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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 66 "Computer Problem 2: The u
nicycle model of an automobile suspension" }}}{EXCHG {PARA 3 "" 0 "" 
{TEXT -1 12 "Introduction" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 20 "Form
ula (21), p.165 " }}{PARA 256 "" 0 "" {XPPEDIT 18 0 "G = F[0]/sqrt((k-
m*omega^2)^2+(c*omega)^2);" "6#/%\"GG*&&%\"FG6#\"\"!\"\"\"-%%sqrtG6#,&
*$,&%\"kGF**&%\"mGF**$%&omegaG\"\"#F*!\"\"\"\"#F**$*&%\"cGF*F5F*\"\"#F
*F7" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 96 "gives the amplitude
 of vibration of the steady-state solution to the damped mass-spring e
quation" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "m*`@@`(D,2)(x)+c*D(x)+k*x \+
= F[0]*cos(omega*t);" "6#/,(*&%\"mG\"\"\"--%#@@G6$%\"DG\"\"#6#%\"xGF'F
'*&%\"cGF'-F,6#F/F'F'*&%\"kGF'F/F'F'*&&%\"FG6#\"\"!F'-%$cosG6#*&%&omeg
aGF'%\"tGF'F'" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 38 "The first
 thing we want to do is plot " }{XPPEDIT 18 0 "G;" "6#%\"GG" }{TEXT 
-1 39 " vesrsus the driving angular frequency " }{XPPEDIT 18 0 "omega;
" "6#%&omegaG" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "Gx := F0/s
qrt((k-m*omega^2)^2+(c*omega)^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>
%#GxG*&%#F0G\"\"\"*$-%%sqrtG6#,**$)%\"kG\"\"#F'\"\"\"*(F/F1%\"mGF1)%&o
megaGF0F'!\"#*&)F3F0F')F5\"\"%F'F1*&)%\"cGF0F'F4F'F1F'!\"\"" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 15 "We now use the " }{TEXT 257 7 "una
pply" }{TEXT -1 17 " command to make " }{XPPEDIT 18 0 "G;" "6#%\"GG" }
{TEXT -1 15 " a function of " }{XPPEDIT 18 0 "omega;" "6#%&omegaG" }
{TEXT -1 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "G := unapp
ly(Gx,omega);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"GGR6#%&omegaG6\"6
$%)operatorG%&arrowGF(*&%#F0G\"\"\"*$-%%sqrtG6#,**$)%\"kG\"\"#F.\"\"\"
*(F6F8%\"mGF8)9$F7F.!\"#*&)F:F7F.)F<\"\"%F.F8*&)%\"cGF7F.F;F.F8F.!\"\"
F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 100 "Later we will want to \+
plot solutions to the mass-spring differential equation so let's enter
 it also" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "diffeq := m*dif
f(x(t),t$2)+c*diff(x(t),t)+k*x(t) = F0*cos(omega*t);" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#>%'diffeqG/,(*&%\"mG\"\"\"-%%diffG6$-%\"xG6#%\"tG-%
\"$G6$F0\"\"#F)F)*&%\"cGF)-F+6$F-F0F)F)*&%\"kGF)F-F)F)*&%#F0GF)-%$cosG
6#*&%&omegaGF)F0F)F)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 47 "Let us us
e the parameters of Example 6, p. 166:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 34 "m := 1; c := 2; k := 26; F0 := 82;" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%\"mG\"\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"c
G\"\"#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"kG\"#E" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#>%#F0G\"##)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 9 "P
lotting " }{XPPEDIT 18 0 "G;" "6#%\"GG" }{TEXT -1 9 " against " }
{XPPEDIT 18 0 "omega;" "6#%&omegaG" }{TEXT -1 27 " gives Figure 2.6.9,
 p. 167" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "plot(G(omega),om
ega=0..10);" }}{PARA 13 "" 1 "" {GLPLOT2D 399 225 225 {PLOTDATA 2 "6%-
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}}}{EXCHG {PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 31 "The resonant angular frequency " }{XPPEDIT 18 0 "omega[1]
;" "6#&%&omegaG6#\"\"\"" }{TEXT -1 94 "  for a damped mass-spring syst
em is defined to be the value of the driving angular frequency " }
{XPPEDIT 18 0 "omega;" "6#%&omegaG" }{TEXT -1 125 " which maximizes th
e amplitude of the steady-state oscillations. We can see form the grap
h above that it is somewhere around " }{XPPEDIT 18 0 "omega = 5;" "6#/
%&omegaG\"\"&" }{TEXT -1 30 ". To find it exactly we solve " }
{XPPEDIT 18 0 "d*G/(d*omega) = 0;" "6#/*(%\"dG\"\"\"%\"GGF&*&F%F&%&ome
gaGF&!\"\"\"\"!" }{TEXT -1 5 " for " }{XPPEDIT 18 0 "omega;" "6#%&omeg
aG" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "solve
(diff(G(omega),omega)=0,omega);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6%\"
\"!,$*$-%%sqrtG6#\"\"'\"\"\"\"\"#,$F%!\"#" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 42 "and thus the resonant angular frequency is" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "omega1 := 2*sqrt(6);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%'omega1G,$*$-%%sqrtG6#\"\"'\"\"\"\"\"#" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 137 "We now want to plot some solution
s to the differential equations corresponding to different values of t
he driving angular frequency, say " }{XPPEDIT 18 0 "omega = omega[1];
" "6#/%&omegaG&F$6#\"\"\"" }{TEXT -1 1 "," }{XPPEDIT 18 0 "omega[1]/2;
" "6#*&&%&omegaG6#\"\"\"\"\"\"\"\"#!\"\"" }{TEXT -1 6 ", and " }
{XPPEDIT 18 0 "2*omega[1];" "6#*&\"\"#\"\"\"&%&omegaG6#\"\"\"F%" }
{TEXT -1 28 " whose numerical values are " }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 48 "evalf(omega1); evalf(omega1/2); evalf(2*omega1);" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+'[z*)*[!\"*" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"+V(*[\\C!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+
s*ezz*!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 83 "The amplitude of th
e steady-state oscillations at these three frequencies should be" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "evalf(G(omega1)); evalf(G(om
ega1/2)); evalf(G(2*omega1));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"++
+++#)!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+N?F#)R!\"*" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#$\"+X11G6!\"*" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 137 "To verify that the predicted apmplitudes of the steady-s
tate solution at these frequencies are correct we solve the differenti
al equation" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "diffeq;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#/,(-%%diffG6$-%\"xG6#%\"tG-%\"$G6$F+\"
\"#\"\"\"-F&6$F(F+F/F(\"#E,$-%$cosG6#*&%&omegaGF0F+F0\"##)" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 28 "with the initial conditions " }{XPPEDIT 
18 0 "x(0) = 6;" "6#/-%\"xG6#\"\"!\"\"'" }{TEXT -1 2 ", " }{XPPEDIT 
18 0 "D(x)(0) = 0;" "6#/--%\"DG6#%\"xG6#\"\"!F*" }{TEXT -1 15 " using \+
Maple's " }{TEXT 256 6 "dsolve" }{TEXT -1 8 " command" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "xsoln1 := dsolve(\{diffeq,x(0)=6,D(
x)(0)=0\},x(t));" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%'xsoln1G/-%\"xG6
#%\"tG,(*&,Z*&-%$sinG6#,$F)\"\"&\"\"\"-F/6#*&,&!\"&F3%&omegaGF3F3F)F3F
3!$I\"*&-%$cosGF0F3-F=6#*&,&F2F3F9F3F3F)\"\"\"F3\"$I\"*&F<FB-F/F?F3!#E
*&F<FB-F=F5F3FC*&F<FBF4FB\"#E*(F.FBFHFB)F9\"\"#FBF3*(F<FBF>FB)F9\"\"$F
BF3*(F<FBFEFBFLFB!\"\"*(F.FBFHFBF9F3\"#5*(F<FBFHFBFLFBF8*(F.FBF>FBF9FB
!#5*(F<FBF>FBFLFBF8*(F<FBF>FBF9FB!#C*(F<FBFEFBF9FBFT*(F<FBFHFBF9FB\"#C
*(F<FBFHFBFOFBFR*(F<FBF4FBFLFBF3*(F<FBF4FBF9FBFT*(F.FBF4FBFLFBF2*(F.FB
F4FBF9FBFZ*(F.FBF4FBFOFBF3*(F.FBF>FBFLFBF3*(F.FBFEFBFLFBF8*(F.FBFEFBF9
FBFZ*(F.FBFEFBFOFBF3*&F.FBFHFBFJ*&F.FBF>FBFJ*&F.FBFEFBFCFB,(*$)F9\"\"%
FBF3*$FLFB!#[\"$w'F3!\"\"#\"#TF2*&*(,(Fio!$.\"\"$i*F3FfoFPF3-%$expG6#,
$F)FRF3F<FBFBFeoF\\pFM*&*(,(FfoFPFio!$&=FcpF3F3FdpFBF.FBFBFeoF\\p#FMF2
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 197 "The form of the solution is \+
more complicated than it need be, but it will do. Recall that lengthy \+
outputs can be surpressed in Maple by using \":\" in lieu of \";\". We
 make the solution a function of " }{XPPEDIT 18 0 "omega;" "6#%&omegaG
" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "xo := un
apply(subs(xsoln1,x(t)),omega):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
90 "and plot the solution for the three values of the driving angular \+
frequncy indicated above" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 75 
"plot(\{xo(omega1),xo(omega1/2),xo(2*omega1)\},t=0..8,color=[red,blue,
black]);" }}{PARA 13 "" 1 "" {GLPLOT2D 399 221 221 {PLOTDATA 2 "6'-%'C
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F`^l$\"1N&\\bA'*H5\"F17$$\"1+++DZ>EAF1$\"12'4\\s)[Q6F17$Fe^l$\"1#f>?Rs
8;\"F17$$\"1+++ba1ZAF1$\"1XR;u'G<<\"F17$Fj^l$\"1*fI(fVwp6F17$F__l$\"1%
*fDd`-Z5F17$Fd_l$\"1&GI#=IdKxFD7$F^`l$\"1li9;u`X'*F.7$Fh`l$!1G'Rh.E-Y%
FD7$$\"1mT&)oH(Q]#F1$!1$e%QlSrE\\FD7$F]al$!1<K.i&[=G&FD7$$\"1Le*)p%4k_
#F1$!1u(H\\([2@bFD7$Fcal$!1U=GqKUTcFD7$$\"1+v$4(f%*[DF1$!1?B+89aTcFD7$
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FD7$Fbbl$!1W1qAKP)\\#FD7$Fgbl$\"11\\b8cNA6FD7$Facl$\"1,=(\\Vvho(FD7$F[
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1z^o&QO#peFD7$F__m$\"109O(=ilW#FD7$Fd_m$\"1rLs^7>smFD7$Fi_m$\"1Pwc5f;^
(*FD7$Fc`m$\"1%H--=Tg1\"F17$F]am$\"1!pA-9+!36F17$Fbam$\"1D&\\j-s(46F17
$Fgam$\"1\"*)*QHne)4\"F17$F\\bm$\"1B7Gs-_u5F17$Fabm$\"1PL#>B+y.\"F17$F
[cm$\"1m=j-K>%z%FD7$Fecm$!1<.h]*Huf$FD7$Fjcm$!1*=Q^H(zGxFD7$F_dm$!1nXi
:zVq5F17$Fddm$!1\\+J6M')f6F17$Fidm$!14MfiNC/7F17$$\"1L3FkKqgWF1$!1GI7
\\X))37F17$F^em$!1yjLq@r,7F17$$\"1*\\7.`J9[%F1$!1ro694z#=\"F17$Fcem$!1
YcjiZI_6F17$F]fm$!1b=S!>=\"G_FD7$Fgfm$\"1.P!3;3hC%FD7$F\\gm$\"1&==XLd[
9)FD7$Fagm$\"1&4oaR!om5F17$$\"1**\\i:TQdZF1$\"1b6mVx)G5\"F17$Ffgm$\"1)
zQC@uu7\"F17$$\"1**\\(o&QLyZF1$\"1iC\")*[1-9\"F17$F[hm$\"1JPAfa(49\"F1
7$$\"1**\\7)f$G*z%F1$\"1T/H*G(zH6F17$F`hm$\"1YG*z?8o5\"F17$$\"1**\\PRL
B?[F1$\"1d@xjzGs5F17$Fehm$\"1X@#)*)pgE5F17$F_im$\"1@fnZ#3#)p$FD7$Fiim$
!1er&*=kYD\\FD7$Fcjm$!1DpaG`^U')FD7$F][n$!1$*py(o5S1\"F17$Fb[n$!1#G_>#
p'f3\"F17$Fg[n$!1*og^)*fP0\"F17$F\\\\n$!1Ar6*e'=!p*FD7$Fa\\n$!17R?#eA$
f$)FD7$F[]n$!1'\\.#o'z+#yF.7$Fe]n$\"19\\e/S\\ntFD7$F_^n$\"1#[\"G-LU<5F
17$Fc_n$\"1c^gX/&*=6F17$$\"1mm\"HDn=V&F1$\"1*o9L@lL6\"F17$$\"1****\\-M
mUaF1$\"1w%3@d!G&4\"F17$$\"1KL3_&fMX&F1$\"1B\"*4Fb)[1\"F17$Fh_n$\"1V-W
\"=2D-\"F17$$\"1KL$3+[e[&F1$\"1Bl$47ix.*FD7$F]`n$\"1tyjBR\"HW(FD7$Fg`n
$!1(fSSVke0&F.7$Faan$!1S=m5&HKE)FD7$Ffan$!1&=K=L\\Ut*FD7$F[bn$!1L;@3Fs
y5F17$$\"1mT5:NW<dF1$!1sb`PO<96F17$F`bn$!1_r1R]fP6F17$$\"1+D\"y8M)QdF1
$!1X?g#*ps[6F17$Febn$!1X>24RWZ6F17$F_cn$!1Wd(>p(Q?5F17$Ficn$!1Ps)Q#*yB
>(FD7$Fcdn$\"1.TtVum97FD7$F]en$\"1qpW(>r;'))FD7$Fben$\"12=Mp)*=75F17$F
gen$\"1,;_a$[`4\"F17$Fafn$\"1hr**4!z@8\"F17$F[gn$\"1>3b(*o<@6F17$F`gn$
\"1v.4zcT(f*FD7$Fegn$\"1C^-J]]%Q'FD7$F_hn$!1'z1V*\\Y$H#FD7$Fihn$!1V+$p
U8E]*FD7$F^in$!1#Q;T5qC0\"F17$Fcin$!10\"*y#*[e26F17$Fhin$!1rAw2i<86F17
$F]jn$!1BAYW],p5F17$Fgjn$!1no><D#[T)FD7$Fa[o$!1YSl3]>^YFD7$F[\\o$\"1r/
0gkbAUFD7$Fe\\o$\"1q'*\\G/:Z5F17$$\"1N$3Fccxn'F1$\"1E)oH5<S3\"F17$Fj\\
o$\"1'=fu]!=46F17$$\"1M$e9ul*)p'F1$\"1lU/_bOA6F17$F_]o$\"1f\\ymkUB6F17
$$\"1N$3-#\\<?nF1$\"1#Q6=Q[B6\"F17$Fd]o$\"1A5'\\\\Z#*3\"F17$$\"1M$e*)4
%QTnF1$\"1a_:D)pV0\"F17$Fi]o$\"17D&\\c)335F17$F^^o$\"1qh9(42e?(FD7$Fc^
o$\"1En[NR+(4$FD7$F]_o$!1=-d<v\"*peFD7$Fa`o$!1Gn;Ut<56F17$$\"1M$3_]@W-
(F1$!1x8.NVhL6F17$Ff`o$!1EjqqHo;6F17$$\"1n;H7o2jqF1$!1vHTs(z*f5F17$F[a
o$!1Ssh/:Ab'*FD7$F`ao$!1eThp_P!y'FD7$Feao$!1g()f'*p!)[HFD7$F_bo$\"1d7
\")R)HSU'FD7$F]do$\"1nB3?qwC6F17$Fgdo$\"1XIeCG<0))FD7$Faeo$\"1y?'*Hz#o
K\"FD7$F[fo$!1SA\"HeEaM(FD7$F_go$!1B[T\"RVY7\"F17$Figo$!1i#GD)yHN\")FD
7$Fcho$!1x#=UtUun\"F.7$F]io$\"1l'4TmVbO)FD7$Fajo$\"1(zoQH$**=6F1-Ffjo6
&FhjoF(F(F(-%+AXESLABELSG6$Q\"t6\"%!G-%%VIEWG6$;F(Fajo%(DEFAULTG" 1 2 
0 1 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 }}}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 109 "Observe that after the oscillations have attained t
heir steady-states, the amplitudes are as predicted above." }}}{EXCHG 
{PARA 3 "" 0 "" {TEXT -1 55 "Problem: The unicycle model of an automob
ile suspension" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 122 "On p. 168, pri
or to Problem 29, it is stated that the motion of a unicycle moving al
ong a corrugated road having equation " }{XPPEDIT 18 0 "y = a*sin(omeg
a*t);" "6#/%\"yG*&%\"aG\"\"\"-%$sinG6#*&%&omegaGF'%\"tGF'F'" }{TEXT 
-1 63 " for the height of the road at the position of the car at time \+
" }{XPPEDIT 18 0 "t;" "6#%\"tG" }{TEXT -1 41 " is governed by the diff
erential equation" }}{PARA 258 "" 0 "" {XPPEDIT 18 0 "m*`@@`(D,2)(x)+c
*D(x)+k*x = c*omega*a*cos(omega*t)+k*a*sin(omega*t);" "6#/,(*&%\"mG\"
\"\"--%#@@G6$%\"DG\"\"#6#%\"xGF'F'*&%\"cGF'-F,6#F/F'F'*&%\"kGF'F/F'F',
&**F1F'%&omegaGF'%\"aGF'-%$cosG6#*&F8F'%\"tGF'F'F'*(F5F'F9F'-%$sinG6#*
&F8F'F>F'F'F'" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 33 "For a car travelling at velocity " }{XPPEDIT 
18 0 "v;" "6#%\"vG" }{TEXT -1 33 " on a road with peaks a distance " }
{XPPEDIT 18 0 "L;" "6#%\"LG" }{TEXT -1 8 " apart  " }{XPPEDIT 18 0 "om
ega = 2*Pi*v/L;" "6#/%&omegaG**\"\"#\"\"\"%#PiGF'%\"vGF'%\"LG!\"\"" }
{TEXT -1 24 " (cf. Example 5, p.163)." }}{PARA 0 "" 0 "" {TEXT -1 44 "
1. a. The differential equation has the form" }}{PARA 259 "" 0 "" 
{TEXT -1 0 "" }{XPPEDIT 18 0 "m*`@@`(D,2)(x)+c*D(x)+k*x = E[0]*cos(ome
ga*t)+F[0]*sin(omega*t);" "6#/,(*&%\"mG\"\"\"--%#@@G6$%\"DG\"\"#6#%\"x
GF'F'*&%\"cGF'-F,6#F/F'F'*&%\"kGF'F/F'F',&*&&%\"EG6#\"\"!F'-%$cosG6#*&
%&omegaGF'%\"tGF'F'F'*&&%\"FG6#F;F'-%$sinG6#*&F@F'FAF'F'F'" }{TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 63 "Recall that the general solution t
o this equation has the form " }{XPPEDIT 18 0 "x = x[c]+x[p];" "6#/%\"
xG,&&F$6#%\"cG\"\"\"&F$6#%\"pGF)" }{TEXT -1 7 " where " }{XPPEDIT 18 
0 "x[c];" "6#&%\"xG6#%\"cG" }{TEXT -1 18 " is transient and " }
{XPPEDIT 18 0 "x[p];" "6#&%\"xG6#%\"pG" }{TEXT -1 43 " is the steady-s
tate part of the solution. " }{TEXT -1 0 "" }{TEXT -1 51 "The appropri
ate guess for a particular solution is " }{XPPEDIT 18 0 "x[p] = A*cos(
omega*t)+B*sin(omega*t);" "6#/&%\"xG6#%\"pG,&*&%\"AG\"\"\"-%$cosG6#*&%
&omegaGF+%\"tGF+F+F+*&%\"BGF+-%$sinG6#*&F0F+F1F+F+F+" }{TEXT -1 60 ". \+
Substitute this into the differential equation, solve for " }{XPPEDIT 
18 0 "A;" "6#%\"AG" }{TEXT -1 6 " and ," }{XPPEDIT 18 0 "B;" "6#%\"BG
" }{TEXT -1 37 " and thereby show that the amplitude " }{XPPEDIT 18 0 
"sqrt(A^2+B^2);" "6#-%%sqrtG6#,&*$%\"AG\"\"#\"\"\"*$%\"BG\"\"#F*" }
{TEXT -1 1 " " }{TEXT -1 22 "of the steady-state is" }}{PARA 260 "" 0 
"" {XPPEDIT 18 0 "G = sqrt(E[0]^2+F[0]^2)/sqrt((k-m*omega^2)^2+(c*omeg
a)^2);" "6#/%\"GG*&-%%sqrtG6#,&*$&%\"EG6#\"\"!\"\"#\"\"\"*$&%\"FG6#F.
\"\"#F0F0-F'6#,&*$,&%\"kGF0*&%\"mGF0*$%&omegaG\"\"#F0!\"\"\"\"#F0*$*&%
\"cGF0F?F0\"\"#F0FA" }{TEXT -1 0 "" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" 
{TEXT -1 86 "Show that for the unicycle model of an automobile suspens
ion this leads to the formula" }}{PARA 261 "" 0 "" {XPPEDIT 18 0 "G = \+
a*sqrt(k^2+(2*c*Pi*v/L)^2)/sqrt((k-m*(2*Pi*v/L)^2)^2+(2*c*Pi*v/L)^2);
" "6#/%\"GG*(%\"aG\"\"\"-%%sqrtG6#,&*$%\"kG\"\"#F'*$*,\"\"#F'%\"cGF'%#
PiGF'%\"vGF'%\"LG!\"\"\"\"#F'F'-F)6#,&*$,&F-F'*&%\"mGF'*$**\"\"#F'F3F'
F4F'F5F6\"\"#F'F6\"\"#F'*$*,\"\"#F'F2F'F3F'F4F'F5F6\"\"#F'F6" }{TEXT 
-1 0 "" }{TEXT -1 0 "" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 66 "f
or the amplitude of the steady-state vibrations versus the speed " }
{XPPEDIT 18 0 "v;" "6#%\"vG" }{TEXT -1 10 " (in m/s) " }{TEXT -1 11 "o
f the car." }}{PARA 0 "" 0 "" {TEXT -1 44 "b. Using the parameters giv
en in Example 5: " }{XPPEDIT 18 0 "m = 800;" "6#/%\"mG\"$+)" }{TEXT 
-1 5 " kg, " }{XPPEDIT 18 0 "c = 3000;" "6#/%\"cG\"%+I" }{TEXT -1 8 " \+
N-s/m, " }{XPPEDIT 18 0 "k = 70000;" "6#/%\"kG\"&++(" }{TEXT -1 6 " N/
m, " }{XPPEDIT 18 0 "a = .5e-1;" "6#/%\"aG$\"\"&!\"#" }{TEXT -1 4 " m,
 " }{XPPEDIT 18 0 "L = 10;" "6#/%\"LG\"#5" }{TEXT -1 148 " m obtain th
e graph of steady-state amplitude versus velocity given in Figure 2.6.
12, p. 169 (see p. 164 for the conversion factor from m/s to mph)." }}
{PARA 0 "" 0 "" {TEXT -1 25 "c. The resonant velocity " }{XPPEDIT 18 
0 "v[1];" "6#&%\"vG6#\"\"\"" }{TEXT -1 45 " is apparently about 30 mph
. Find it exactly." }}{PARA 0 "" 0 "" {TEXT -1 49 "d. Use Maple's dsol
ve command to plot the motion " }{XPPEDIT 18 0 "x(t);" "6#-%\"xG6#%\"t
G" }{TEXT -1 42 " of a unicycle corresponding to speeds of " }
{XPPEDIT 18 0 "v[1];" "6#&%\"vG6#\"\"\"" }{TEXT -1 2 ", " }{XPPEDIT 
18 0 "v[1]/2;" "6#*&&%\"vG6#\"\"\"\"\"\"\"\"#!\"\"" }{TEXT -1 5 " and \+
" }{XPPEDIT 18 0 "2*v[1];" "6#*&\"\"#\"\"\"&%\"vG6#\"\"\"F%" }{TEXT 
-1 29 " assuming initial conditions " }{XPPEDIT 18 0 "x(0) = 0;" "6#/-
%\"xG6#\"\"!F'" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "D(x)(0) = 0;" "6#/--%
\"DG6#%\"xG6#\"\"!F*" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 38 "2. The value of the damping parameter \+
" }{XPPEDIT 18 0 "c;" "6#%\"cG" }{TEXT -1 50 " which would make the sy
stem critically damped is " }{XPPEDIT 18 0 "c[cr] = 2*sqrt(m*k);" "6#/
&%\"cG6#%#crG*&\"\"#\"\"\"-%%sqrtG6#*&%\"mGF*%\"kGF*F*" }{TEXT -1 100 
" which for the given mass and spring constant has a value of about 15
000 N-s/m. Thus for the value, " }{XPPEDIT 18 0 "c = 3000;" "6#/%\"cG
\"%+I" }{TEXT -1 104 ", given above the suspension is (very) underdamp
ed. Plot the steady-state amplitude versus velocity for " }{XPPEDIT 
18 0 "c = 10000;" "6#/%\"cG\"&++\"" }{TEXT -1 17 " (underdamped),  " }
{XPPEDIT 18 0 "c = c[cr];" "6#/%\"cG&F$6#%#crG" }{TEXT -1 26 " (critic
ally damped) and  " }{XPPEDIT 18 0 "c = 30000;" "6#/%\"cG\"&++$" }
{TEXT -1 124 " (overdamped). Discuss the advantages and disadvantages \+
of different amounts of damping in relation to the speed of the car." 
}}}}{MARK "29 1 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 }