homework 13 – GAUTHIER, JOSEPH – Due: Dec 4 2007, 1:00 am
1
Question 1, chap 17, sect 2.
part 1 of 1
10 points
A rifle is fired in a valley with parallel ver
tical walls.
The echo from one wall is heard
1
.
37 s after the rifle was fired. The echo from
the other wall is heard 3
.
64 s after the first
echo.
The velocity of sound is
v
= 343 m
/
s
.
How wide is the valley?
Correct answer:
1094
.
17
m (tolerance
±
1
%).
Explanation:
Let :
t
1
= 1
.
37 s
,
t
2
= 3
.
64 s
,
and
v
= 343 m
/
s
.
Let the distance from the closer wall be
d
1
and the distance from the farther wall be
d
2
.
The first echo is heard after traveling a
distance of 2
d
1
=
v t
1
.
The second echo
is heard after traveling a distance of 2
d
2
=
v
(
t
1
+
t
2
). Each time the sound traveled from
your position to the wall and back to you.
2
d
=
v t
1
+
v
(
t
1
+
t
2
) =
v
(2
t
1
+
t
2
)
Thus the distance
d
between these walls is
d
=
v
bracketleftBig
2
t
1
+
t
2
bracketrightBig
2
=
(343 m
/
s)
bracketleftBig
2 (1
.
37 s) + (3
.
64 s)
bracketrightBig
2
=
1094
.
17 m
.
Question 2, chap 17, sect 2.
part 1 of 1
10 points
A rock band plays at a 72 dB sound level.
How
many
times
greater
is
the
sound
pressure from another rock band playing at
127 dB?
Correct answer: 562
.
341
(tolerance
±
1 %).
Explanation:
Let :
β
1
= 72 dB
and
β
2
= 127 dB
.
For a given sound level, every increase of
20 dB corresponds to a pressure amplitude 10
times larger than the original pressure ampli
tude, so
n
= 10
(
β
2
−
β
1
)
/
20
= 10
(127 dB
−
72 dB)
/
(20 dB)
=
562
.
341
.
Question 3, chap 17, sect 2.
part 1 of 1
10 points
A stereo speaker is placed between two ob
servers who are 121 m apart, along the line
connecting them.
If one observer records an intensity level of
24
.
6 dB, and the other records an intensity
level of 29
.
6 dB, how far is the speaker from
the closer observer?
Correct answer:
43
.
5521
m (tolerance
±
1
%).
Explanation:
Let :
β
1
= 24
.
6 dB
,
β
2
= 29
.
6 dB
,
and
L
= 121 m
.
I
∝
r
−
2
, so
I
2
I
1
=
r
2
1
r
2
2
.
β
1
= 10 log
parenleftbigg
I
1
I
0
parenrightbigg
and
β
2
= 10 log
parenleftbigg
I
2
I
0
parenrightbigg
,
so
β
2
−
β
1
= 10 log
parenleftbigg
I
2
I
1
parenrightbigg
= 10 log
parenleftbigg
r
1
r
2
parenrightbigg
2
= 20 log
parenleftbigg
r
1
r
2
parenrightbigg
β
2
−
β
1
20
= log
parenleftbigg
r
1
r
2
parenrightbigg
r
1
r
2
= 10
(
β
2
−
β
1
)
/
20
r
1
= 10
(
β
2
−
β
1
)
/
20
r
2
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homework 13 – GAUTHIER, JOSEPH – Due: Dec 4 2007, 1:00 am
2
L
=
r
2
+
r
1
,
=
r
2
+ 10
(
β
2
−
β
1
)
/
20
r
2
,
so
r
2
=
L
1 + 10
(
β
2
−
β
1
)
/
20
=
121 m
1 + 10
(29
.
6 dB
−
24
.
6 dB)
/
(20 dB)
=
43
.
5521 m
.
Question 4, chap 17, sect 2.
part 1 of 1
10 points
A vacuum cleaner has a measured sound
level of 70
.
5 dB.
What is the intensity of this sound
in
W
/
m
2
?
Correct answer: 1
.
12202
×
10
−
5
W
/
m
2
(tol
erance
±
1 %).
Explanation:
Let :
β
= 70
.
5 dB
.
β
= 10 log
parenleftbigg
I
I
0
parenrightbigg
β
10
= log
parenleftbigg
I
I
0
parenrightbigg
I
=
I
0
10
β/
10
= (1
×
10
−
12
W
/
m
2
) 10
(70
.
5 dB)
/
(10 dB)
=
1
.
12202
×
10
−
5
W
/
m
2
Question 5, chap 17, sect 3.
part 1 of 1
10 points
An open pipe 0
.
54 m in length is placed
vertically in a cylindrical bucket with a cross
sectional area of 0
.
069 m
2
.
Water is poured
into the bucket until a tuning fork of fre
quency 812 Hz, placed over the pipe, produces
resonance.
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 Fall '07
 Gilbert
 Decibel, Correct Answer, Kilogram

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