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88. (a) The frequency is
f
= 1/
T
= 1/4 Hz, so
v
=
f
λ
= 5.0 cm/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. 11 the simple harmonic motion relation
u
m
=
y
m
ω
=
y
m
2
π
f
, we have
1
5.0
2
3.2 cm.
4
mm
yy
§·
=π
¡
=
¨¸
©¹
(c) As already noted,
f
= 0.25 Hz.
(d) Since
k
= 2
π
/
λ
, we have
k
= 10
π
rad/m. There must be a sign difference between the
t
and
x
terms in the argument in order for the wave to travel to the right. The figure shows
that at
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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, Simple Harmonic Motion

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