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{SECT 0 {PARA 256 "" 0 "" {TEXT 256 10 "MATH 2280 " }}{PARA 258 "" 0 "
" {TEXT 264 22 "PROJECT 2: EARTHQUAKES" }}{PARA 259 "" 0 "" {TEXT 270 
10 "March 2001" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 240 "This is the Earthquake project on pages 333-334 of Edwar
ds-Penney.  You are mostly on your own for this project, but here is a
 small example of a spring system worked out on Maple, so that you can
 get an idea about useful commands to use.  " }{TEXT 265 1 " " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 204 "Let's st
art with Example 1 on page 323 of Edwards-Penney.  Initially it is an \+
unforced system with two masses and two springs, as you can see from t
he description on page 323.  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" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "with(linalg)
:with(plots):with(DEtools): #tools for project" }}}{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 also be written as " }
{TEXT 261 6 "x''=Ax" }{TEXT -1 176 ", and the eigenvectors of A determ
ine fundamental modes, and the corresponding negative eigenvalues are \+
the (opposites) of the squares of the corresponding angular frequencie
s:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "eigenvects(A);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6$7%!$+\"\"\"\"<#-%'vectorG6#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 fund
amental modes correspond to the masses moving in opposite directions (
with equal amplitudes and angular  frequency 10) and in parallel direc
tions (with amplitude ratio of two and angular frequency 5).  " }}
{PARA 0 "" 0 "" {TEXT -1 164 "     Now, let's consider the forced syst
em with force vector equal to cos(wt)[0,50], i.e. the second mass is b
eing forced periodically.  In other words, the system " }{TEXT 262 11 
"Mx''=Kx + F" }{TEXT -1 269 ", where F=cos(wt)[0,50] ; this is Example
 3 on  page 433.  We follow the method described on that page to find \+
a particular solution to the forced oscillation problem, of the form g
iven by equation (31).  Here is the Maple version of the details summa
rized in the text:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 452 "F0:=e
valm(inverse(M)&*vector([0,50]));\n    #The F0 in the normalized equat
ion (32), page 329\nIden:=array(1..2,1..2,identity);\n    #the 2 by 2 \+
identity matrix\nAleft:=omega->evalm(A + omega^2*Iden);\n    #the matr
ix function multiplying\n    #c on the left side of (32)\nc:=omega->ev
alm(-inverse(Aleft(omega))&*F0);\n    #the solution vector c(omega) to
 (32),\n    #obtained by multiplying both sides of equation\n    #(34)
 on the left, by the inverse to Aleft" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#>%#F0G-%'vectorG6#7$\"\"!\"#]" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%
%IdenG-%&arrayG6&%)identityG;\"\"\"\"\"#F)7\"" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%&AleftGR6#%&omegaG6\"6$%)operatorG%&arrowGF(-%&evalmG
6#,&%\"AG\"\"\"*&)9$\"\"#F1%%IdenGF1F1F(F(F(" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"cGR6#%&omegaG6\"6$%)operatorG%&arrowGF(-%&evalmG6#,
$-%#&*G6$-%(inverseG6#-%&AleftG6#9$%#F0G!\"\"F(F(F(" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 37 "c(omega); #see equation (35) page 329" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'vectorG6#7$,$*&\"\"\"F),(\"%+DF)*&
\"$D\"F))%&omegaG\"\"#F)!\"\"*$)F/\"\"%F)F)F1\"%]7,$*&,&!#vF)*$F.F)F)F
)F*F1!#]" }}}{PARA 11 "" 1 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 636 "The vector c(w) above, times the oscillation cos(wt), is a par
ticular solution to the forced oscillation problem we are considering.
  If we assume that our actual problem has a small amount of damping, \+
then we expect that this particular solution is very close to the stea
dy periodic solution to the damped problem.  See the discussion on pag
e 330.  We can study resonance phenomena for these slightly damped pro
blems by plotting the maximum amplitude of the steady state solutions \+
to the undamped problems.  That would be the maximum absolute value of
 c1 and c2 above.  Use the Maple command ``norm'' to measure this maxi
mum amplitude:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "norm(c(ome
ga));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$maxG6$,$*&\"\"\"F(-%$absG6
#,(\"%+DF(*&\"$D\"F()%&omegaG\"\"#F(!\"\"*$)F1\"\"%F(F(F3\"%]7,$-F*6#*
&,&!#vF(*$F0F(F(F(F,F3\"#]" }}}{PARA 0 "" 0 "" {TEXT -1 229 "(If you w
anted the actual amplitude of the solution when written in amplitude-p
hase form you would want the square root of c1^2 + c2^2, but that func
tion and the one above both blow up similarly near the two natural fre
quencies.)" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 85 "plot(norm(c(om
ega)),omega=0..15,amplitude=0..15,\n       numpoints=200,color=`black`
);" }}{PARA 13 "" 1 "" {GLPLOT2D 352 182 182 {PLOTDATA 2 "6&-%'CURVESG
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SLABELSG6$Q&omega6\"Q*amplitudeFf`p-%'COLOURG6&%$RGBGF)F)F)-%%VIEWG6$;
F(F^`pF_ap" 1 2 0 1 10 0 2 6 1 4 2 1.000000 45.000000 45.000000 0 0 "C
urve 1" }}}}{PARA 0 "" 0 "" {TEXT -1 183 "This is qualitatively the pi
cture on page 330, figure 5.3.10.  Notice the peaks at angular frequen
cy 5 and 10, corresponding to approximate resonance near the two funda
mental modes.  " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 92 "      We can get a plot of resonance as a function of per
iod by recalling that 2*Pi/T=omega:" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 102 "res:=T->norm(c(2*Pi/T));\nplot(res(period),period=0.
1..3,amplitude=0..15,\nnumpoints=200,color=`black`);" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#>%$resGR6#%\"TG6\"6$%)operatorG%&arrowGF(-%%normG6#-
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!\"\"Fe_p;$Fg_pFg_p$\"#:Fg_p" 1 2 0 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 THE EARTHQUAKE PROJECT:  " 
}}{PARA 0 "" 0 "" {TEXT -1 89 "(1)  Students are often confused by the
 forcing term in equation (2) of page 334, namely " }}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 27 "E*(omega)^2*cos(omega*t)*b;" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#**%\"EG\"\"\")%&omegaG\"\"#F%-%$cosG6#*&F'F%%\"tGF
%F%%\"bGF%" }}}{PARA 0 "" 0 "" {TEXT -1 869 "where b is the transpose \+
of  [1,1,1,1,1,1,1].  They ask, ``how can the earthquake be forcing al
l seven stories, it seems like it's just shaking the bottom one.''  We
ll, the students are correct, but so is Edwards-Penney.  The authors t
alk about an ``opposite inertial force'' being the reason for this for
cing term and  here's one way to think about it. Maybe your instructor
 can help you more if it's still confusing.  Anyhow, think of the grou
nd as the zeroth story.  In the rest frame it is shaking with oscillat
ion Ecos(wt).  And so its acceleration is its second time derivative, \+
namely -E*w^2*cos(wt).  If you write down the inhomogeneous system of \+
EIGHT second order DE's for the accelerations of stories zero thru sev
en, 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 solution 8-vector to this sys
tem " }{TEXT 266 1 "y" }{TEXT -1 78 "(t), then see what the shaking lo
oks like to someone on the ground by letting " }}{PARA 0 "" 0 "" 
{TEXT 267 1 "x" }{TEXT -1 4 "(t)=" }{TEXT 268 1 "y" }{TEXT -1 194 "(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 components will satisfy
 equation (2) on 334, exactly as the authors claim." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 301 "(2)  For large matrice
s the eigenvect command won't work well unless you enter at least one \+
decimal number; if all entries are rational numbers (expressed without
 decimal points), Maple tries to find the eigenvalues and eigenvectors
 algebraically and exactly, instead of numerically, and often fails. \+
" }{TEXT 269 74 "Make sure at least one of your matrix entries has a d
ecimal point in it.  " }}}{MARK "30 3" 163 }{VIEWOPTS 1 1 0 1 1 1803 
1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }