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{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT -1 25 "Equa\347\365es de Segund
a Ordem" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 79 "J\341 vimos que para resolver uma ed
o de segunda ordem com coeficientes constantes" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 256 "" 0 "" {XPPEDIT 18 0 "a*diff(y(t),`$`(t,2))
+b*diff(y(t),t)+c*y(t) = 0;" "6#/,(*&%\"aG\"\"\"-%%diffG6$-%\"yG6#%\"t
G-%\"$G6$F.\"\"#F'F'*&%\"bGF'-F)6$-F,6#F.F.F'F'*&%\"cGF'-F,6#F.F'F'\"
\"!" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 23 "\351 importante analisar a" }{TEXT 256 23 " equa\347\343o
 caracter\355stica" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "
" {XPPEDIT 18 0 "a*lambda^2+b*lambda+c=0" "6#/,(*&%\"aG\"\"\"*$%'lambd
aG\"\"#F'F'*&%\"bGF'F)F'F'%\"cGF'\"\"!" }{TEXT -1 1 "." }}{PARA 0 "" 
0 "" {TEXT -1 62 "O Maple ingenuamente resolve  a edo acima da seguint
e maneira:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "restart;with(
plots):with(DETools):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "ed
o:=a*diff(y(t),t,t)+b*diff(y(t),t)+c*y(t)=0;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%$edoG/,(*&%\"aG\"\"\"-%%diffG6$-%\"yG6#%\"tG-%\"$G6$F
0\"\"#F)F)*&%\"bGF)-F+6$F-F0F)F)*&%\"cGF)F-F)F)\"\"!" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 12 "dsolve(edo);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/-%\"yG6#%\"tG,&*&%$_C1G\"\"\"-%$expG6#,$**\"\"#!\"\"%
\"aGF2,&%\"bGF2*$,&*$)F5F1F+F+*(\"\"%F+%\"cGF+F3F+F2#F+F1F+F+F'F+F+F+F
+*&%$_C2GF+-F-6#,$**F1F2,&F5F+F6F+F+F3F2F'F+F2F+F+" }}}{PARA 11 "" 1 "
" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 129 "Isso estaria ce
rto, se n\343o fosse o problema quando o radicando acima se anula. A
\355 a equa\347\343o caracter\355sitca s\363 tem uma raiz dupla." }}
{PARA 0 "" 0 "" {TEXT -1 86 "O Maple at\351 sabe qual a resposta, mas \+
temos que avisar pra ele que estamos nesse caso:" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 15 "b:=sqrt(4*a*c);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%\"bG,$*&\"\"#\"\"\"*&%\"cGF(%\"aGF(#F(F'F(" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 12 "dsolve(edo);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#/-%\"yG6#%\"tG,&*&%$_C1G\"\"\"-%$expG6#,$*(*&%\"cGF+%\"aGF+#F+\"
\"#F3!\"\"F'F+F6F+F+*(%$_C2GF+F,F+F'F+F+" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 166 "Voltando ao caso em que temos duas ra\355zes distintas, \+
vimos em sala que as exponenciais complexas que aparecem podem ser tro
cadas por suas partes reais e imagin\341rias. " }}{PARA 0 "" 0 "" 
{TEXT -1 81 "Um exemplo cl\341ssico de uma tal edo \351 o sistema mass
a-mola, tamb\351m conhecido como " }{TEXT 257 19 "oscilador harm\364ni
co" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 195 "Temos uma mola com
 constante k  presa a uma massa m. Supondo a resist\352ncia do ar e o \+
atrito com a superf\355cie desprez\355veis, a edo que descreve o movim
ento, de acordo com a segunda lei de Newton \351:" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 256 "" 0 "" {XPPEDIT 18 0 "F=m*diff(x(t),`$`(t,2
))" "6#/%\"FG*&%\"mG\"\"\"-%%diffG6$-%\"xG6#%\"tG-%\"$G6$F.\"\"#F'" }
{TEXT -1 3 " = " }{XPPEDIT 18 0 "-k*x(t)" "6#,$*&%\"kG\"\"\"-%\"xG6#%
\"tGF&!\"\"" }{TEXT -1 1 "." }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 30 "edo:=m*diff(x(t),t,t)=-k*x(t);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%$edoG/*&%\"mG\"\"\"-%%diffG6$-%\"xG6#%\"tG-%\"$G
6$F/\"\"#F(,$*&%\"kGF(F,F(!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 12 "dsolve(edo);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"xG6#%\"tG
,&*&%$_C1G\"\"\"-%$sinG6#*(%\"mG#!\"\"\"\"#%\"kG#F+F3F'F+F+F+*&%$_C2GF
+-%$cosGF.F+F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 97 "Nesse caso, a r
esposta \351 uma combina\347\343o linear de seno e cosseno ambos com a
 mesma frequ\352ncia, a " }{TEXT 261 26 "frequ\352ncia angular natural
" }{TEXT -1 11 " do sistema" }}{PARA 256 "" 0 "" {XPPEDIT 18 0 "omega=
sqrt(k/m)" "6#/%&omegaG-%%sqrtG6#*&%\"kG\"\"\"%\"mG!\"\"" }{TEXT -1 1 
"." }}{PARA 0 "" 0 "" {TEXT -1 85 "Identidades trigonom\351tricas tamb
\351m nos dizem que podemos escrever a resposta na forma" }}{PARA 256 
"" 0 "" {XPPEDIT 18 0 "x(t)=A*cos(omega*t+t[0])" "6#/-%\"xG6#%\"tG*&%
\"AG\"\"\"-%$cosG6#,&*&%&omegaGF*F'F*F*&F'6#\"\"!F*F*" }{TEXT -1 1 ".
" }}{PARA 0 "" 0 "" {TEXT -1 99 "De fato, basta expandir a express\343
o acima e resolver para A e t[0] em fun\347\343o das constantes c1 e c
2" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "expand(A*cos(omega*t+t
0));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*(%\"AG\"\"\"-%$cosG6#*&%&om
egaGF&%\"tGF&F&-F(6#%#t0GF&F&*(F%F&-%$sinGF)F&-F2F.F&!\"\"" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "%=alpha*cos(omega*t)+beta*sin(omega
*t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,&*(%\"AG\"\"\"-%$cosG6#*&%&o
megaGF'%\"tGF'F'-F)6#%#t0GF'F'*(F&F'-%$sinGF*F'-F3F/F'!\"\",&*&%&alpha
GF'F(F'F'*&%%betaGF'F2F'F'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 11 "As \+
fun\347\365es " }{XPPEDIT 18 0 "cos(omega*t)" "6#-%$cosG6#*&%&omegaG\"
\"\"%\"tGF(" }{TEXT -1 3 " e " }{XPPEDIT 18 0 "sin(omega*t)" "6#-%$sin
G6#*&%&omegaG\"\"\"%\"tGF(" }{TEXT -1 120 " s\343o linearmente indepen
dentes (por que?) e portanto a igualdade acima na verdade \351 um sist
ema (n\343o linear) para A e t0." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 47 "solve(\{A*cos(t0)=alpha,A*sin(t0)=beta\},\{A,t0\});" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#<$/%\"AG-%'RootOfG6#,(*$)%%betaG\"\"
#\"\"\"!\"\"*$)%&alphaGF-F.F/*$)%#_ZGF-F.F./%#t0G-%'arctanG6$*&F,F.F&F
/*&F2F.F&F/" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 135 "Olha para a expre
ss\343o acima e se conven\347a de que voc\352 saberia fazer isso na m
\343o (quem \351 a raiz do polin\364mio em Z para A, por exemplo?)." }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 29 "Vamos ve
r um exemplo concreto" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "m:
=1;k:=2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"mG\"\"\"" }}{PARA 11 "
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{MPLTEXT 1 0 66 "# largando com velocidade positiva\ndsolve(\{edo,x(0)
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F(F(-%$cosGF-F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "gsol:=pl
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72 "De fato, como esper\341vamos, a solu\347\343o \351 um cosseno desl
ocado. Para ver a " }{TEXT 258 9 "amplitude" }{TEXT -1 18 " A, basta c
alcular" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "A:=sqrt((sqrt(2)
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3!HJlR$o')o5F,7$Fc`m$\"3>?yRAzmF(*F@7$Fh`m$\"3g(3l/h-VZ)F@7$F]am$\"3X&
)>91RH@qF@-Fcam6&FeamFfamF(F(-%+AXESLABELSG6%%\"tGQ!6\"-%%FONTG6#%(DEF
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46.000000 0 0 "Curve 1" "Curve 2" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
54 "As curvas azul e vermelha n\343o se encaixam pois h\341 uma " }
{TEXT 259 4 "fase" }{TEXT -1 116 " n\343o nula na solu\347\343o de nos
sa edo. Por exemplo, quando t=0, a solu\347\343o vale A*cos(t0), o que
 nos permite calcular t0:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
24 "A*cos(t0)=subs(t=0,sol);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,$*&#
\"\"\"\"\"#F'*&\"\"'F&-%$cosG6#%#t0GF'F'F',&*&F&F'*&-%$sinG6#\"\"!F'F(
F&F'F'-F,F4F'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "evalf(solv
e(%,t0));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+(3(zah!#5" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 193 "Olhando para a figura, vemos que queremo
s uma fase t0 ligeiramente negativa (queremos deslocar a curva azul pa
ra direita), portanto a resposta acima n\343o \351 a que queremos! Vam
os fazer um gr\341fico" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "p
lot(A*cos(t0)-subs(t=0,sol),t0=-1..1);" }}{PARA 13 "" 1 "" {GLPLOT2D 
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0 "Curve 1" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 33 "Agora sim, vamos a
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20 "6#$!+(3(zah!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "t0:=%;
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#t0G$!+(3(zah!#5" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "plot(A*cos(omega*t+t0),t=0..10,colo
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{TEXT -1 10 "Definindo " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "
omega:=sqrt(k/m);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&omegaG*$*&%\"k
G\"\"\"%\"mG!\"\"#F(\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
16 "mu:=alpha/(2*m);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#muG,$*(\"\"
#!\"\"%&alphaG\"\"\"%\"mGF(F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 15 "
as ra\355zes ficam" }}{PARA 256 "" 0 "" {XPPEDIT 18 0 "-mu+sqrt(mu^2-o
mega^2);" "6#,&%#muG!\"\"-%%sqrtG6#,&*$F$\"\"#\"\"\"*$%&omegaGF+F%F," 
}{TEXT -1 7 "   e   " }{XPPEDIT 18 0 "-mu-sqrt(mu^2-omega^2);" "6#,&%#
muG!\"\"-%%sqrtG6#,&*$F$\"\"#\"\"\"*$%&omegaGF+F%F%" }{TEXT -1 1 "." }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 142 "O caso visto no laborat\363rio \+
de f\355sica, com um bloco de madeira no ar corresponde a um coeficien
te de resist\352ncia ordens de grandeza menor que " }{XPPEDIT 18 0 "4*
k*m" "6#*(\"\"%\"\"\"%\"kGF%%\"mGF%" }{TEXT -1 11 ", ou seja, " }
{XPPEDIT 18 0 "mu;" "6#%#muG" }{TEXT -1 1 " " }{TEXT 260 9 "bem menor
" }{TEXT -1 8 " do que " }{XPPEDIT 18 0 "omega" "6#%&omegaG" }{TEXT 
-1 105 ". Nesse caso, as ra\355zes ser\343o complexas e teremos uma os
cila\347\343o modulada por uma exponencial decrescente." }}{PARA 0 "" 
0 "" {TEXT -1 20 "Um exemplo concreto:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 21 "m:=1;k:=2;alpha:=1/3;" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%\"mG\"\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"kG\"\"#" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&alphaG#\"\"\"\"\"$" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "omega:=sqrt(k/m);" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#>%&omegaG*$\"\"##\"\"\"F&" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 16 "mu:=alpha/(2*m);" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6#>%#muG#\"\"\"\"\"'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "# r
esolve largando do repouso\ndsolve(\{edo,x(0)=3,D(x)(0)=0\});" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"xG6#%\"tG,&*&#\"\"$\"#r\"\"\"*(F,
#F-\"\"#-%$expG6#,$*&\"\"'!\"\"F'F-F7F--%$sinG6#,$*(F6F7F,F/F'F-F-F-F-
F-*(F+F-F1F--%$cosGF:F-F-" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
12 "sol:=rhs(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG,&*&#\"\"$
\"#r\"\"\"*(F)#F*\"\"#-%$expG6#,$*&\"\"'!\"\"%\"tGF*F4F*-%$sinG6#,$*(F
3F4F)F,F5F*F*F*F*F**(F(F*F.F*-%$cosGF8F*F*" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 39 "gsol:=plot(sol,t=0..15):\ndisplay(gsol);" }}{PARA 
13 "" 1 "" {GLPLOT2D 734 352 352 {PLOTDATA 2 "6%-%'CURVESG6$7jq7$$\"\"
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be))F;7$$\"3;+]i:$4wb'F2$!3'=Wz_:W*o&*F;7$$\"3CDc^`!)y(f'F2$!3cq;06Awv
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LeF;7$$\"3q*****\\xJLu)F2$\"3QEB5%*)HVY'F;7$$\"3W*\\(o/L*=#))F2$\"3rzF
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hXnF;7$$\"3W++v$*ydd!*F2$\"3_OlTpVFsmF;7$$\"3>+D\"G`-'4#*F2$\"3P]u32%o
&3jF;7$$\"3#***\\(=<F;O*F2$\"3#e#>%[%=fzcF;7$$\"35***\\i0A#*p*F2$\"3KN
6B^hQaNF;7$$\"3*)****\\2mD+5!#;$\"3wpccQDSy6F;7$$\"3*****\\i0XE.\"Ffel
$!3:[T2&Q$))>8F;7$$\"3%**\\(o/Q*>1\"Ffel$!3wo'R)e%)e]JF;7$$\"31](=<xO!
y5Ffel$!3?Rs?Qw*[(QF;7$$\"3=++vQ(zS4\"Ffel$!3yy>!y3+\"oVF;7$$\"3/v$4'H
ti,6Ffel$!3UlQZw;7;XF;7$$\"33](o/#\\<46Ffel$!3o\"3n:e-(4YF;7$$\"37D\"G
8^An6\"Ffel$!3oIVw\"[V\"\\YF;7$$\"3***\\(=-,FC6Ffel$!3A)4**>9Y`j%F;7$$
\"3.+Dc;*[+9\"Ffel$!3b&\\CNtw;W%F;7$$\"33+v$4tFe:\"Ffel$!3i0W;\"zoL/%F
;7$$\"3)*\\P41WDr6Ffel$!3)z')*fcff%[$F;7$$\"3!****\\73\"o'=\"Ffel$!3;U
D='33Gz#F;7$$\"3-+voz;)*=7Ffel$!3id&HJt[r3\"F;7$$\"31+++&*44]7Ffel$\"3
?(p`,2q,&fF/7$$\"35+]7jZ!>G\"Ffel$\"3(Hvnoy'fI?F;7$$\"35]i:I*zwH\"Ffel
$\"3Ca*)GA7>dDF;7$$\"34+v=(4bMJ\"Ffel$\"3\\(G801QL$HF;7$$\"38]PMP3&zK
\"Ffel$\"3I]ogQYAPJF;7$$\"3;++]xlWU8Ffel$\"3ANzg)pNK?$F;7$$\"3;+DcJ.1f
8Ffel$\"3P9;q#>)[9JF;7$$\"39+]i&3ucP\"Ffel$\"3_97#\\eLS'GF;7$$\"3-+DJD
O`!R\"Ffel$\"3`mQ_ar%4_#F;7$$\"3\"******\\;$R09Ffel$\"3/HB*zoBi3#F;7$$
\"38+v=-*zqV\"Ffel$\"3<G;j71D<'*F/7$$\"33+D\"G:3uY\"Ffel$!30G!p.KN,\"=
F/7$$\"#:F)$!33aJ$p5\"*pC\"F;-%'COLOURG6&%$RGBG$\"*++++\"!\")F(F(-%+AX
ESLABELSG6$%\"tGQ!6\"-%%VIEWG6$;F(Fg]m%(DEFAULTG" 1 2 0 1 10 0 2 9 1 
4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 58 "Da mesma forma que antes, podemos escrever a solu
\347\343o como " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" 
{XPPEDIT 18 0 "x(t)=A*exp(-mu*t)*cos(omega*t+t0)" "6#/-%\"xG6#%\"tG*(%
\"AG\"\"\"-%$expG6#,$*&%#muGF*F'F*!\"\"F*-%$cosG6#,&*&%&omegaGF*F'F*F*
%#t0GF*F*" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 70 "(tente achar A e t0 de forma an\341loga ao que fiz
emos no caso anterior)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 85 "Sobrepondo o gr\341fico da exponencial com a amplitu
de certa, vemos melhor o que ocorre." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 59 "# f\363rmula para A \351 parecida\nA:=sqrt((3/71*71^(
1/2))^2+3^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"AG,$*(\"#=\"\"\"
\"#r!\"\"\"$U\"#F(\"\"#F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
61 "gexp:=plot([A*exp(-mu*t),-A*exp(-mu*t)],t=0..15,color=green):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "display(gsol,gexp);" }}
{PARA 13 "" 1 "" {GLPLOT2D 763 316 316 {PLOTDATA 2 "6'-%'CURVESG6$7jq7
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\\8^F,\"F27$Fd`l$\"3EnM9g.C*f*F;7$F^al$\"3Lv2<2)G%Q\"*F;7$Fcal$\"35#H(
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AuF;7$F\\cl$\"322ZZW%)\\NqF;7$F`dl$\"3%pCD(G>]wmF;7$Fjdl$\"3GBh'*\\/gY
jF;7$F_el$\"3%*R.(z<k$**fF;7$Fdel$\"3'pD*fG?f.dF;7$Fjel$\"3F$\\T]CqQS&
F;7$F_fl$\"3/86;:G!f9&F;7$Fifl$\"35=(*y.i%z([F;7$F]hl$\"3K22>NSdQYF;7$
Fghl$\"3xaTX)H34S%F;7$Fail$\"3QA>QqPK!=%F;7$Ffil$\"3)zn?_:H7'RF;7$F[jl
$\"3_q=&efx5w$F;7$F`jl$\"3q$QC$o*[oc$F;7$Fjjl$\"3;Vcq[h8%Q$F;7$Fd[m$\"
3nl4Q!H0XA$F;7$F^\\m$\"3))**ob[$*y]IF;7$Fh\\m$\"3*4VHTjfL!HF;7$F]]m$\"
3)e+C[23Sv#F;7$Fb]m$\"3\">fiYrg#=EF;7$Fg]m$\"3yoa2B7$)zCF;-F\\^m6&F^^m
F(F_^mF(-F$6$7S7$F($!3j!3o)*)G0@IF27$FI$!3;FgK$4K3'GF27$FS$!3\"e&H6hTN
GFF27$Fao$!3i,*RZf$o'e#F27$Fep$!3U#Q'ppE]^CF27$Fiq$!3+tI<Q\"zRK#F27$Fc
r$!3V(*oV*=#o6AF27$F]s$!3E\\\\/'3?65#F27$F[u$!3OQ,IJqe#*>F27$Feu$!3UT$
*ou;)**)=F27$F_v$!3KAyBK#\\**y\"F27$Fiv$!3\")fU(yTNiq\"F27$Fcw$!3/)R-'
\\*omh\"F27$F]x$!3=2F8M]YJ:F27$Fgx$!35=Zp3pg`9F27$Fez$!3%Q2Lh'*QjQ\"F2
7$F_[l$!3I70h(\\!R58F27$Fi[l$!31gZ%oo;$\\7F27$Fc\\l$!3e\"fcnaX==\"F27$
F]]l$!3n>\"[Vzo^7\"F27$Fb]l$!3Og91Yu5m5F27$F\\^l$!3eBe!\\8^F,\"F27$Fd`
l$!3EnM9g.C*f*F;7$F^al$!3Lv2<2)G%Q\"*F;7$Fcal$!35#H(o^'[hm)F;7$Fhal$!3
7t'G>Y39?)F;7$F]bl$!3u?KGo*)=<yF;7$Fbbl$!3EK<nBTTAuF;7$F\\cl$!322ZZW%)
\\NqF;7$F`dl$!3%pCD(G>]wmF;7$Fjdl$!3GBh'*\\/gYjF;7$F_el$!3%*R.(z<k$**f
F;7$Fdel$!3'pD*fG?f.dF;7$Fjel$!3F$\\T]CqQS&F;7$F_fl$!3/86;:G!f9&F;7$Fi
fl$!35=(*y.i%z([F;7$F]hl$!3K22>NSdQYF;7$Fghl$!3xaTX)H34S%F;7$Fail$!3QA
>QqPK!=%F;7$Ffil$!3)zn?_:H7'RF;7$F[jl$!3_q=&efx5w$F;7$F`jl$!3q$QC$o*[o
c$F;7$Fjjl$!3;Vcq[h8%Q$F;7$Fd[m$!3nl4Q!H0XA$F;7$F^\\m$!3))**ob[$*y]IF;
7$Fh\\m$!3*4VHTjfL!HF;7$F]]m$!3)e+C[23Sv#F;7$Fb]m$!3\">fiYrg#=EF;7$Fg]
m$!3yoa2B7$)zCF;Fhgm-%+AXESLABELSG6%%\"tGQ!6\"-%%FONTG6#%(DEFAULTG-%%V
IEWG6$;F(Fg]mFian" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 
45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" }}}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 95 "Ex.: Ache o primeiro tempo a partir do qual a oscila
\347\343o da massa n\343o passa, em m\363dulo, de 0.95." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 258 "Finalmente, se noss
o sistema massa mola estiver imerso digamos em \341gua em vez de ar, e
sperar\355amos que o bloco parasse bem antes. Na verdade, se o radican
do for nulo ou real, n\343o haver\341 oscila\347\343o nenhuma. Vamos v
er um exemplo em que o radicando seja positivo:" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 52 "m:=1;k:=2;alpha:=3;omega:=sqrt(k/m);mu:=alpha/
(2*m);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"mG\"\"\"" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%\"kG\"\"#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&
alphaG\"\"$" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&omegaG*$\"\"##\"\"\"
F&" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#muG#\"\"$\"\"#" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "edo:=m*diff(x(t),t,t)=-k*x(t)-alpha
*diff(x(t),t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$edoG/-%%diffG6$-%
\"xG6#%\"tG-%\"$G6$F,\"\"#,&*&F0\"\"\"F)F3!\"\"*&\"\"$F3-F'6$F)F,F3F4
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "# resolve largando do r
epouso como no ar\ndsolve(\{edo,x(0)=3,D(x)(0)=0\});" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#/-%\"xG6#%\"tG,&*&\"\"'\"\"\"-%$expG6#,$F'!\"\"F+F+*
&\"\"$F+-F-6#,$*&\"\"#F+F'F+F0F+F0" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 12 "sol:=rhs(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$so
lG,&*&\"\"'\"\"\"-%$expG6#,$%\"tG!\"\"F(F(*&\"\"$F(-F*6#,$*&\"\"#F(F-F
(F.F(F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "gsol:=plot(sol,t
=0..15):\ndisplay(gsol);" }}{PARA 13 "" 1 "" {GLPLOT2D 783 312 312 
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$\"35+](=xZ&\\iF2$\"3gy*)*fRxw:\"F\\r7$$\"3;+]i:$4wb'F2$\"3nJ3Y^^g4&)!
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`t7$$\"3I++Dchf#\\(F2$\"3+#[W_qNAM$F`t7$$\"3))**\\(of2L#yF2$\"3;08iQ@H
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*F2$\"3'**e9uZe(zOFdv7$$\"3*)****\\2mD+5!#;$\"3!>,B4D`pr#Fdv7$$\"3****
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hy!#A7$$\"33+v$4tFe:\"Fgw$\"3F0-;t'RRt&F^y7$$\"3!****\\73\"o'=\"Fgw$\"
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4bMJ\"Fgw$\"3k?[/xCY&=\"F^y7$$\"3;++]xlWU8Fgw$\"3I[oz*yT6())!#B7$$\"39
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[l7$$\"38+v=-*zqV\"Fgw$\"3st#*eF\"RMW$Fb[l7$$\"33+D\"G:3uY\"Fgw$\"3Yh>
J;GgUDFb[l7$$\"#:F)$\"33mAGUOTN=Fb[l-%'COLOURG6&%$RGBG$\"*++++\"!\")F(
F(-%+AXESLABELSG6$%\"tGQ!6\"-%%VIEWG6$;F(Fh\\l%(DEFAULTG" 1 2 0 1 10 
0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 50 "H\341 um caso lim\355trofe, em que temos \+
uma raiz dupla:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "m:=1;k:=
2;alpha:=sqrt(4*k*m);omega:=sqrt(k/m);mu:=alpha/(2*m);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%\"mG\"\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%
\"kG\"\"#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&alphaG,$*&\"\"#\"\"\"F
'#F(F'F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&omegaG*$\"\"##\"\"\"F&
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#muG*$\"\"##\"\"\"F&" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "edo:=m*diff(x(t),t,t)=-k*x(t)-alpha
*diff(x(t),t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$edoG/-%%diffG6$-%
\"xG6#%\"tG-%\"$G6$F,\"\"#,&*&F0\"\"\"F)F3!\"\"*(F0F3F0#F3F0-F'6$F)F,F
3F4" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "# resolve largando d
o repouso como no ar\ndsolve(\{edo,x(0)=3,D(x)(0)=0\});" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#/-%\"xG6#%\"tG,&*&\"\"$\"\"\"-%$expG6#,$*&\"\"##
F+F1F'F+!\"\"F+F+**F*F+F1F2F,F+F'F+F+" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 177 "Agora temos um termo com um fator t, que para valores pr
\363ximos a zero tem alguma influ\352ncia, mas que perde logo para o f
ator exponencial decrescente. Um gr\341fico para vermos isso" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "sol:=rhs(%);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%$solG,&*&\"\"$\"\"\"-%$expG6#,$*&\"\"##F(F.%\"tG
F(!\"\"F(F(**F'F(F.F/F)F(F0F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 85 "gsoltpequeno:=plot(sol,t=0..1):\ngsoltgrande:=plot(sol,t=0..5)
:\ndisplay(gsoltpequeno);" }}{PARA 13 "" 1 "" {GLPLOT2D 668 378 378 
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{TEXT -1 131 "Vemos acima que, de fato, inicialmente a posi\347\343o n
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