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Homework Assignment #5 Solutions  1
ECE 473/TAM 413
Homework Assignment #5 Solutions
Hour Exam 1: Wednesday, March 5, 2007, 7:008:30 pm
Hour exams are open book (the text for the course) with one sheet of notes (you can use
both sides). Your class notes and lecture notes are not permitted. You can use calculators.
1. What SPL (re:
20
μ
Pa
rms
) would be necessary in air (at 20˚C) to cause an rms temperature
fluctuation of 1˚K? Provide your answer in dB. For your information, the answer to Kinsler et
al’s problem 5.9.2a is
!
T
T
o
=
" #
1
"
$
%
&
’
(
)
p
P
o
$
%
&
’
(
)
.
p
=
P
o
!
T
T
o
"
#
$
%
&
’
( )
1
(
"
#
$
%
&
’
=
101.3kPa
1˚K
293.16˚K
"
#
$
%
&
’
1.402
)
1
1.402
"
#
$
%
&
’
=
1205 Pa (rms)
SPL
=
20
!
log
10
1205 Pa
20
μ
Pa
"
#
$
%
&
’
=
155.6dB
2. Problem 5.12.8 in Kinsler et al. Use sea water at 13˚C.
a)
P
o
=
200 kPa
I
=
p
2
2
!
o
c
o
=
200,000 Pa
( )
2
2 1026 kg / m
3
( )
1500 m / s
( )
=
12,995 W / m
2
b)
ref:1
μ
bar
=
0.1Pa
!
RMS value!
SPL
=
20
!
log
10
p
p
ref
"
#
$
%
&
’
=
20
!
log
10
200,000 Pa
!
"
#
$
%
&
’
=
123dB
c)
s
o
=
k
!
o
=
kU
o
"
=
P
o
#
c
2
=
200,000 Pa
1026 kg / m
3
( )
1500 m / s
( )
2
=
8.66
$
10
%
5
d) Recall the expression
p
=
!
gh
. In the case of water depth h,
!
=
1026 kg / m
3
and
g
=
9.8m / s
2
. Also, recall that the pressure at the surface of the sea (sea level) is 1 atm = 101
kPa. Therefore, the additional pressure from water is 99 kPa (200101 kPa). Thus,
h
=
p
!
g
=
99 kPa
1026 kg / m
3
( )
9.8 m / s
2
( )
=
9.8 m
Thus, the water pressure of
P
o
=
200 kPa
is at a depth of 9.8 m
3. Calculate the missing values for a spherical outgoing wave propagating in the two liquids.
distance from source (m)
0.1
0.1
frequency (kHz)
3
30
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View Full DocumentHomework Assignment #5 Solutions  2
angular frequency (kr/s)
18.850
188.496
propagation speed (m/s)
1500
1500
ambient temperature (˚C)
20
20
wave number (1/m)
12.57
125.66
wavelength (m)
0.500
0.050
equilibrium density (kg/m
3
)
1000
1000
ratio of specific heats (
γ
)
1.01
1.01
equilibrium pressure (kPa)
100
100
isothermal bulk modulus (GPa)
2.228
2.228
adiabatic bulk modulus (GPa)
2.250
2.250
peak particle displacement amplitude (μm)
6.984
0.548
peak particle velocity amplitude (mm/s)
131.6
103.3
peak particle acceleration amplitude (m/s
2
)
2481.4
19477.6
peak condensation amplitude (x10
6
)
87.76
68.89
peak acoustic pressure amplitude (kPa)
154.510
154.510
peak excess density amplitude (g/m
3
)
87.8
68.9
SPL (re: 1 μPa
rms
)
220.8
220.8
characteristic acoustic impedance (Mrayl)
1.500
1.500
average potential energy density (mJ/m
3
)
2653
2653
average kinetic energy density (mJ/m
3
)
4332
2669
average energy density (mJ/m
3
)
6985
5322
average acoustic intensity (W/m
2
)
7958
7958
total power of source (kW)
1.0
1.0
4. For a gas, develop an expression for the ratio of the discontinuity distance L
d
to the shock
distance D, that is, L
d
/D.
L
d
=
1
!
Mk
=
2c
o
2
"
+
1
( )
#
U
o
and
D
=
2
!"
B/ A
P
o
p
c
+
p
r
=
!
c
o
/f
( )
! #
1
P
o
p
o
L
d
D
=
! "
1
( )
!
+
1
( )
#
5. a) A 20kHz sonar source produces sound underwater in fresh water (assume B/A = 5 for
water and 1.1 atmosphere ambient pressure at the depth of the source at 20˚C). If fully
developed shock wave formation occurs (the discontinuity distance) at a distance of 1 m, then
determine the SPL (re: 20 μPa rms) of the source. b) If the sonar source operates at 2 MHz
(typically of diagnostic ultrasound frequencies) with the same SPL, then determine the distance
that shock formation would occur.
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