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Exercises 4.8
It follows that
period =
2
π
ω
=
2
π
4
=
π
2
,
natural frequency =
ω
2
π
=
2
π
.
A general solution to (4.10), given in (4) on page 211 in the text, becomes
y
(
t
)=
C
1
cos
ωt
+
C
2
sin
ωt
=
C
1
cos 4
t
+
C
2
sin 4
t.
We Fnd
C
1
and
C
2
from the initial conditions.
y
(0) = (
C
1
cos 4
t
+
C
2
sin 4
t
)
±
±
t
=0
=
C
1
=
−
1
/
2
,
y
0
(0) = (
−
4
C
1
sin 4
t
+4
C
2
cos 4
t
)
±
±
t
=0
=4
C
2
=2
⇒
C
1
=
−
1
/
2
,
C
2
=1
/
2
.
Thus, the solution to the initial value problem is
y
(
t
)=
−
1
2
cos 4
t
+
1
2
sin 4
t
=
√
2
2
sin
²
4
t
−
π
4
³
,
where we have used formulas (6) rewriting the solution in form (5), page 211 in the text. The
amplitude of the motion therefore is
√
2
/
2.
Setting
y
= 0 in the above solution, we Fnd values of
t
when the mass passes through the
point of equilibrium.
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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