D O C . 3 9 P R O P A G A T I O N O F S O U N D 1 8 9
By solving the equation system (13), (14), (17) one obtains , , and as
functions of . And for the quotient one gets
.
(18)
Next, one gets from (18) and (10), because of the equilibrium condition
:
, (19)
where we put
, (20)
, (21)
. (22)
With (19) and (8) our problem is completely solved. We next get
[p. 385]
ΔT Δn1 ΔV
Δ(pV)
Δ(pV)
ΔV
----------------
ΔpV
ΔV
----------- –
p +-------------
κDn1
T
R n1 n2) + ( κ–--------
4n2
V
jω⎞
κ2⎠
----- - +
⎝
⎛
+
n2
V⎠
----⎞
-
⎝
⎛
2
+ RD n1 n2) + ( CRT] – [
C⎝ κ
4n2
V
-------- –
jω⎞
κ2
-----⎠ - +
⎛
κD2n
RT
2
----------------1
+
----------------------------------------------------------------------------------------------------------------------------------------------------------------------- =
κ1----
n1
V
-
κ2⎝
n2
V⎠
----⎞
-
⎛
2
=
π
Δ
---
p
ρ
-- -
1
κ1A jRω +
κ1B jcω +
-------------------------- - + =
c
C
n1 n2 +
---------------- -
c1n1 c2n2 +
n1 n2 +
--------------------------- - = =
A 2
D
T
--- -
c⎞
–
⎝ ⎠
⎛
n1
n1 n2 +
---------------- - R 1
4----
n1⎞
n2⎠
-
–
⎝
⎛
+ =
B
D
2
RT
2
--------- -
n1
n1 n2 +
---------------- - c 1
4----
n1⎞
n2⎠
-
–
⎝
⎛
+ =