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48. (a) The nodes are located from vanishing of the spatial factor sin 5
πx
= 0 for which the solutions are
5
πx
=0
,π,
2
π,
3
π,.
..
=
⇒
x
=0
,
1
5
,
2
5
,
3
5
,...
so that the values of
x
lying in the allowed range are
x
=0,
x
=0
.
20 m, and
x
=0
.
40 m.
(b) 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.
(c) Comparing the given function with Eq. 1745 through Eq. 1747, we obtain
y
1
=0
.
020 sin(5
πx
−
40
πt
)a
n
d
y
2
=0
.
020 sin(5
πx
+40
πt
)
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

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