161
S
OUND AND
H
EARING
16.1.
I
DENTIFY
and
S
ET
U
P
:
Eq.(15.1) gives the wavelength in terms of the frequency. Use Eq.(16.5) to relate the
pressure and displacement amplitudes.
E
XECUTE
:
(a)
/
(344 m/s)/1000 Hz
0.344 m
v
f
λ
=
=
=
(b)
max
p
BkA
=
and
Bk
is constant gives
max1
1
max2
2
/
/
p
A
p
A
=
8
5
max2
2
1
2
max1
30 Pa
1.2
10
m
1.2
10
m
3.0
10
Pa
p
A
A
p
−
−
−
⎛
⎞
⎛
⎞
=
=
×
=
×
⎜
⎟
⎜
⎟
×
⎝
⎠
⎝
⎠
(c)
max
2
/
p
BkA
BA
π
λ
=
=
max
2
constant
p
BA
λ
π
=
=
so
max1
1
max2
2
p
p
λ
λ
=
and
2
max1
2
1
3
max2
3.0
10
Pa
(0.344 m)
6.9 m
1.5
10
Pa
p
p
λ
λ
−
−
⎛
⎞
⎛
⎞
×
=
=
=
⎜
⎟
⎜
⎟
×
⎝
⎠
⎝
⎠
/
(344 m/s)/6.9 m
50 Hz
f
v
λ
=
=
=
E
VALUATE
:
The pressure amplitude and displacement amplitude are directly proportional. For the same
displacement amplitude, the pressure amplitude decreases when the frequency decreases and the wavelength
increases.
16.2.
I
DENTIFY
:
Apply
max
p
BkA
=
and solve for
A
.
S
ET
U
P
:
2
k
π
λ
=
and
,
v
f
λ
=
so
2
f
k
v
π
=
and
2
.
fBA
p
v
π
=
E
XECUTE
:
2
12
max
9
(3.0
10
Pa) (1480 m s)
3.21
10
m.
2
2
(2.2
10
Pa) (1000 Hz)
p
v
A
π
Bf
π
−
−
×
=
=
=
×
×
E
VALUATE
:
Both
v
and
B
are larger, but
B
is larger by a much greater factor, so
/
v B
is a lot smaller and
therefore
A
is a lot smaller.
16.3.
I
DENTIFY
:
Use Eq.(16.5) to relate the pressure and displacement amplitudes.
S
ET
U
P
:
As stated in Example 16.1 the adiabatic bulk modulus for air is
5
1.42
10
Pa.
B
=
×
Use Eq.(15.1) to
calculate
λ
from
f
, and then
2
/
.
k
π
λ
=
E
XECUTE
:
(a)
150 Hz
f
=
Need to calculate
k
:
/
v
f
λ
=
and
2
/
k
π
λ
=
so
2
/
(2
rad)(150 Hz)/344 m/s
2.74 rad/m.
k
f
v
π
π
=
=
=
Then
3
max
(1.42
10
Pa)(2.74 rad/m)(0.0200
10
m)
7.78 Pa.
p
BkA
5
−
=
=
×
×
=
This is below the pain threshold of 30 Pa.
(b)
f
is larger by a factor of 10 so
2
/
k
f
v
π
=
is larger by a factor of 10, and
max
p
BkA
=
is larger by a factor of
10.
max
77.8 Pa,
p
=
above the pain threshold.
(c)
There is again an increase in
f
,
k
, and
max
p
of a factor of 10, so
max
778 Pa,
p
=
far above the pain threshold.
E
VALUATE
:
When
f
increases,
λ
decreases so
k
increases and the pressure amplitude increases.
16.4.
I
DENTIFY
:
Apply
max
.
p
BkA
=
2
2
,
f
k
v
π
π
λ
=
=
so
max
2
.
fBA
p
v
π
=
S
ET
U
P
:
344 m/s
v
=
E
XECUTE
:
3
max
5
6
(344 m/s)(10.0 Pa)
3.86
10
Hz
2
2
(1.42
10
Pa)(1.00
10
m)
vp
f
BA
π
π
−
=
=
=
×
×
×
E
VALUATE
:
Audible frequencies range from about 20 Hz to about 20,000 Hz, so this frequency is audible.
16.5.
I
DENTIFY
:
.
v
f
λ
=
Apply Eq.(16.7) for the waves in the liquid and Eq.(16.8) for the waves in the metal bar.
S
ET
U
P
:
In part (b) the wave speed is
4
1.50 m
3.90
10
s
d
v
t
−
=
=
×
16
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162
Chapter 16
E
XECUTE
:
(a)
Using Eq.(16.7),
2
2
(
λ
)
,
B
v
ρ
f
ρ
=
=
so
[
]
2
3
10
(8 m)(400 Hz)
(1300 kg
m
)
1.33
10
Pa.
B
=
=
×
(b)
Using Eq.(16.8),
2
2
2
4
3
10
(
)
(1.50 m)
(3.90
10
s)
(6400 kg
m
)
9.47
10
Pa.
Y
v
ρ
L t
ρ
−
⎡
⎤
=
=
=
×
=
×
⎣
⎦
E
VALUATE
:
In the liquid,
3200 m/s
v
=
and in the metal,
3850 m/s.
v
=
Both these speeds are much greater than
the speed of sound in air.
16.6.
I
DENTIFY
:
/ .
v
d t
=
Apply Eq.(16.7) to calculate
B
.
S
ET
U
P
:
3
3
3.3
10
kg/m
ρ
=
×
E
XECUTE
:
(a)
The time for the wave to travel to Caracas was 9 min
39 s
579 s
=
and the speed was
4
1.08
10
m/s.
×
Similarly, the time for the wave to travel to Kevo was 680 s for a speed of
4
1.28
10
m/s,
×
and the
time to travel to Vienna was 767 s for a speed of
4
1.26
10
m/s.
×
The average for these three measurements is
4
1.21
10
m/s.
×
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 Spring '08
 Jabbour
 Materials Science And Engineering

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