52. The nodes are located from vanishing of the spatial factor sin 5
π
x
= 0 for which the
solutions are
123
50
,
,
2
,
3
,
0
,
,
,
,
555
xx
π= π ππ
¡
=
!"
(a) The smallest value of
x
which corresponds to a node is
x
= 0.
(b) The second smallest value of
x
which corresponds to a node is
x
= 0.20 m.
(c) The third smallest value of
x
which corresponds to a node is
x
= 0.40 m.
(d) Every point (except at a node) is in simple harmonic motion of frequency
f
=
ω
/2
π
=
40
π
/2
π
= 20 Hz. Therefore, the period of oscillation is
T
= 1/
f
= 0.050 s.
(e) Comparing the given function with Eq. 16–58 through Eq. 16–60, we obtain
12
0.020sin(5
40 )
and
0.020sin(5
40 )
yx
t
t
=π
−
π
+
π
for the two traveling waves. Thus, we infer from these that the speed is
v
=
/
k
= 40
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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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