Chapter 5
For the mass
m
2
there are two forces acting on it: the force due to the spring with constant
k
2
is
−
k
2
y
; and the force due to the spring with constant
k
1
is
k
1
(
y
−
x
). So we get
m
2
y
0
=
k
1
(
x
−
y
)
−
k
2
y.
So the system is
m
1
x
0
=
k
1
(
y
−
x
)
,
m
2
y
0
=
−
k
1
(
y
−
x
)
−
k
2
y,
or, in operator form,
(
m
1
D
2
+
k
1
)
[
x
]
−
k
1
y
=0
,
−
k
1
x
+
±
m
2
D
2
+(
k
1
+
k
2
)
²
[
y
]=0
.
With
m
1
=1
,
m
2
=2,
k
1
=4
,and
k
2
=10
/
3, we get
(
D
2
+4)[
x
]
−
4
y
,
−
4
x
+(2
D
2
+22
/
3) [
y
,
(5.28)
with initial conditions:
x
(0) =
−
1
,x
0
(0) = 0
,y
(0) = 0
0
(0) = 0
.
Multiplying the second equation of the system given in (5.28) by 4, applying (2
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

Click to edit the document details