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{SECT 0 {PARA 256 "" 0 "" {TEXT 256 12 "MATH 2280-1 " }}{PARA 258 "" 
0 "" {TEXT 264 27 "MAPLE PROJECT : EARTHQUAKES" }}{PARA 259 "" 0 "" 
{TEXT 270 13 "March 6, 2006" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 318 "This handout might help for the Earthquake pro
ject in section 5.3 of the text, on pages 326-328.  You are mostly on \+
your own for this project, but here is a small example of a spring sys
tem worked out on Maple, so that you can get an idea about useful comm
ands to use.  This is the example we did by hand on Friday...." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 195 "This is \+
example 1 on page 317 of Edwards-Penney.  Initially it is an unforced \+
system with two masses and two springs, as you can see from the descri
ption on page 317.  We can write the system as " }{TEXT 257 2 "Mx" }
{TEXT 271 2 "''" }{TEXT 272 3 "=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 vector.  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 pr
oject" }}}{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 1 "x" }{TEXT 273 2 "''" }{TEXT 274 3 
"=Ax" }{TEXT -1 199 "; as we discussed on Friday, the eigenvectors of \+
A determine fundamental modes, and the corresponding negative eigenval
ues are the (opposites) of the squares of the corresponding angular fr
equencies:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "eigenvectors(A
);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$7%!$+\"\"\"\"<#-%'vectorG6#7$!\"
\"F%7%!#DF%<#-F(6#7$F%\"\"#" }}}{PARA 0 "" 0 "" {TEXT -1 286 "Therefor
e, the natural frequencies of this system are the 10 and 5, and the tw
o fundamental modes correspond to the masses moving in opposite direct
ions (with equal amplitudes and angular  frequency 10) and in parallel
 directions (with amplitude ratio of two and angular frequency 5).  " 
}}{PARA 0 "" 0 "" {TEXT -1 164 "     Now, let's consider the forced sy
stem with force vector equal to cos(wt)[0,50], i.e. the second mass is
 being forced periodically.  In other words, the system " }{TEXT 262 
11 "Mx''=Kx + F" }{TEXT -1 307 ", where F=cos(wt)[0,50] ; this is Exam
ple 3 on  page 323, and we also discussed this on Friday.  We follow t
he method described on that page to find a particular solution to the \+
forced oscillation problem, of the form given by equation (31).  Here \+
is the Maple version of the details summarized in the text:" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 452 "F0:=evalm(inverse(M)&*vector([0,50
]));\n    #The F0 in the normalized equation (32), page 323\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 multiplying\n    \+
#c on the left side of (32)\nc:=omega->evalm(-inverse(Aleft(omega))&*F
0);\n    #the solution vector c(omega) to (32),\n    #obtained by mult
iplying both sides of equation\n    #(32) on the left, by the inverse \+
to Aleft" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#F0G-%'vectorG6#7$\"\"!
\"#]" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%IdenG-%&arrayG6&%)identityG
;\"\"\"\"\"#F)7\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&AleftGf*6#%&om
egaG6\"6$%)operatorG%&arrowGF(-%&evalmG6#,&%\"AG\"\"\"*&)9$\"\"#F1%%Id
enGF1F1F(F(F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"cGf*6#%&omegaG6\"
6$%)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 323" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'v
ectorG6#7$,$*&\"%]7\"\"\",(\"%+DF**&\"$D\"F*)%&omegaG\"\"#F*!\"\"*$)F0
\"\"%F*F*F2F*,$*(\"#]F*,&\"#vF2*$F/F*F*F*F+F2F2" }}}{PARA 11 "" 1 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 636 "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 periodic solution to the damp
ed problem.  See the discussion on page 324.  We can study resonance p
henomena for these slightly damped problems by plotting the maximum am
plitude 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 maximum amplitude:" }}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 15 "norm(c(omega));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#-%$maxG6$,$*&\"%]7\"\"\"-%$absG6#,(\"%+DF)*&\"$D\"F))%&
omegaG\"\"#F)!\"\"*$)F2\"\"%F)F)F4F),$*&\"#]F)-F+6#*&,&\"#vF4*$F1F)F)F
)F-F4F)F)" }}}{PARA 0 "" 0 "" {TEXT -1 100 "Another way to measure the
 size of c(omega) is to take its Euclidean magnitude, which is the com
mand" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "norm(c(omega),2);" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&\"#]\"\"\",&*&\"$D'F&-%$absG6#,(
\"%+DF&*&\"$D\"F&)%&omegaG\"\"#F&!\"\"*$)F2\"\"%F&F&!\"#F&*$)-F+6#*&,&
\"#vF4*$F1F&F&F&F-F4F3F&F&#F&F3F&" }}}{PARA 0 "" 0 "" {TEXT -1 436 "(Y
ou will use the first command in the Earthquake project, which perhaps
 makes the most sense since it will be measuring the maximum amplitude
 that any floor oscillates.)  The following picture illustrates that t
he maximum amplitude of the particular solution blows up when omega is
 near the two natural angular frequences.  Thus, in the slightly dampe
d problem, one would experience practical resonance in the steady peri
odic solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 85 "plot(norm(
c(omega)),omega=0..15,amplitude=0..15,\n       numpoints=200,color=`bl
ack`);" }}{PARA 13 "" 1 "" {GLPLOT2D 352 182 182 {PLOTDATA 2 "6&-%'CUR
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+@x9F__l$\"3lSg\"zHb]8$F67$$\"3)Gf=(Qa'\\[\"F__l$\"3#>g1Gd3z3$F67$$\"3
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SLABELSG6$Q&omega6\"Q*amplitudeFf`p-%'COLOURG6&%$RGBGF)F)F)-%%VIEWG6$;
F(F^`pF_ap" 1 2 0 1 10 0 2 6 1 4 2 1.000000 45.000000 45.000000 0 0 "C
urve 1" }}}}{PARA 0 "" 0 "" {TEXT -1 208 "This is qualitatively the pi
cture on page 323, figure 5.3.10, although they plotted the Euclidean \+
magnitude of c(omega) rather than the maximum amplitude.  Notice how w
e get Maple to label the axes as desired" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 92 "      We can get a plot of resonan
ce as a function of period by recalling that 2*Pi/T=omega:" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 86 "plot(norm(c(2*Pi/period)),period=0.
1..3,amplitude=0..15,\nnumpoints=200,color=`black`);" }}{PARA 13 "" 1 
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!\"\"Fe_p;$Fg_pFg_p$\"#:Fg_p" 1 2 0 1 10 0 2 6 1 4 2 1.000000 
45.000000 45.000000 0 0 "Curve 1" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT 263 38 "COMMENTS FOR THE EARTHQUAKE PROJECT:  " 
}}{PARA 0 "" 0 "" {TEXT -1 89 "(1)  Students are often confused by the
 forcing term in equation (2) of page 327, namely " }}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 27 "E*(omega)^2*cos(omega*t)*b;" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#**%\"EG\"\"\")%&omegaG\"\"#F%-%$cosG6#*&F'F%%\"tGF
%F%%\"bGF%" }}}{PARA 0 "" 0 "" {TEXT -1 852 "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  there's a detailed discussion of this on page 4 of Frid
ay's notes.   here's a brief summary. Tink of the ground as the zeroth
 story.  In the rest frame it is shaking with oscillation Ecos(wt).  A
nd so its acceleration is its second time derivative, namely -E*w^2*co
s(wt).  If you write down the inhomogeneous system of EIGHT second ord
er 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 solution 8-vector to this system " }{TEXT 266 
1 "y" }{TEXT -1 79 "(t), then see what the shaking looks like to someo
ne on the ground by letting  " }{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 o
ther seven components will satisfy equation (2) on 327, exactly as the
 authors claim." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT 275 20 "Very important note:" }}{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 entries are rational number
s (expressed without decimal points), Maple tries to find the eigenval
ues and eigenvectors algebraically and exactly, instead of numerically
, and often fails. " }{TEXT 269 74 "Make sure at least one of your mat
rix entries has a decimal point in it.  " }}}{MARK "32 0" 0 }
{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }