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μ
1
=
ρ
1
AL
1
/
L
1
=
1
A
and
11
/.
A
ν
τρ
=
A similar expression holds for the wave speed in the steel section:
22
/.
vA
τρ
=
We note that the crosssectional area and the tension are the same for the
two sections. The equality of the frequencies for the two sections now leads to
11 1 2 2 2
,
nL
nL
ρρ
=
where
A
has been canceled from both sides. The ratio of the
integers is
()
()
33
22
2
33
1
11
0.866m
7.80 10 kg/m
2.50.
0.600m
2.60 10 kg/m
L
n
n
L
ρ
ρ
×
==
=
×
The smallest integers that have this ratio are
n
1
= 2 and
n
2
= 5. The frequency is
()
11
1
1
/2
/2
/
.
f
nv
L
n
L
A
τρ
==
The tension is provided by the hanging block and is
τ
=
mg
, where
m
is the mass of the
block. Thus,
()
()
()
()
(
)
2
33
6
2
11
10.0kg 9.80m/s
2
324Hz.
2
2 0.600m
2.60 10 kg/m
1.00 10 m
nm
g
f
LA
ρ
−
==
=
××
(b) The standing wave pattern has two loops in the aluminum section and five loops in
the steel section, or seven loops in all. There are eight nodes, counting the end points.
59. (a) The frequency of the wave is the same for both sections of the wire. The wave
speed and wavelength, however, are both different in different sections. Suppose there
are
n
1
loops in the aluminum section of the wire. Then,
L
1
=
n
1
λ
1
n
1
v
1
/2
f
,
where
λ
1
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This note was uploaded on 04/16/2011 for the course PHYSICS 191262 taught by Professor Najafzadeh during the Spring '09 term at The Petroleum Institute.
 Spring '09
 NAJAFZADEH
 mechanics

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