Quiz 6
Name:
Math 240  Calculus III
March 17, 2009
Note:
In order to receive full credit, you must show work that justifies your answer.
Suppose the position at time
t
of an object in motion is given by
x
(
t
), satisfying
x

ω
2
x
=

g
sin(
ωt
)
with initial conditions
x
(0) = 0 and
x
(0) =
v
0
. Note that
ω
and
g
are fixed constants.
(a) (7 points) Find the equation of motion
x
(
t
).
Solution
: The complementary solution is
x
c
=
c
1
e
ωt
+
c
2
e

ωt
.
The particular solution has the form
x
p
=
A
sin(
ωt
) +
B
cos(
ωt
). Substituting
x
p
into
the differential equation, we have:
x
p

ω
2
x
p
=

2
ω
2
(
A
sin(
ωt
) +
B
cos(
ωt
)) =

g
sin(
ωt
)
.
Thus,
B
= 0 and

2
Aω
2
=

g
, so
A
=
g
2
ω
2
.
The general solution is then
x
g
(
t
) =
c
1
e
ωt
+
c
2
e

ωt
+
g
2
ω
2
sin(
ωt
).
The initial condition
x
(0) = 0 implies 0 =
c
1
+
c
2
, or
c
1
=

c
2
.
The initial condition
x
(0) =
v
0
implies
v
0
=
ωc
1

ωc
2
+
g
2
ω
=

2
ωc
2
+
g
2
ω
.
So
c
2
=
g

2
ωv
0
4
ω
2
=

c
1
, and the equation of motion is
x
(
t
) =
2
ωv
0

g
4
ω
2
e
ωt
+
g

2
ωv
0
4
ω
2
e

ωt
+
g
2
ω
2
sin(
ωt
)
.
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 Spring '07
 Cremins
 Calculus, Linear Algebra, Algebra, Sin

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