Version PREVIEW – Homework 2 – Savrasov – (39825)
1
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Displacement on a Wave
001
10.0 points
Calculate the displacement at position
17
.
5 m and time 1
.
48 s For a sound wave in air
described by cos(
x, t, φ
)
,
where
φ
= 0 when
x
= 0 and
t
= 0
.
The wavelength is 0
.
0639 m,
the speed oF sound is 343 m
/
s, the density oF
the air is 1
.
35 kg
/
m
3
, and the pressure ampli
tude is 0
.
262 Pa.
Correct answer:
−
7
.
84692
×
10
−
9
m.
Explanation:
Let :
x
= 17
.
5 m
,
t
= 1
.
48 s
,
λ
= 0
.
0639 m
,
v
= 343 m
/
s
,
ρ
= 1
.
35 kg
/
m
3
,
and
Δ
P
max
= 0
.
262 Pa
.
The wave number is
k
=
2
π
λ
, its angular
Frequency is
ω
= 2
π f
=
2
π v
λ
, the maximum
amplitude is
s
m
=
Δ
P
max
ρ v ω
=
Δ
P
max
λ
ρ v
·
2
π v
=
(0
.
262 Pa) (0
.
0639 m)
(1
.
35 kg
/
m
3
) (343 m
/
s)
·
2
π
(343 m
/
s)
= 1
.
67765
×
10
−
8
m
,
and the angle is
θ
=
k x
−
ω t
=
2
π x
λ
−
2
π v t
λ
=
2
π
λ
(
x
−
v t
)
=
2
π
0
.
0639 m
(17
.
5 m
−
343 m
/
s 1
.
48 s)
=
−
48194
.
7 rad
,
so the displacment is
s
=
s
m
sin(
k x
−
ω t
)
= (1
.
67765
×
10
−
8
m)
×
sin(
−
48194
.
7 rad)
=
−
7
.
84692
×
10
−
9
m
.
Explosion in the Atmosphere
002
10.0 points
Sound waves in air are absorbed at a rate
oF approximately 9
.
52 dB
/
km. An explosive
charge is detonated at a height oF several kilo
meters in the atmosphere. At a distance oF
564 m From the explosion the acoustic pres
sure reaches a maximum oF 7
.
16 N
/
m
2
.
What will be the sound level 6
.
28 km From
the explosion? Consider the atmosphere to be
homogeneous over the distances considered.
The density oF the air is 1
.
24 kg
/
m
3
and the
speed oF sound is 343 m
/
s.
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 Fall '09
 RichardScalettar
 Work, Wavelength, 0.000486093 W, 1.24638 m, 3 7.26 W

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