
then 
> 
70 |
3 Solutions of Acoustic Wave Equation |
To determine the propagation direction |
of a wave, we can trace one point |
(a constant value of pressure p) in the wave from (x1, t1) to (x2, t2).
For the forward wave p+ above, where the signs of ω and k are different, we have:
ct2 x2 ¼ ct1 x1
! cðt2 t1Þ ¼ x2 x1
Therefore, if t2 > t1, then x2 > x1, and the wave is a forward wave traveling from left (x1) to right (x2).
For the backward wave p above, where the signs of ω and k are the same, we have:
ct2 þ x2 ¼ ct1 þ x1
! cðt2 t1Þ ¼ ðx2 x1Þ
Therefore, if t2 > t1, then x1 > x2, and the wave is a backward wave traveling from right (x1) to left (x2).
In this course, the solutions of the acoustic wave equation are represented as harmonic waves. Using Fourier analysis, any traveling waves can be represented by the summation of a series of harmonic waves.
However, another common approach for analyzing the wave equation is by using the function in moving wave form, f(x + ct) or f(x ct). Any moving wave function f(x + ct) or f(x ct) is a solution of the acoustic wave equation. This can be easily proved by substituting the function to the wave equation and using the following relations:
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∂ |
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f 0ðx ctÞ, |
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∂2 |
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f ðx ctÞ ¼ |
and |
∂x2 f ðx ctÞ ¼ f }ðx ctÞ |
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∂x |
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∂ |
f x |
ct |
Þ ¼ |
c f |
0 |
ð |
x |
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ct |
, |
and |
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∂2 |
f |
x |
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ct |
Þ ¼ |
c2 f } |
x |
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ct |
Þ |
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∂t ð |
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Þ |
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∂t2 ð |
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ð |
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All BTWs are the special cases (i.e., harmonic waves) of these moving wave functions. At a fixed location, say x ¼ 0, any arbitrary wave function, p(t), can be decomposed into a series of harmonic waves:
N=2 1
X
pðtÞ ¼ Ao þ 2 ½Ak cos ðωktÞ Bk sin ðωktÞ& þ AN=2
k¼1
where: