Husain, Zeena – Homework 13 – Due: Nov 25 2003, 4:00 am – Inst: H L Berk
1
This printout should have 19 questions, check
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The due time is Central time.
001
(part 1 oF 1) 10 points
Sound with a Frequency oF 437 Hz travels
through steel.
A wavelength oF 11
.
44 m is
measured.
What is the speed oF the sound in steel?
Correct answer: 4999
.
28 m
/
s.
Explanation:
v
=
fλ
= (437 Hz)(11
.
44 m)
= 4999
.
28 m
/
s
.
002
(part 1 oF 1) 10 points
A bat can detect small objects such as an
insect whose size is approximately equal to
one wavelength oF the sound the bat makes.
IF bats emit a chirp at a Frequency oF 25 kHz,
and iF the speed oF sound in air is 328 m
/
s,
what is the smallest insect a bat can detect?
Correct answer: 13
.
12 mm.
Explanation:
±rom
λ
=
v
f
,
we get
λ
=
(328 m
/
s)
(25000 Hz)
= 13
.
12 mm
.
003
(part 1 oF 1) 10 points
Karen shouts across a canyon and hears an
echo 6
.
2 s later.
Assume:
The speed oF sound is 343 m
/
s.
How wide is the canyon?
Correct answer: 1063
.
3 m.
Explanation:
Karen hears the sound aFter it had traveled
across the canyon and back, so
2
d
=
v t
d
=
v t
2
=
(343 m
/
s)(6
.
2 s)
2
= 1063
.
3 m
.
004
(part 1 oF 1) 10 points
The speed oF sound in air is
v
s
=
s
γ
P
ρ
the constant
γ
= 1
.
4,
P
is the air pressure,
and
ρ
is the density oF air.
Calculate the speed oF sound For
P
=
0
.
651 atm and
ρ
= 1
.
12 kg
/
m
3
.
Correct answer: 287
.
111 m
/
s.
Explanation:
Applying the Formula
v
=
s
γP
ρ
=
s
(1
.
4)(0
.
651 atm)(101300 Pa
/
atm)
1
.
12 kg
/
m
3
= 287
.
111 m
/
s
.
005
(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 7
.
72
◦
C, reaches the
opposite end oF the rod 9
.
32 ms later than
the sound transmitted through the rod. The
speed oF sound in copper is 3500 m
/
s and the
speed oF sound in air at 7
.
72
◦
C is 331 m
/
s.
What is the length oF the rod?
Correct answer: 3
.
40714 m.
Explanation:
The di²erence in the times oF travel oF
sound through the copper rod and the air
is
t
=
L
µ
1
v
air

1
v
Cu
¶
,
where
L
is the distance traveled (
i.e.
, the
length oF the rod). Then
L
=
t
•
v
air
v
Cu
v
Cu

v
air
‚
= (0
.
00932 s)
×
•
(331 m
/
s)(3500 m
/
s)
(3500 m
/
s)

(331 m
/
s)
‚
= 3
.
40714 m
.
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View Full DocumentHusain, Zeena – Homework 13 – Due: Nov 25 2003, 4:00 am – Inst: H L Berk
2
006
(part 1 of 1) 10 points
An experimenter wishes to generate in air a
sound wave that has a displacement ampli
tude equal to 3
×
10

6
m. The pressure am
plitude is to be limited to 0
.
267 Pa. Use the
equilibrium density of air to be 1
.
29 kg
/
m
3
and the speed of sound in air to be 343 m
/
s.
What
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 Fall '08
 Turner
 Work, Wavelength, Correct Answer, Standing wave

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