Name________________________
Student I.D.___________________
Math 2280-001
Quiz 9 SOLUTIONS
March 27, 2015
1a) Use the methods we've been discussing to find the general solution to the system of differential
equations
x1 ## t
=
x2 ## t
1
x1
2 K4
x2
K3
.
(8 points)
Hint: This second order system of DE's could be modeling a two-mass, three-spring system without
damping and so it will have solutions that oscillate.
We find the eigenvalues and eigenvectors of the matrix. Then get solutions cos w t v, sin w t v with
w=
Kl . (why?)
det
K3 K l
2
1
2
K4 K l
= l C 7 l C 10 = l C 5
lC2
so the eigenvalues are l =K5,K2.
El =K2 : Solving (A C 2 I v = 0 is the system
K1
so v =
1
1
0
2 K2
0
is an eigenbasis.
1
El =K5 :
so v =
1
K2
0
2 1
0
is an eigenbasis.
For l =K2, w =
x1 t
x2 t
2 1
2 ; for l =K5, w =
= c1 cos
5 . So,
2 t C c2 sin
2t
1
1
C c3 cos
5 t C c4 sin
5t
1
K2
.
(The first two terms represent the slower in-phase mode, and the last two terms represent the faster out of
phase mode.)
1
, k = 2, k2 = 1, k3 = 1. Show that the displacements x1 t , x2 t of the two
2 1
masses from equilibrium in the configuration below satisfy the system in part (a), i.e.
x1 ## t =K3 x1 C x2
x2 ## t = 2 x1 K 4 x2
1b) Set m1 = 1, m2 =
Hint: Use Newton's second law that mass times acceleration equals net forces.
(2 points)
m1 x1 ## t =Kk1 x1 C k2 x2 K x1 =K k1 C k2 x1 C k2 x2
0 x1 ## t =K3 x1 C x2 .
m2 x2 ## t =Kk2 x2 K x1 K k3 x2 = k2 x1 K k2 C k3 x2
1
0 x2 ## t = x1 K 2 x2
2
0 x2 ## t = 2 x1 K 4 x2 .