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{SECT 0 {PARA 256 "" 0 "" {TEXT 256 10 "MATH 2250 " }}{PARA 259 "" 0 "
" {TEXT 265 22 "PROJECT 3: EARTHQUAKES" }}{PARA 260 "" 0 "" {TEXT 275 
14 "November, 2000" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 282 "Your final project for Math 2250 this semester is the Ea
rthquake project on pages 437-438 of Edwards-Penney.  You are mostly o
n your own for this project, but here is a small example of a spring s
ystem worked out on Maple, so that you can get an idea about useful co
mmands to use.  " }{TEXT 266 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 204 "Let's start with example 1 on page 427 o
f Edwards-Penney.  Initially it is an unforced system with two masses \+
and two springs, as you can see from the description on page 427.  We \+
can write the system as " }{TEXT 258 7 "Mx''=Kx" }{TEXT -1 8 ", where \+
" }{TEXT 259 1 "M" }{TEXT -1 25 " is the ``mass matrix'', " }{TEXT 
260 1 "K" }{TEXT -1 31 " is the ``spring matrix'', and " }{TEXT 261 1 
"x" }{TEXT -1 69 " is the displacement vector.  Following the book's n
otation, we enter" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "with(li
nalg):with(plots):with(DEtools): #tools for project" }}{PARA 7 "" 1 "
" {TEXT -1 45 "Warning, the name adjoint has been redefined\n" }}}
{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 258 "" 1 "" {TEXT -1 39 "Then the system can also be wr
itten as " }{TEXT 262 6 "x''=Ax" }{TEXT -1 176 ", and the eigenvectors
 of A determine fundamental modes, and the corresponding negative eige
nvalues are the (opposites) of the squares of the corresponding angula
r frequencies:" }}{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 "Theref
ore, the natural frequencies of this system are the 10 and 5, and the \+
two fundamental modes correspond to the masses moving in opposite dire
ctions (with equal amplitudes and angular  frequency 10) and in parall
el directions (with amplitude ratio of two and angular frequency 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 second mass \+
is being forced periodically.  In other words, the system " }{TEXT 
263 11 "Mx''=Kx + F" }{TEXT -1 296 ", where F=cos(wt)[0,50] discussed \+
on page 433.  We follow the method described on that page to find a pa
rticular solution to the forced oscillation problem, of the form given
 by equation (31).  The details of this computation are explained in e
xample 3 of the text, and here is the Maple version:" }}{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:=array(1..
2,1..2,identity);\n    #the 2 by 2 identity matrix\nAleft:=omega->eval
m(A + omega^2*Iden);\n    #the matrix function on the left side of (32
)\nc:=omega->evalm(-inverse(Aleft(omega))&*F0);\n     #the vector c(om
ega) in (32)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#F0G-%'vectorG6#7$\"
\"!\"#]" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%IdenG-%&arrayG6&%)identi
tyG;\"\"\"\"\"#F)7\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&AleftGR6#%&
omegaG6\"6$%)operatorG%&arrowGF(-%&evalmG6#,&%\"AG\"\"\"*&)9$\"\"#F1%%
IdenGF1F1F(F(F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"cGR6#%&omegaG6
\"6$%)operatorG%&arrowGF(-%&evalmG6#,$-%#&*G6$-%(inverseG6#-%&AleftG6#
9$%#F0G!\"\"F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "c(ome
ga); #see equation (35) page 433" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%
'vectorG6#7$,$*&\"\"\"F),(\"%+DF)*&\"$D\"F))%&omegaG\"\"#F)!\"\"*$)F/
\"\"%F)F)F1\"%]7,$*&,&!#vF)*$F.F)F)F)F*F1!#]" }}}{PARA 11 "" 1 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 603 "The vector c(w) above, t
imes the oscillation cos(wt), is a particular solution to the forced o
scillation problem we are considering.  If we assume that our actual p
roblem has a small amount of damping, then we expect that this particu
lar solution is very close to the steady state solution to the damped \+
problem.  See the dsiscussion on page 434.  We can study resonance phe
nomena for these slightly damped problems by plotting the maximum ampl
itude of the steady state solutions to the undamped problems, much lik
e you did in the Tacoma Narrows project.  Use ``norm'' to measure this
 maximum amplitude:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "norm(
c(omega));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$maxG6$,$-%$absG6#*&,&
!#v\"\"\"*$)%&omegaG\"\"#F-F-F-,(\"%+DF-*&\"$D\"F-F/F-!\"\"*$)F0\"\"%F
-F-F6\"#],$*&F-F--F(6#F2F6\"%]7" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 77 "plot(norm(c(omega)),omega=0..15,y=0..15,\n       numpoints=200
,color=`black`);" }}{PARA 13 "" 1 "" {GLPLOT2D 282 153 153 {PLOTDATA 
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1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{PARA 0 "" 0 "" {TEXT 
-1 142 "This is the picture on page 434.  Notice the peaks at angular \+
frequency 5 and 10, corresponding to resonance with the two fundamenta
l modes.  " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 92 "      We can get a plot of resonance as a function of period by
 recalling that 2*Pi/T=omega:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 83 "res:=T->norm(c(2*Pi/T));\nplot(res(T),T=0.1..3,y=0..15,numpoints
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{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 264 38 "COMMENTS
 FOR THE EARTHQUAKE PROJECT:  " }}{PARA 0 "" 0 "" {TEXT -1 89 "(1)  St
udents are often confused by the forcing term in equation (2) of page \+
438, namely " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "E*(omega)^2*
cos(omega*t)*b;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#**%\"EG\"\"\")%&ome
gaG\"\"#F%-%$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 all seven stories, it seems like it's j
ust shaking the bottom one.''  Well, the students are correct, but so \+
is Edwards-Penney.  The authors talk about an ``opposite inertial forc
e'' being the reason for this forcing term and  here's one way to thin
k about it. Maybe your instructor can help you more if it's still conf
using.  Anyhow, think of the ground as the zeroth story.  In the rest \+
frame it is shaking with oscillation Ecos(wt).  And so its acceleratio
n 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 accel
erations 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 solution 8-vector to this system " }{TEXT 267 1 "y" }{TEXT -1 78 "
(t), then see what the shaking looks like to someone on the ground by \+
letting " }}{PARA 0 "" 0 "" {TEXT 268 1 "x" }{TEXT -1 4 "(t)=" }{TEXT 
269 1 "y" }{TEXT -1 213 "(t)-E*cos(wt)*[1,1,1,1,1,1,1,1].  Then the ze
roth story component of x(t) will be identically zero, and the other s
even components will satisfy equation (2) on the bottom of page 303, e
xactly as the authors claim." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 301 "(2)  For large matrices the eigenvect command \+
won't work well unless you enter at least one decimal number; if all e
ntries are rational numbers (expressed without decimal points), Maple \+
tries to find the eigenvalues and eigenvectors algebraically and exact
ly, instead of numerically, and often fails. " }{TEXT 270 74 "Make sur
e at least one of your matrix entries has a decimal point in it.  " }}
}{MARK "3 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 
33 1 1 }