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22. Eq. 1512 gives the angular velocity:
100 N/m
7.07rad/s.
2.00 kg
k
m
ω
==
=
Energy methods (discussed in §154) provide one method of solution. Here, we use
trigonometric techniques based on Eq. 153 and Eq. 156.
(a) Dividing Eq. 156 by Eq. 153, we obtain
=+
v
x
t
−
ωω
φ
tan
b
g
so that the phase (
t
+
) is found from
()
(
)
11
3.415 m/s
tan
tan
.
7.07 rad/s 0.129 m
v
t
x
ωφ
−−
§·
+=
=
¨¸
©¹
With the calculator in radians mode, this gives the phase equal to –1.31 rad. Plugging this
back into Eq. 153 leads to0.129m
cos( 1.31)
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 Spring '08
 Any
 Physics, Energy

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