P18_100.PDF
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Chapter 18 - 18.100
100. (a) Since the speed of sound is lower in air than in water, the speed of sound in the air-water mixture
is lower than in pure water (see Table 18-1). Frequencyis proportional to the speed of sound (see
Eq. 18-39 and Eq. 18-41), so the decrease in speed is “heard” due to the accompanying decrease in
frequency.
(b) This follows from Eq. 18-3 and Eq. 18-2 (with ∆’s replaced byderivatives). Thus,
v
2
=
ρ
B
=
V
dV
=
ρ
V
dV
dp
.
dp
(c) Returning to the ∆ notation, and letting the absolute values be “understood,” we write ∆
V
=
∆
V
w
+∆
V
a
as indicated in the problem. Subject to the approximations mentioned in the problem,
our equation becomes
v
2
=
ρ
w
∆
V
w
∆
p
+
∆
V
a
∆
p
=
ρ
w
V
w
∆
V
w
∆
p
+
ρ
w
ρ
a
V
a
V
w
ρ
a
V
a
∆
V
a
∆
p
.
V
w
In a pure water system or a pure air system, we would have
1
v
w
=
ρ
w
V
w
∆
V
w
∆
p
or
1
v
a
=
ρ
a
V
a
∆
V
a
∆
p
.
Substituting these into the above equation, and using the notation
r
=
V
a
/V
w
, we arrive at
1
v
2
=
1
v
w
+
ρ
w
ρ
a
r
v
a
=
⇒
v
=
1
1
/v
w
+
r
(
ρ
w
/ρ
a
)
/v
a
.
(d) Dividing our result in the previous part by
v
w
and using the fact that the wave speed is proportional
to the frequency, we find
v
w
=
f
shift
v
=
1
v
w
1
/v
w
+
r
(
ρ
w
/ρ
a
)
/v
a
=
1
1+
r
(
ρ
w
/ρ
a
)(
v
w
/v
a
)
2
f
which becomes the expression shown in the problem when we plug in
ρ
w
= 1000kg
/
m
3
,
ρ
a
=
1
.
21kg
/
m
3
,
v
w
= 1482 m/s and
v
a
= 343 m/s, and round to three significant figures.
(e) The graph of
f
shift
/f
versus
r
is shown below.
1
0.8
frequency_ratio
0.6
0.4
0.2
volume_ratio
(f) From the graph (or more accuratelybysolving the equation itself) we find
r
=5
.
2
0
0.001
0.002
0.003
0.004
×
10
−
4
corresponds
to
f
shift
/f
=1
/
3.
1
ρ
1
Plik z chomika:
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Inne pliki z tego folderu:
P18_003.PDF
(62 KB)
P18_004.PDF
(58 KB)
P18_001.PDF
(61 KB)
P18_002.PDF
(62 KB)
P18_005.PDF
(60 KB)
Inne foldery tego chomika:
chap01
chap02
chap03
chap04
chap05
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