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{SECT 0 {PARA 256 "" 0 "" {TEXT 256 11 "MATH 2250-3" }}{PARA 258 "" 0 
"" {TEXT 264 23 "SPRINGS AND EARTHQUAKES" }}{PARA 259 "" 0 "" {TEXT 
269 17 "November 17, 2003" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 353 "Your final Maple project for Math 2250 this semes
ter is the Earthquake project on pages 437-438 of Edwards-Penney.  The
  project is at our home page http://www.math.utah.edu/~korevaar/2250f
all03/2250maple.html.  In these notes we will work through the book ex
amples from section 7.4, using illustrative Maple commands, as a warmu
p for your project work." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 245 "Let's start with example 1 on page 427 of Edwards
-Penney, which we have also been working by hand.  Initially it is an \+
unforced system with two masses and two springs, as you can see from t
he description on page 427.  We can write the system as " }{TEXT 257 
7 "Mx''=Kx" }{TEXT -1 8 ", where " }{TEXT 258 1 "M" }{TEXT -1 25 " is \+
the ``mass matrix'', " }{TEXT 259 1 "K" }{TEXT -1 31 " is the ``spring
 matrix'', and " }{TEXT 260 1 "x" }{TEXT -1 69 " is the displacement v
ector.  Following the book's notation, we enter" }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 46 "with(linalg):with(plots):with(DEtools): #tools
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 83 "M:=matrix([[2,0],[0,1]]
);\nK:=matrix([[-150,50],[50,-50]]);\nA:=evalm(inverse(M)&*K);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"MG-%'matrixG6#7$7$\"\"#\"\"!7$F+\"
\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"KG-%'matrixG6#7$7$!$]\"\"#
]7$F+!#]" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"AG-%'matrixG6#7$7$!#v
\"#D7$\"#]!#]" }}}{PARA 257 "" 1 "" {TEXT -1 39 "Then the system can a
lso be written as " }{TEXT 261 6 "x''=Ax" }{TEXT -1 176 ", and the eig
envectors of A determine fundamental modes, and the corresponding nega
tive eigenvalues are the (opposites) of the squares of the correspondi
ng angular frequencies:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "e
igenvects(A);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$7%!$+\"\"\"\"<#-%'vec
torG6#7$!\"\"F%7%!#DF%<#-F(6#7$F%\"\"#" }}}{PARA 0 "" 0 "" {TEXT -1 
286 "Therefore, the natural frequencies of this system are the 10 and \+
5, and the two fundamental modes correspond to the masses moving in op
posite directions (with equal amplitudes and angular  frequency 10) an
d in parallel directions (with amplitude ratio of two and angular freq
uency 5).  " }}{PARA 0 "" 0 "" {TEXT -1 164 "     Now, let's consider \+
the forced system with force vector equal to cos(wt)[0,50], i.e. the s
econd mass is being forced periodically.  In other words, the system \+
" }{TEXT 262 11 "Mx''=Kx + F" }{TEXT -1 322 ", where F=cos(wt)[0,50] d
iscussed on page 433.  We follow the method described on that page to \+
find a particular solution to the forced oscillation problem, of the f
orm given by equation (31).  The details of this computation are expla
ined in example 3 of the text and in our handwritten notes and here is
 a Maple version:" }}{PAGEBK }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 325 "F0:=evalm(inverse(M)&*vector([0,50
]));\n    #The F0 in the normalized equation (30), page 433\nIden:=arr
ay(1..2,1..2,identity);\n    #the 2 by 2 identity matrix\nAleft:=omega
->evalm(A + omega^2*Iden);\n    #the matrix function on the left side \+
of (32)\nc:=omega->evalm(-inverse(Aleft(omega))&*F0);\n     #the vecto
r c(omega) in (32)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#F0G-%'vectorG
6#7$\"\"!\"#]" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%IdenG-%&arrayG6&%)
identityG;\"\"\"\"\"#F)7\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&Aleft
Gf*6#%&omegaG6\"6$%)operatorG%&arrowGF(-%&evalmG6#,&%\"AG\"\"\"*&)9$\"
\"#F1%%IdenGF1F1F(F(F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"cGf*6#%&
omegaG6\"6$%)operatorG%&arrowGF(-%&evalmG6#,$-%#&*G6$-%(inverseG6#-%&A
leftG6#9$%#F0G!\"\"F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
37 "c(omega); #see equation (35) page 433" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#-%'vectorG6#7$,$*&\"%]7\"\"\",(\"%+DF**&\"$D\"F*)%&omeg
aG\"\"#F*!\"\"*$)F0\"\"%F*F*F2F*,$*(\"#]F*,&\"#vF2*$F/F*F*F*F+F2F2" }}
}{PARA 11 "" 1 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 612 "The ve
ctor c(w) above, times the oscillation cos(wt), is a particular soluti
on to the forced oscillation problem we are considering.  If we assume
 that our actual problem has a small amount of damping, then we expect
 that this particular solution is very close to the steady state solut
ion to the damped problem.  See the discussion on page 434.  We can st
udy resonance phenomena for these slightly damped problems by plotting
 the maximum amplitude of the steady state solutions to the undamped p
roblems, much like you did in the Tacoma Narrows project.  Use ``norm,
 infinity'' to measure this maximum amplitude:" }}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 24 "norm(c(omega),infinity);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#-%$maxG6$,$*&\"#]\"\"\"-%$absG6#*&,&\"#v!\"\"*$)%&omega
G\"\"#F)F)F),(\"%+DF)*&\"$D\"F)F2F)F0*$)F3\"\"%F)F)F0F)F),$*&\"%]7F)-F
+6#F5F0F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 77 "plot(norm(c(om
ega)),omega=0..15,y=0..15,\n       numpoints=200,color=`black`);" }}
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{PARA 0 "" 0 "" {TEXT -1 288 "This is the picture on page 434.  It mea
sures the maximum amount that one of the masses will be displaced from
 equilibrium, for the particular solution we derived.  Notice the peak
s are at angular frequency 5 and 10, reflecting the resonance which wi
ll occur at the natural frequencies.  " }}{PAGEBK }{PARA 0 "" 0 "" 
{TEXT -1 91 "     We can get a plot of resonance as a function of peri
od by recalling that 2*Pi/T=omega:" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 83 "res:=T->norm(c(2*Pi/T));\nplot(res(T),T=0.1..3,y=0..1
5,numpoints=200,color=`black`);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$
resGf*6#%\"TG6\"6$%)operatorG%&arrowGF(-%%normG6#-%\"cG6#,$*(\"\"#\"\"
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1 10 0 2 6 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 263 38 "COMMENTS FOR T
HE EARTHQUAKE PROJECT:  " }}{PARA 0 "" 0 "" {TEXT -1 89 "(1)  Students
 are often confused by the forcing term in equation (2) of page 438, n
amely " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "E*(omega)^2*cos(om
ega*t)*b;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#**%\"EG\"\"\")%&omegaG\"
\"#F%-%$cosG6#*&F'F%%\"tGF%F%%\"bGF%" }}}{PARA 0 "" 0 "" {TEXT -1 795 
"where b is the transpose of  [1,1,1,1,1,1,1].  They ask, ``how can th
e earthquake be forcing all seven stories, it seems like it's just sha
king the bottom one.''  Well, the students are correct, but so is Edwa
rds-Penney.  The authors talk about an ``opposite inertial force'' bei
ng the reason for this forcing term and  here's one way to think about
 it: Think of the ground as the zeroth story.  In the rest frame it is
 shaking with oscillation Ecos(wt).  And so its acceleration is its se
cond time derivative, namely -E*w^2*cos(wt).  If you write down the in
homogeneous system of EIGHT second order DE's for the accelerations of
 stories zero thru seven, the  forcing (well, accelerating) term is -E
*w^2*cos(wt)*[1,0,0,0,0,0,0,0], as you would expect.  Call the solutio
n 8-vector to this system " }{TEXT 265 1 "y" }{TEXT -1 78 "(t), then s
ee what the shaking looks like to someone on the ground by letting " }
}{PARA 0 "" 0 "" {TEXT 266 1 "x" }{TEXT -1 4 "(t)=" }{TEXT 267 1 "y" }
{TEXT -1 213 "(t)-E*cos(wt)*[1,1,1,1,1,1,1,1].  Then the zeroth story \+
component of x(t) will be identically zero, and the other seven compon
ents will satisfy equation (2) on the bottom of page 303, exactly as t
he authors claim." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 379 "(2)  For large matrices the eigenvect command won't work
 well unless Maple knows you want a decimal approximation:  If all ent
ries are rational numbers (expressed without decimal points), Maple tr
ies to find the eigenvalues and eigenvectors algebraically and exactly
, instead of numerically, and often fails.  The way around this is to \+
either ask for evalf(eigenvects(A)), or to " }{TEXT 268 74 "Make sure \+
at least one of your matrix entries has a decimal point in it.  " }}}
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1 1 }