{VERSION 4 0 "SUN SPARC SOLARIS" "4.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 261 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 262 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 263 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 265 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 266 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{PSTYLE "Normal " -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Text Output" -1 2 1 {CSTYLE "" -1 -1 "Courier" 1 10 0 0 255 1 0 0 0 0 0 1 3 0 3 0 }1 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Warning" 2 7 1 {CSTYLE "" -1 -1 "" 0 1 0 0 255 1 0 0 0 0 0 0 1 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple O utput" 0 11 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 3 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 11 12 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }1 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 } {PSTYLE "Maple Plot" 0 13 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 256 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 257 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 258 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 11 259 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }1 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 11 260 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }1 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 261 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {PARA 256 "" 0 "" {TEXT -1 11 "Math 2250-3" }}{PARA 257 "" 0 " " {TEXT -1 20 "Wednesday November 1" }}{PARA 261 "" 0 "" {TEXT 266 13 "FINAL VERSION" }}{PARA 258 "" 0 "" {TEXT -1 21 "Solving DE's in Maple " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 331 "Thes e notes discuss material from sections 5.5 and 5.6 of the text, in the context of Maple. They may be helpful when you try to do the Maple \+ project, and are online at our Maple page, go to www.math.utah.edu/~k orevaar/2250fall00, click on the Maple projects and handout link, and \+ then on the 2250Nov1.html or 2250Nov1.mws link." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 256 43 "Part 1: checking homewor k problems on Maple" }}{PARA 0 "" 0 "" {TEXT -1 312 " Here's a hom ework problem worked out in Maple, #37 from section 5.5. This problem was a pain to work by hand. For more information on the commands whi ch are used, use the Help files. I found the syntax for putting highe r order derivatives and initial conditions into \"dsolve\" under the t opic \"dsolve IC\"." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "with( DEtools):\n #DE library of commands" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 144 "\nf:=x->1+x*exp(x);\n #right hand side\ndeqtn:= di ff(y(x),x,x,x) -2*diff(y(x),x,x) \n + diff(y(x),x) =f(x);\n #dif ferential equation from #37" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fGR 6#%\"xG6\"6$%)operatorG%&arrowGF(,&\"\"\"F-*&9$F--%$expG6#F/F-F-F(F(F( " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&deqtnG/,(-%%diffG6$-%\"yG6#%\"x G-%\"$G6$F-\"\"$\"\"\"*&\"\"#F2-F(6$F*-F/6$F-F4F2!\"\"-F(6$F*F-F2,&F2F 2*&F-F2-%$expGF,F2F2" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 181 "an s37:=dsolve(\{deqtn,y(0)=0,D(y)(0)=0,D(D(y))(0)=1\},y(x));\n #getting the expression for second derivative right\n #took a little fiddling . But it still only took a few\n #minutes." }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&ans37G/-%\"yG6#%\"xG,.*&)F)\"\"$\"\"\"-%$expGF(F.#F. \"\"'*&#F.\"\"#F.*&)F)F5F.F/F.F.!\"\"*(F-F.F)F.F/F.F.*&\"\"%F.F/F.F8F) F.F;F." }}}{EXCHG {PARA 0 "" 0 "" {TEXT 262 57 "Part 2: Undamped, forc ed oscillators, w and w0 different:" }{TEXT -1 390 " If the forcing f requency is not equal to the natural frequency the general solution to will be the superposition of two cos-sin terms, one corresponding to \+ the particular solution with angular frequency w, and the other being \+ the general solution to the homogeneous problem, with angular frequenc y w0. When w and w0 as well as the corresponding amplitudes are close , the system exhibits " }{TEXT 264 7 "beating" }{TEXT -1 59 ". Music ians tune their instruments using this phenomenon." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "restart:with(DEtools):with(plots):\n #I w ant to undefine m, x0 etc." }}{PARA 7 "" 1 "" {TEXT -1 50 "Warning, th e name changecoords has been redefined\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 132 "deqtn1a:=diff(x(t),t,t) + w0^2*x(t) = (F0/m)*cos(w*t );\n #this is the case w not equal to w0,\n #I divided the model e quation by m" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(deqtn1aG/,&-%%diffG 6$-%\"xG6#%\"tG-%\"$G6$F-\"\"#\"\"\"*&)%#w0GF1F2F*F2F2*&*&%#F0GF2-%$co sG6#*&%\"wGF2F-F2F2F2%\"mG!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 115 "dsolve(deqtn1a,x(t));\n #general solution, equal to particul ar\n #solution plus general homogeneous eqtn solution. " }}{PARA 11 " " 1 "" {XPPMATH 20 "6#/-%\"xG6#%\"tG,(*&%$_C1G\"\"\"-%$cosG6#*&%#w0GF+ F'F+F+F+*&%$_C2GF+-%$sinGF.F+F+*&*&%#F0GF+-F-6#*&%\"wGF+F'F+F+F+*&%\"m GF+,&*$)F0\"\"#F+!\"\"*$)F;FAF+F+F+FBFB" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 85 "sol4:=dsolve(\{deqtn1a,x(0)=0,D(x)(0)=0\},x(t));\n # a nice choice of initial conditions" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #>%%sol4G/-%\"xG6#%\"tG,&*&*&%#F0G\"\"\"-%$cosG6#*&%#w0GF.F)F.F.F.*&% \"mGF.,&*$)F3\"\"#F.!\"\"*$)%\"wGF9F.F.F.F:F.*&*&F-F.-F06#*&F=F.F)F.F. F.*&F5F.F6F.F:F:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "sol5:=s ubs(\{w=45,w0=55,F0=50,m=.1\},sol4);\n #Example 2 page 342" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%sol5G/-%\"xG6#%\"tG,&-%$cosG6#,$F)\"#b$!+ ++++]!#5*&$\"+++++]F2\"\"\"-F,6#,$F)\"#XF6F6" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 194 "actual:=plot(rhs(sol5),t=0..3, color=black):\ne nvel1:=plot(sin(5*t),t=0..3,color=black):\nenvel2:=plot(-sin(5*t),t=0. .3,color=black):\ndisplay(\{actual, envel1,envel2\});\n #the picture on page 342." }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6 '-%'CURVESG6$7h`n7$$\"\"!F)F(7$$\"3s******\\i9Rl!#>$!3uI33_$[!*4%F-7$$ \"3:+vVtdN]sF-$!3og$*Ro@()[;!#=7$$\"3f+](oHl:'zF-$!3SMs^WV\\%)GF57$$\" 3.,DJ?[xs')F-$!3#o/&*G$>A3RF57$$\"32++vVV)RQ*F-$!3'pDdRK\"y?XF57$$\"3= v$4Ys'yh&*F-$!30]uF]X/!f%F57$$\"3!*[(oa5*eR(*F-$!3^fD%Q(Qt@YF57$$\"3,C \"Gj[\"R<**F-$!35+m`T!pVh%F57$$\"3\"*\\(=nQ>&45F5$!3\\(elVH5oc%F57$$\" 3$\\i!*G')z]/\"F5$!3!fDy6-j$[VF57$$\"3&**\\i!R.k!3\"F5$!3V*[**Rg3Z'RF5 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To be consistent with the example above, we consider the dif ference" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "h:=g-f;\nh(t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"hG,&%\"gG\"\"\"%\"fG!\"\"" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#,&*&%#C1G\"\"\"-%$cosG6#,&*&%#w1GF&%\" tGF&F&%\"bG!\"\"F&F&*&%#C0GF&-F(6#,&*&%#w0GF&F-F&F&%\"aGF/F&F/" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 359 "T hen if the two amplitudes C0 and C1 are close, and if the two frequenc ies w0 and w1 are close, there will be long time intervals in which h= g-f essentially equals twice one of them, followed by long time inter vals in which their difference is essentially zero. In the example a bove, C0=C1=1/2 and a=b=0. In fact, if we set a=b=0 and C1=C2=C, the n we get" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "h1(t):=t->C*(co s(w1*t)-cos(w0*t));\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>-%#h1G6#%\"t GRF&6\"6$%)operatorG%&arrowGF)*&%\"CG\"\"\",&-%$cosG6#*&%#w1GF/9$F/F/- F26#*&%#w0GF/F6F/!\"\"F/F)F)F)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 157 "Mathematically, beating is explained by defining the average angu lar frequency w=(w0 + w1)/2, and half the difference dw=(w1-w0)/2, so \+ that h1(t) is given by" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "h 1:=t->C*(cos((w+dw)*t)-cos((w-dw)*t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#h1GR6#%\"tG6\"6$%)operatorG%&arrowGF(*&%\"CG\"\"\",&-%$cosG6# *&,&%\"wGF.%#dwGF.F.9$F.F.-F16#*&,&F5F.F6!\"\"F.F7F.F " 0 "" {MPLTEXT 1 0 19 "expand(h1(t),trig);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*(%\"CG\" \"\"-%$sinG6#*&%\"tGF&%\"wGF&F&-F(6#*&F+F&%#dwGF&F&!\"#" }}}{PARA 260 "" 1 "" {TEXT -1 375 "So we see that this difference of cosines is act ually a product of sines. If dw is small, then we can consider 2Csin (t*dw) as a slowly varying amplitude (with large period T=2*Pi/dw), a nd this will create the \"envelope\" for the more rapidly oscillating \+ sin(tw) term. In our example the envelope was created by sin(5t), and the rapid oscillations were created by sin(50t)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 257 71 "Part 3: Resonance in undamped , forced harmonic oscillators, when w=w0:" }{TEXT -1 322 " In class w e used variation of parameters to solve the forced oscillator, with no damping, in the case that the driving frequency w exactly equals the \+ natural frequency w0. Well, we found a particular solution. Converti ng our variables from \"x\" to \"t\", and our function from \"y(x)\" t o \"x(t)\", we were considering the DE" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 89 "deqtn1:=diff(x(t),t,t) + w0^2*x(t) = (F0/m)*cos(w0*t) ;\n #I have again written k/m = w0^2" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'deqtn1G/,&-%%diffG6$-%\"xG6#%\"tG-%\"$G6$F-\"\"#\"\"\"*&)%#w0GF1 F2F*F2F2*&*&%#F0GF2-%$cosG6#*&F5F2F-F2F2F2%\"mG!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "sol1:=dsolve(deqtn1,x(t));\n #general s olution" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%sol1G/-%\"xG6#%\"tG,**&% $_C1G\"\"\"-%$cosG6#*&%#w0GF-F)F-F-F-*&%$_C2GF--%$sinGF0F-F-*&*(#F-\" \"#F-%#F0GF-F.F-F-*&%\"mGF-)F2F:F-!\"\"F-*&**F9F-F;F-F5F-F)F-F-*&F2F-F =F-F?F-" }}}{PARA 0 "" 0 "" {TEXT -1 67 "You see there the particular \+ solution which we found, namely xp(t):" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "xp:=t->(1/2)*F0*t*sin(w0*t)/(w0*m);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#xpGR6#%\"tG6\"6$%)operatorG%&arrowGF(,$*&*(%#F0G \"\"\"9$F0-%$sinG6#*&%#w0GF0F1F0F0F0*&F6F0%\"mGF0!\"\"#F0\"\"#F(F(F(" }}}{PARA 0 "" 0 "" {TEXT -1 95 "If we wanted to solve the general ini tial value problem, we would have set things up this way:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "sol2:=dsolve(\{deqtn1,x(0)=x0,D(x)( 0)=v0\},x(t));\nsimplify(rhs(%));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#> %%sol2G/-%\"xG6#%\"tG,**&*&,&%#F0G!\"\"**\"\"#\"\"\"%#x0GF2%\"mGF2)%#w 0GF1F2F2F2-%$cosG6#*&F6F2F)F2F2F2*&F4F2F5F2F/#F2F1*&*&%#v0GF2-%$sinGF9 F2F2F6F/F2*&*(F " 0 "" {MPLTEXT 1 0 43 "sol3:=subs(\{F0=1,m=1,w0=3,x0=0,v0=0\},sol2); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%sol3G/-%\"xG6#%\"tG,$*&-%$sinG6 #,$F)\"\"$\"\"\"F)F1#F1\"\"'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT 265 31 "Graphing solutions from dsolve :" }{TEXT -1 554 " If we want to graph this x(t) we cannot just try to plot x(t), because it hasn't actually been defined yet. We have to e xtract it from the right hand side of sol3. One way is manually type in the formula. Or, use your mouse to highlight it (in this case t he 1/6*t*sin(3*t)) and then paste it into the \"plot\" command. Or, \+ plot \"rhs(sol1)\". Alternately, we can make an intermediate step i n which we define the solution function. I prefer this approach, espe cially if I'm going to work with the solution later or if I want to cr eate a template." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "with(plo ts):\n #plotting library" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 147 "f1:=t1->subs(t=t1,rhs(sol3));\n #O.K. so this is weird. You hav e to do it\n #so that Maple knows what the actual variable \n #is \+ in your formula" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#f1GR6#%#t1G6\"6$ %)operatorG%&arrowGF(-%%subsG6$/%\"tG9$-%$rhsG6#%%sol3GF(F(F(" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "f1(6);\n #check my answer " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$sinG6#\"#=" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 69 "plot(f1(t),t=0..15, color=black);\n #A cla ssic picture of resonance." }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6&-%'CURVESG6#7^y7$$\"\"!F)F(7$$\"3')*****\\7t&pK!#=$\"3u '>\")Guk#GX!#>7$$\"3=+](=<#*>p%F-$\"3C%49n')zgr(F07$$\"3$****\\(=7T9hF -$\"3W,reQZ()Q)*F07$$\"3:]P%['fK9lF-$\"31d%**yTao+\"F-7$$\"3O+v$4rSU\" pF-$\"3RwOb#Rw$45F-7$$\"3e]7.da:9tF-$\"3-*3Q.\"=*o*)*F07$$\"3p**\\7.-2 9xF-$\"3gh`/s_ek%*F07$$\"3-*\\7`p**Q^)F-$\"3uho?SbDkyF07$$\"3X****\\(= HPJ*F-$\"3'G:;+?5fG&F07$$\"3;+++v\"*R#4\"!#<$!3'HU$e76NhCF07$$\"3;++DJ aU`7Ffn$!34Tpx;zc67F-7$$\"32+D\"yN'o89Ffn$!3QlF-7$$\"3!***\\PM0'\\-# 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If the driving frequency w1 is allowed to vary, and approaches the natural frequency w0, then the beating packets will stretch out since T=2*Pi/dw will approach infini ty. The beating picture will transform into the resonance picture. F or example, consider the following initial value problem, which is lik e the one we just , except that the driving angular velocity is 3.1 in stead of 3:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "restart:with( DEtools):with(plots):" }}{PARA 7 "" 1 "" {TEXT -1 50 "Warning, the nam e changecoords has been redefined\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 76 "deqtn:=diff(x(t),t,t) + 9*x(t)=cos(3.1*t):\ndsolve(\{ deqtn,x(0)=0,D(x)(0)=0\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"xG6 #%\"tG,&-%$cosG6#,$F'#\"#J\"#5#!$+\"\"#h*&#\"$+\"F2\"\"\"-F*6#,$F'\"\" $F6F6" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 118 "plot(-100/61*cos( 31/10*t)+100/61*cos(3*t),\n t=0..15, color=black);\n #this will lo ok very close to previous picture" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6&-%'CURVESG6#7^y7$$\"\"!F)F(7$$\"3')*****\\7t&pK !#=$\"3&e$*>0,y>]%!#>7$$\"3=+](=<#*>p%F-$\"39eZ3oa6;wF07$$\"3$****\\(= 7T9hF-$\"3S]sH\"4`92q%*F07$$\"3p**\\ 7.-29xF-$\"3dr.)>WQ-(*)F07$$\"3-*\\7`p**Q^)F-$\"3G^eFi_tJsF07$$\"3X*** *\\(=HPJ*F-$\"3;*oI3N!pBXF07$$\"3;+++v\"*R#4\"!#<$!3K.$oQ1!R%Q$F07$$\" 3;++DJaU`7Ffn$!3qr%>l%yN$H\"F-7$$\"32+D\"yN'o89Ffn$!3Rq*)[mj@K@F-7$$\" 3(***\\P%GZRd\"Ffn$!3@4\"Hv!4dnDF-7$$\"3'**\\7G)G46;Ffn$!3H4GI12$\\e#F -7$$\"3%****\\7[Q#[;Ffn$!3g,&4q>#RmDF-7$$\"3\"**\\(ozSQ&o\"Ffn$!3$fnK6 E\"p5DF-7$$\"3!***\\7y'HDs\"Ffn$!3#pw=$H`4zL#Ffn$\"3i4o])>Zk'GF-7$$\"3/+](ofHq\\#Ffn$\"3fAN/2ZtnRF-7$ $\"3U+vo/'zm`#Ffn$\"3Ab\\c=P%*=TF-7$$\"3O++]7'Hjd#Ffn$\"3)f.O$=&*y6UF- 7$$\"3J+DJ?'zfh#Ffn$\"3s,M_$RQIC%F-7$$\"3D+]7G'Hcl#Ffn$\"3#eze=We.@%F- 7$$\"39++vV'H\\t#Ffn$\"3g,6$Q&yu[RF-7$$\"3-+]Pf'HU\"GFfn$\"3'G-$o/OIGM F-7$$\"3$)*\\PfGpt(HFfn$\"36-\\?$oR\"R;F-7$$\"33++]7*309$Ffn$!3o/x-SMM `yF07$$\"3*)**\\P%Q-UG$Ffn$!3Wf4c\"*e/\"*HF-7$$\"39++Dce*yU$Ffn$!3=^'G .`;\\z%F-7$$\"3E+D1k#y(3NFfn$!3_PB-5*>uZ&F-7$$\"3P+](=ng'*e$Ffn$!3H')) G$Q*[VC_F-7$$\"35]P4YoTgWFfn$\"3![Py1ErhC'F-7$$\"34+voz\"*[JX Ffn$\"3t:;JqWI+qF-7$$\"3juV[Y`-nXFfn$\"3*HW&G^HXisF-7$$\"32]7G8:c-YFfn $\"3s&3=hLz=W(F-7$$\"3]D\"y+o(4QYFfn$\"33Fo!ybi^`(F-7$$\"3/+](o%QjtYFf n$\"37n+_'fz)RvF-7$$\"31+vo/wlU[Ffn$\"3ULkb>\">'QjF-7$$\"32++]i8o6]Ffn $\"33w%z]\"G[KLF-7$$\"3H++Dcm'[:&Ffn$!3rB&[j'yu,7F07$$\"3i******\\>0)H &Ffn$!3AVu>(o`Au$F-7$$\"3a*\\Pf[5YY&Ffn$!3Aup#4)33VsF-7$$\"3Y**\\(=-p6 j&Ffn$!3w&z-fZ.51*F-7$$\"3dCc^`r-ocFfn$!3u&\\L9L;w;*F-7$$\"3r\\i:&G&)[ q&Ffn$!3w:iOZ&\\$e\"*F-7$$\"3$[(oz;MuTdFfn$!3iTh_N;'=.*F-7$$\"3&**\\P% [:gydFfn$!3!>V=0w7$)y)F-7$$\"3?](=<\"yJ_eFfn$!3Q**>LwzDezF-7$$\"3c**** *\\2Mg#fFfn$!3K'yR'*>='*p'F-7$$\"3q\\(=#*\\7p+'Ffn$!3mwf'fV#*3*[F-7$$ \"3$)*\\PM#4z(3'Ffn$!3Cr7)G[mDt#F-7$$\"3'*\\ilZ$p'ohFfn$!3;z\"z%H!=U\\ $F07$$\"35+](=xZ&\\iFfn$\"3Y4])pLTl6#F-7$$\"37+D\"y:jlK'Ffn$\"3UH^T*)e #QS%F-7$$\"38++vV&yNS'Ffn$\"3QC\"f!*4`&*\\'F-7$$\"39+voHRf!['Ffn$\"3cC [@Qq`\"G)F-7$$\"3:+]i:$4wb'Ffn$\"3lfb8$poCk*F-7$$\"3CDc^`!)y(f'Ffn$\"3 )pvp2mib,\"Ffn7$$\"3Y\\iS\"zmzj'Ffn$\"3@3grE%)3_5Ffn7$$\"3ch:Ngh0emFfn $\"3T7@-Y#oX1\"Ffn7$$\"3auoHHb9ymFfn$\"3H&=\"\\$f%4t5Ffn7$$\"3_(=U#)*[ B)p'Ffn$\"30FB.!=7w2\"Ffn7$$\"3k*\\(=nUK=nFfn$\"3\\$fhzq!3y5Ffn7$$\"3# )\\(oHu\"o)z'Ffn$\"3AI4hZZ>R5Ffn7$$\"3,++v=#R!zoFfn$\"3;zQoT%)4j$*F-7$ $\"3T]iSmC$G&pFfn$\"3#ps\"3U2B&*yF-7$$\"3!3]iSrDm-(Ffn$\"33Zm>k.4$*fF- 7$$\"3J](=<'*=/5(Ffn$\"3wi3GurgXPF-7$$\"3p+]P4A@urFfn$\"3(*y?Ps#>8E\"F -7$$\"3Q]P4'>3QD(Ffn$!3#)QhM0n^.&f)oF-7$$\"3I++Dchf#\\(Ffn$!3jxSW+UD-\"*F-7$ $\"3(*\\iS;SFvvFfn$!3)3uO.\\ix3\"Ffn7$$\"3l*\\il(=&zl(Ffn$!3?'4(>*4`#* >\"Ffn7$$\"3si:gT8iywFfn$!3-J+6\"p0d@\"Ffn7$$\"3#\\iSm!3H*p(Ffn$!32AU6 =JQF7Ffn7$$\"37(oz;Fg*>xFfn$!3k!\\2:;9UB\"Ffn7$$\"3@](=ntH1u(Ffn$!3'>L \"zen9O7Ffn7$$\"3S7yv,#*HhxFfn$!3B0/q4$[JB\"Ffn7$$\"3guozm'o>y(Ffn$!3l %oA[!f?D7Ffn7$$\"3oPf$=8QE!yFfn$!3a]u]vkK77Ffn7$$\"3))**\\(of2L#yFfn$! 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The final answer shows that for small dw (and small dw*t), x1(t) i s the first order approximation for x2(t). In other words the beating picture looks like the resonance picture, for t in a fixed interval. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 260 45 "Part 4: Forcing damped harmonic oscillators." }{TEXT -1 595 " Wh en c>0 (damping), the homogeneous part of the general solution dies ou t, no matter what the initial conditions are. Therefore this part of \+ the solution is called TRANSIENT. The part of the solution which is a linear combination of cos(wt) and sin(wt) (and which we get by the me thod of undetermined coefficients as a particular solution), persists, and is called the STEADY STATE solution. No matter what the initial \+ conditions are, you will always end up with the same steady state solu tion for a given spring system! (WHY?) Let's study these phenomena, \+ by doing Example 6, page 347. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "restart:with(DEtools):with(p lots):" }}{PARA 7 "" 1 "" {TEXT -1 50 "Warning, the name changecoords \+ has been redefined\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 97 "deq tn:=diff(x(t),t,t) +2*diff(x(t),t) + 26*x(t) = 82*cos(4*t);\ndsolve(\{ deqtn, x(0)=6,D(x)(0)=0\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&deqt nG/,(-%%diffG6$-%\"xG6#%\"tG-%\"$G6$F-\"\"#\"\"\"*&F1F2-F(6$F*F-F2F2*& \"#EF2F*F2F2,$-%$cosG6#,$F-\"\"%\"##)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"xG6#%\"tG,**&-%$expG6#,$F'!\"\"\"\"\"-%$sinG6#,$F'\"\"&F/!\"$* &F*F/-%$cosGF2F/F/*&F4F/-F86#,$F'\"\"%F/F/*&F=F/-F1F;F/F/" }}}{PARA 0 "" 0 "" {TEXT -1 290 "So we see the transient solution (the terms with exp(t) in them), and the steady periodic solution. We could convert \+ the steady state solution into Ccos(4t-a), if we wanted to, as is done on page 348. For our graph it is enough to notice that the amplitude of the steady state solution is" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "C:=sqrt(5^2+4^2);\nevalf(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#>%\"CG*$-%%sqrtG6#\"#T\"\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+ PU7.k!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 130 "Here's a plot of th e solution, decomposed into transient and steady state pieces. Notice we got the steady-state amplitude right." }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 242 "xtrans:=t->-3*exp(-t)*sin(5*t)+exp(-t)*cos(5*t):\n xsp:=t->5*cos(4*t)+4*sin(4*t):\nplot1:=plot(xtrans(t),t=0..4,color=bla ck):\nplot2:=plot(xsp(t),t=0..4,color=black):\nplot3:=plot(xtrans(t)+x sp(t),t=0..4,color=black):\ndisplay(\{plot1,plot2,plot3\});" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6'-%'CURVESG6$7jq7$$\"\" !F)$\"\"\"F)7$$\"3HLLL3x&)*3\"!#>$\"3%=V*30plg#)!#=7$$\"3dmmm;arz@F/$ \"3&[Jkf%yeLlF27$$\"3')*****\\7t&pKF/$\"3-H0#*QF!Q#[F27$$\"39LLLL3VfVF /$\"3L\\Px$p]h8$F27$$\"3s******\\i9RlF/$!3P)*[-r3HU:F/7$$\"3Hmmmm;')=( )F/$!3tXGbsjQ-LF27$$\"3JLLeR?ah5F2$!3e/Q3z*=4!fF27$$\"3,++]7z>^7F2$!3t 7$[$eAi[$)F27$$\"3qmmT&y`3W\"F2$!3]t\"p/amG1\"!#<7$$\"3RLLLe'40j\"F2$! 3K`twY\"QEF\"Fen7$$\"3qmm\"H_(zV=F2$!3LMW\")feA&[\"Fen7$$\"3.++](Q&3d? 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d#*yPFxeFen7$Fbgl$!3uh6vwT>\"R&Fen7$Fggl$!3[:,*pUjrr%Fen7$F\\hl$!3$ysD -**GF!RFen7$Fahl$!3Wphk`IojIFen7$Ffhl$!3KjG(\\@Fen7$F[il$!3W#)zmEK A\"4\"Fen7$F`il$!3H%[Hh2!\\5%)F`]m7$Feil$\"3;,=yS9sD5Fen7$Fjil$\"3%fbs Zc[d-#Fen7$F^cn$\"3C1@A#*Gk9IFen7$F_jl$\"3X)3z$G'Q'Fen7$F`en$\"3Q H;P#**)***R'Fen7$Feen$\"38SKQ6*)y-kFen7$Fjen$\"3'QJHvJkXR'Fen7$F^[m$\" 3pdaJf$R`P'Fen7$Fbfn$\"3!e)=x2A<$='Fen7$Fc[m$\"3kzT63RU8eFen7$Fjfn$\"3 tJ$y@a/=G&Fen7$Fh[m$\"3G(*4Sns)4g%Fen7$Fbgn$\"3^XXN&ez-'QFen7$F]\\m$\" 3'4+)HG^YFIFen7$Fjgn$\"3/i(=tk-a)>Fen7$Fb\\m$\"3k$yO_KC:\"))F27$Fbhn$! 3hSFen7$Fb]m$!3w\\2-*)p%*pZFen7$Fjin$!3*= TQQ?eoV&Fen7$Fg]m$!31%=x#o1\"*RfFenF[^m-%+AXESLABELSG6%Q\"t6\"Q!6\"%(D EFAULTG-%%VIEWG6$;F(Fg]mF`cp" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" }}}}{PARA 0 "" 0 "" {TEXT 261 20 "Practical resonance." }{TEXT -1 724 " A damped sys tem can't resonate completely, but different driving frequencies will \+ effect the amplitude of the steady state solution. For small \"c\" on e expects the steady-state amplitude to get large with driving frequen cy near to that of the natural frequency of the system. This principa l can be a useful tool in engineering: find the natural frequencies of systems in the theoretical case of c=0, and be sure to damp frequenci es near those in the actual physical problem. For our forced oscillat or it is not too hard to calculate the amplitude of the steady state s olution, as the text does on page 346. Doing this by hand is a good e xercise; you may get more out of it than you get out of letting Maple \+ do the work:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "restart:with(linalg):with(plots):" }}{PARA 7 "" 1 "" {TEXT -1 80 "Warning, the protected names norm and trace have been \+ redefined and unprotected\n" }}{PARA 7 "" 1 "" {TEXT -1 50 "Warning, t he name changecoords has been redefined\n" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 31 "xp:=t->A*cos(w*t) + B*sin(w*t);" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#>%#xpGR6#%\"tG6\"6$%)operatorG%&arrowGF(,&*&%\"AG\"\" \"-%$cosG6#*&%\"wGF/9$F/F/F/*&%\"BGF/-%$sinGF2F/F/F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "eqtn1:=m*diff(xp(t),t,t)+c*diff(xp( t),t) + k*xp(t)=F0*cos(w*t);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%&eqt n1G/,(*&%\"mG\"\"\",&*(%\"AGF)-%$cosG6#*&%\"wGF)%\"tGF)F))F1\"\"#F)!\" \"*(%\"BGF)-%$sinGF/F)F3F)F5F)F)*&%\"cGF),&*(F,F)F8F)F1F)F5*(F7F)F-F)F 1F)F)F)F)*&%\"kGF),&*&F,F)F-F)F)*&F7F)F8F)F)F)F)*&%#F0GF)F-F)" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 63 "Equating coefficients for coswt an d sinwt leads to the system: " }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 82 "what:=solve(\{-m*A*w^2 + c*B *w + k*A = F0,\n -m*B*w^2 -c*A*w + k*B =0\},\{A,B\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%whatG<$/%\"AG,$*&*&,&*&%\"mG\"\"\")%\"wG \"\"#F.F.%\"kG!\"\"F.%#F0GF.F.,**&)%\"cGF1F.F/F.F.*&)F-F1F.)F0\"\"%F.F .**F1F.F-F.F/F.F2F.F3*$)F2F1F.F.F3F3/%\"BG*&*(F8F.F0F.F4F.F.F5F3" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "what[1];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%\"AG,$*&*&,&*&%\"mG\"\"\")%\"wG\"\"#F+F+%\"kG!\"\"F+% #F0GF+F+,**&)%\"cGF.F+F,F+F+*&)F*F.F+)F-\"\"%F+F+**F.F+F*F+F,F+F/F+F0* $)F/F.F+F+F0F0" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "rhs(what[ 1]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&*&,&*&%\"mG\"\"\")%\"wG\" \"#F)F)%\"kG!\"\"F)%#F0GF)F),**&)%\"cGF,F)F*F)F)*&)F(F,F))F+\"\"%F)F)* *F,F)F(F)F*F)F-F)F.*$)F-F,F)F)F.F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 101 "amplitude:=simplify(sqrt(rhs(what[1])^2 + rhs(what[2 ])^2));\n #this is the formula (21) on page 346." }}{PARA 11 "" 1 " " {XPPMATH 20 "6#>%*amplitudeG*$-%%sqrtG6#*&*$)%#F0G\"\"#\"\"\"F.,**&) %\"cGF-F.)%\"wGF-F.F.*&)%\"mGF-F.)F4\"\"%F.F.**F-F.F7F.F3F.%\"kGF.!\" \"*$)F;F-F.F.F " 0 "" {MPLTEXT 1 0 45 "ex amp6:=subs(\{F0=82,c=2,k=26,m=1\},amplitude);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'examp6G*&-%%sqrtG6#\"%Cn\"\"\"-F'6#*&F*F*,(*$)%\"wG \"\"#F*!#[*$)F1\"\"%F*F*\"$w'F*!\"\"F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "amp:=t->subs(w=t,examp6);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$ampGR6#%\"tG6\"6$%)operatorG%&arrowGF(-%%subsG6$/%\" wG9$%'examp6GF(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "amp(t );" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&-%%sqrtG6#\"%Cn\"\"\"-F%6#*&F( F(,(*$)%\"tG\"\"#F(!#[*$)F/\"\"%F(F(\"$w'F(!\"\"F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "plot(amp(w),w=0..20, color=black);\n #prac tical resonance, page 348" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6&-%'CURVESG6#7]o7$$\"\"!F)$\"3>ah%Q:YQ:$!#<7$$\"39LLLL3V fV!#=$\"30U%3fkd_<$F,7$$\"3%pmm;H[D:)F0$\"3-Ww')*G,*HKF,7$$\"3MLLLe0$= C\"F,$\"3cEA*ym^bL$F,7$$\"3iLLL3RBr;F,$\"331>v!eJt\\$F,7$$\"3imm;zjf)4 #F,$\"3HV$QaXwss$F,7$$\"3ULL$e4;[\\#F,$\"3%z_\"f:gY?SF,7$$\"3!)****\\i 'y]!HF,$\"3k!ebp]6KV%F,7$$\"3'pmm\"HdA(e)oF,7$$\"3/,+]7l$*yVF,$\"3B%eBWg&H&Q(F,7$$\"3$QLL$3y_qXF ,$\"3EaT2B6,IyF,7$$\"3$4+](=5PyYF,$\"3:pg#4%*zG-)F,7$$\"3;nm;HU@'y%F,$ \"3U&**[4WG::)F,7$$\"3)F,7$$\"3QLLeRu0%*[F,$ \"3=#>#zTZ!**>)F,7$$\"3Rn;zW!zz%\\F,$\"3aT$e0rr/>)F,7$$\"3]+++]1!>+&F, $\"3a2')\\o]vd\")F,7$$\"3S++++FZ=_F,$\"3*[r'fWI\\-yF,7$$\"3J+++]Z/NaF, $\"3'R+B854H<(F,7$$\"3!)******\\?vVcF,$\"3:5N*G1T%\\kF,7$$\"3;+++]$fC& eF,$\"3![\"*QP5*)es&F,7$$\"3Qnm\"Hd&)>/'F,$\"3)pO9\\\"H5@^F,7$$\"3qLL$ ez6:B'F,$\"39rx>-\"zSe%F,7$$\"3P++D1o(oX'F,$\"3x/c5C$F,7$$\"3Mnmmm#pS1( F,$\"3%y28'o^T`HF,7$$\"3<++]i`A3vF,$\"3LH)=q;6,U#F,7$$\"3Rmmmm(y8!zF,$ \"3N%*3G7P*[1#F,7$$\"3K,+]i.tK$)F,$\"3.Eo))4>hi>OF07 $$\"3\"****\\i!*3`i\"Fcx$\"3zub6PJR6MF07$$\"3_LLL$*zym;Fcx$\"3%Q*[N$3a \"GKF07$$\"3fLL$3N1#4Fcx$\"3k#**** fMO')Q#F07$$\"3A++v.Uac>Fcx$\"3dAzgnrY%G#F07$$\"#?F)$\"3m9a5y,3!=#F0-% 'COLOURG6&%$RGBGF)F)F)-%+AXESLABELSG6$Q\"w6\"Q!6\"-%%VIEWG6$;F(Fe_l%(D EFAULTG" 1 2 0 1 10 0 2 6 1 4 2 1.000000 45.000000 45.000000 0 0 "Curv e 1" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "solve(diff(amp(w),w )=0,w);\n #find w value at max amplitude" }}{PARA 11 "" 1 "" {XPPMATH 20 "6%\"\"!,$*$-%%sqrtG6#\"\"'\"\"\"\"\"#,$F%!\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "evalf(2*sqrt(6));evalf(amp(2*sqrt(6 )));\n #practical resonance" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+'[ z*)*[!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+++++#)!\"*" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "evalf(sqrt(26));\n #natural frequency of undamped system " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+ 9&>!*4&!\"*" }}}{PARA 0 "" 0 "" {TEXT -1 4 "END!" }}}{MARK "66 0" 455 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }