oldhomewk 36 – VALENCIA, DANIEL – Due: Apr 28 2008, 4:00 am
1
Question 1, chap 17, sect 3.
part 1 of 1
10 points
Given:
The speed of sound in air is 344 m
/
s.
A student uses an audio oscillator of ad
justable frequency to measure the depth of a
water well. Two successive resonant frequen
cies are heard at 99 Hz and 121 Hz.
What is the depth of the well?
Correct answer:
7
.
81818
m (tolerance
±
1
%).
Explanation:
Basic Concept:
For a tube closed at one
end, resonances occur at
λ
=
4
L
2
n

1
,
where
n
= 1
,
2
,
3
,
· · ·
.
Solution:
Call
L
the depth of the well and
v
s
the speed of sound. Then for some integer
n
L
= (2
n

1)
λ
1
4
= (2
n

1)
v
s
4
f
1
,
(1)
then
n
=
2
f
1
v
s
parenleftbigg
L
+
v
s
4
f
1
parenrightbigg
.
(2)
Presuming that
f
2
> f
1
, let
n
=
n
+ 1 for the
next resonant frequency
f
2
L
= [2 (
n
+ 1)

1]
λ
2
4
= (2
n
+ 1)
v
s
4
f
2
,
(3)
then
n
=
2
f
2
v
s
parenleftbigg
L

v
s
4
f
2
parenrightbigg
.
(4)
Eliminating
n
from Eqs. (2) and (4), we obtain
f
1
parenleftbigg
L
+
v
s
4
f
1
parenrightbigg
=
f
2
parenleftbigg
L

v
s
4
f
2
parenrightbigg
(
f
2

f
1
)
L
=
v
s
2
from where we can obtain
L
L
=
v
s
2 [
f
2

f
1
]
(5)
=
(344 m
/
s)
2 [(121 Hz)

(99 Hz)]
= 7
.
81818 m
.
From Eq. (4), we have
n
=
2
f
2
L
v
s

1
2
=
2 (121 Hz) (7
.
81818 m)
344 m
/
s

1
2
= 5
.
To check this result, using Eq. (1) and (3),
we have
L
= [2
n

1]
v
s
4
f
1
= [2 (5)

1]
(344 m
/
s)
4 (99 Hz)
= 7
.
81818 m
,
and
L
= [2
n
+ 1]
v
s
4
f
2
= [2 (5) + 1]
(344 m
/
s)
4 (121 Hz)
=
7
.
81818 m
.
The wavelengths are
λ
1
=
v
s
f
1
= 3
.
47475 m
λ
2
=
v
s
f
2
= 2
.
84298 m
.
Question 2, chap 17, sect 2.
part 1 of 1
10 points
A copper rod is given a sharp compressional
blow at one end.
The sound of the blow,
traveling through air at

4
.
24
◦
C, reaches the
opposite end of the rod 7
.
22 ms later than
the sound transmitted through the rod. The
speed of sound in copper is 3800 m
/
s and the
speed of sound in air at

4
.
24
◦
C is 331 m
/
s.
What is the length of the rod?
Correct answer:
2
.
61785
m (tolerance
±
1
%).
Explanation:
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oldhomewk 36 – VALENCIA, DANIEL – Due: Apr 28 2008, 4:00 am
2
The difference in the times of travel of
sound through the copper rod and the air
is
t
=
L
parenleftbigg
1
v
air

1
v
Cu
parenrightbigg
,
where
L
is the distance traveled (
i.e.
, the
length of the rod). Then
L
=
t
bracketleftbigg
v
air
v
Cu
v
Cu

v
air
bracketrightbigg
= (0
.
00722 s)
×
bracketleftbigg
(331 m
/
s) (3800 m
/
s)
(3800 m
/
s)

(331 m
/
s)
bracketrightbigg
= 2
.
61785 m
.
Question 3, chap 17, sect 2.
part 1 of 3
10 points
Compare sound intensities in terms of the
difference in decibels for two sound waves.
Denote the angular frequency, the amplitude
and the intensity of the first sound wave to be
ω
1
,
A
1
, and
I
1
, respectively, and those of the
second to be
ω
2
,
A
2
, and
I
2
, respectively.
If
ω
2
= 2
ω
1
and
A
2
=
A
1
, find
β
2

β
1
,
where
β
is the sound level in decibels.
1.
β
2

β
1
= 15
2.
β
2

β
1
= 9
3.
β
2

β
1
= 3
4.
β
2

β
1
= 6
correct
5.
β
2

β
1
= 12
6.
β
2

β
1
= 1
7.
β
2

β
1
= 4
8.
β
2

β
1
= 2
Explanation:
Let :
ω
2
= 2
ω
1
and
A
2
=
A
1
.
The sound level in decibels is
β
= 10 log
parenleftbigg
I
I
0
parenrightbigg
,
so
β
2

β
1
= 10 log
parenleftbigg
I
2
I
0
parenrightbigg

10 log
parenleftbigg
I
1
I
0
parenrightbigg
= 10 log
parenleftbigg
I
2
I
1
parenrightbigg
.
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 Spring '08
 Turner
 Frequency, Correct Answer, Daniel

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