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54. (a) The frequency is
f
=1
/T
=1
/
4Hz
,so
v
=
fλ
=5
.
0cm/s
.
(b) We refer to the graph to see that the maximum transverse speed (which we will refer to as
u
m
)is
5
.
0 cm/s. Recalling from Ch
.
12 the simple harmonic motion relation
u
m
=
y
m
ω
=
y
m
2
πf
,wehave
5
.
0=
y
m
µ
2
π
1
4
¶
=
⇒
y
m
=3
.
2cm
.
(c) As already noted,
f
=0
.
25 Hz.
(d) Since
k
=2
π/λ
,w
ehav
e
k
=10
π
rad/m. There must be a sign diFerence between the
t
and
x
terms in the argument in order for the wave to travel to the right. The ±gure shows that at
x
=0,
the transverse velocity function is 0
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 Fall '08
 SPRUNGER
 Physics, Simple Harmonic Motion

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