{VERSION 5 0 "IBM INTEL NT" "5.0" } {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 Input" 2 19 "" 0 1 255 0 0 1 0 0 0 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 "" 1 16 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Out put" -1 11 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 3 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Plot" -1 13 1 {CSTYLE " " -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 1 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 256 13 "Wave equation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 267 "Classical mechan ics texts discuss the wave equation as an example of continuous many-p article systems where the particles are connected by springs. One deri vation can be found in Cassiday and Fowles (6th ed., chapter 11.5/6) t he summary of which is given in the page " }{TEXT 19 11 "WaveEqn.jpg" }{TEXT -1 97 " . An example of a finite system of masses coupled by sp rings to simulate vibrations is given in " }{TEXT 19 9 "Glass.mws" } {TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 110 "The wave equation conta ins a constant, namely the square of the propagation speed for the wav es which we call " }{TEXT 19 1 "c" }{TEXT -1 296 ". In classical mecha nics (transverse vibrations of a continuous string) the square of the \+ speed equals the string tension divided by the linear mass density. Th e wave equation plays an important role in electromagnetism, where it \+ is derived from Maxwell's equation for the electromagnetic fields." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 195 "Our inte rest is in applying the Fourier series to solve the wave equation in o ne dimension for a plucked string. We have used this technique in the \+ analysis of the one-dimensional heat equation (" }{TEXT 19 11 "HeatEqn .mws" }{TEXT -1 30 "), and explored it further in " }{TEXT 19 17 "Four ierSeries.mws" }{TEXT -1 2 ". " }{TEXT 19 6 "u(x,t)" }{TEXT -1 203 " d enotes the vertical (transverse) displacement of the string as a funct ion of location and time. The assumption is that no longitudinal motio n occurs, which can be achieved in the small-amplitude limit." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 365 "The equa tion (in mechanics) inherits the two time derivatives from Newton's eq uation. The two spatial derivatives arise as a consequence of the harm onic force experienced by a mass point from its two neighbours: the di fference of relative displacements can be expressed as a difference of first derivatives, and thus as a second derivative with respect to lo cation." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "WE:=diff(u(x,t),t$2) = c^2*d iff(u(x,t),x$2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#WEG/-%%diffG6$- %\"uG6$%\"xG%\"tG-%\"$G6$F-\"\"#*&)%\"cGF1\"\"\"-F'6$F)-F/6$F,F1F5" }} }{EXCHG {PARA 0 "" 0 "" {TEXT -1 33 "We choose a unit system in which \+ " }{TEXT 19 9 "0 < x < 1" }{TEXT -1 7 " , and " }{TEXT 19 3 "c=1" } {TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "sol:=pdsol ve(WE);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG/-%\"uG6$%\"xG%\"tG, &-%$_F1G6#,&*&%\"cG\"\"\"F*F2F2F)F2F2-%$_F2G6#,&F0F2F)!\"\"F2" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 8 "Maple's " }{TEXT 19 7 "pdsolve" } {TEXT -1 315 " engine realizes that the solutions to the wave equation can be expressed as a sum of a left-travelling and a right-travelling wave solution. While this is interesting in its own right (waves on t he string can cross, standing wave solutions are possible) this is not necessarily directly useable to find the answer." }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 5 "c:=1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\" cG\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "pdsolve(WE,HINT =f(x)*g(t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'&whereG6$/-%\"uG6$% \"xG%\"tG*&-%\"fG6#F*\"\"\"-%\"gG6#F+F07#<$/-%%diffG6$F--%\"$G6$F*\"\" #*&&%#_cG6#F0F0F-F0/-F86$F1-F;6$F+F=*&F?F0F1F0" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 284 "If we ask Maple for a solution in product form we a re reminded of how separation of variables connects the spatial and te mporal parts in the solution. The structure of the ODEs is identical, \+ but we usually have a boundary-value problem in space, and an initial- value problem in time. " }{TEXT 19 3 "_c1" }{TEXT -1 28 " is the separ ation constant." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 "Let us fix the string at the endpoints." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "ODE_x:=diff(f(x),x$2)=c1*f(x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&ODE_xG/-%%diffG6$-%\"fG6#%\"xG-%\"$G6$F, \"\"#*&%#c1G\"\"\"F)F3" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 124 "This i s a differential-equation eigenvalue problem which defines a Fourier b asis. The condition that the string be fixed at " }{TEXT 19 3 "x=0" } {TEXT -1 135 " is used to select sine-solutions only, i.e., it removes one integration constant. The other boundary condition selects the ei genvalue " }{TEXT 19 2 "c1" }{TEXT -1 252 ". The amplitude of the sine -wave solutions is arbitrary, because the problem is homogeneous (an e igenvalue problem always is). The eigenvalues are negative (for sine/c osine type solutions) and correspond to the negative of the square of \+ the wavenumber " }{TEXT 19 1 "k" }{TEXT -1 8 " in the " }{TEXT 19 8 "s in(k*x)" }{TEXT -1 29 " solutions. Let us find them." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "sol_x:=dsolve(\{ODE_x,f(0)=0\});" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%&sol_xG/-%\"fG6#%\"xG,&*&%$_C2G\"\" \"-%$expG6#*&%#c1G#F-\"\"#F)F-F-!\"\"*&F,F--F/6#,$F1F5F-F-" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 119 "This reminds us of the fact that the tri g solutions can be expressed as complex exponentials. Those are less i ntuitive." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "assume(c1<0); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "sol_x:=dsolve(\{ODE_x,f (0)=0\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&sol_xG/-%\"fG6#%\"xG,$ *&%$_C1G\"\"\"-%$sinG6#*&,$%$c1|irG!\"\"#F-\"\"#F)F-F-F4" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "Now demand that the solution vanish at th e right boundary, i.e., at " }{TEXT 19 3 "x=1" }{TEXT -1 1 ":" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "solve(sin(k*1)=0,k);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 66 "This solution is uninteresting, as it yields the trivial \+ solution " }{TEXT 19 6 "f(x)=0" }{TEXT -1 57 ". How do we make Maple g ive us the interesting solutions?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "_EnvAllSolutions := true:\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "solve(sin(k*1)=0,k);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&%#PiG\"\"\"%%_Z1|irGF%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "OK, this we know from high school. Integer multiples of \+ " }{TEXT 19 2 "Pi" }{TEXT -1 15 " will work for " }{TEXT 19 1 "k" } {TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "k_n:=n->n* Pi;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$k_nGf*6#%\"nG6\"6$%)operator G%&arrowGF(*&9$\"\"\"%#PiGF.F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "fB:=n->sin(k_n(n)*x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#fBGf*6#%\"nG6\"6$%)operatorG%&arrowGF(-%$sinG6#*&-%$k_nG6#9$ \"\"\"%\"xGF4F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "The separa tion constants:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "c_n:=n-> -k_n(n)^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$c_nGf*6#%\"nG6\"6$%)o peratorG%&arrowGF(,$*$)-%$k_nG6#9$\"\"#\"\"\"!\"\"F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 37 "Are these basis functions normalized?" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "IP:=(f1,f2)->int(expand(f1* f2),x=0..1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#IPGf*6$%#f1G%#f2G6 \"6$%)operatorG%&arrowGF)-%$intG6$-%'expandG6#*&9$\"\"\"9%F5/%\"xG;\" \"!F5F)F)F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "IP(fB(1),fB( 1));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6##\"\"\"\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "IP(fB(2),fB(2));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6##\"\"\"\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "IP(fB(5),fB(5));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6##\"\"\"\"\"#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 75 "This is no proof, but evidence \+ that we should redefine our basis functions:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "fBn:=n->sqrt(2)*sin(k_n(n)*x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$fBnGf*6#%\"nG6\"6$%)operatorG%&arrowGF(*&-%%sqrtG 6#\"\"#\"\"\"-%$sinG6#*&-%$k_nG6#9$F1%\"xGF1F1F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "IP(fBn(1),fBn(3));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "IP( fBn(3),fBn(3));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 97 "Orthonormality of the basis seems to be s atisfied; check this out by trying different values for " }{TEXT 19 2 "n1" }{TEXT -1 2 ", " }{TEXT 19 2 "n2" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 118 "The next question is: how do these basis states evolve in time? The separation constant spe cifies their time behavior:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "ODE_t:=diff(g(t),t$2)=c1*g(t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# >%&ODE_tG/-%%diffG6$-%\"gG6#%\"tG-%\"$G6$F,\"\"#*&%$c1|irG\"\"\"F)F3" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "dsolve(ODE_t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"gG6#%\"tG,&*&%$_C1G\"\"\"-%$sinG6#*&,$% $c1|irG!\"\"#F+\"\"#F'F+F+F2*&%$_C2GF+-%$cosGF.F+F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 143 "Two initial conditions are required to specify the temporal behaviour of each mode (usually the initial displacement , and the intial velocity)." }}{PARA 0 "" 0 "" {TEXT -1 558 "For the c ase of a plucked string we can argue that we have a known initial conf iguration of the string, and zero time derivative for this configurati on at the beginning. This would imply that we are interested in cosine -type temporal solutions only. Another intial condition would be to be in the equilibrium configuration with an intial velocity profile brou ght on by, e.g., a hammer hitting the string at some time. This would \+ call for sine-type solutions only. The general case would be covered b y having a superposition of both types, as expressed above." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 35 "Each mode has a different value of " }{TEXT 19 2 "c1" }{TEXT -1 54 ", and therefore o scillates with a different frequency." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 189 "The actual superposition of modes re quired is obtained by projecting the initial configuration onto the Fo urier basis. Suppose we displace the string at time zero such that it \+ has a height " }{TEXT 19 1 "a" }{TEXT -1 20 " in the middle (for " } {TEXT 19 5 "x=1/2" }{TEXT -1 28 "), and is linear in-between." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "a:=1/10;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"aG#\"\"\"\"#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "f0:=piecewise(x<1/2,2*a*x,x>1/2,2*(a-a*x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#f0G-%*PIECEWISEG6$7$,$*&\"\"&!\"\"%\"xG\" \"\"F.2F-#F.\"\"#7$,&#F.F+F.*&F+F,F-F.F,2F0F-" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "plot(f0,x=0..1);" }}{PARA 13 "" 1 "" {GLPLOT2D 947 161 161 {PLOTDATA 2 "6%-%'CURVESG6$7U7$$\"\"!F)F(7$$\"3emmm;arz@!# >$\"3)HLLL$3VfV!#?7$$\"3[LL$e9ui2%F-$\"3Inmm\"H[D:)F07$$\"3nmmm\"z_\"4 iF-$\"3TLLLe0$=C\"F-7$$\"3[mmmT&phN)F-$\"3ILLL3RBr;F-7$$\"3CLLe*=)H\\5 !#=$\"3imm;zjf)4#F-7$$\"3gmm\"z/3uC\"FC$\"3MLL$e4;[\\#F-7$$\"3%)***\\7 LRDX\"FC$\"3$)****\\i'y]!HF-7$$\"3]mm\"zR'ok;FC$\"3:LL$ezs$HLF-7$$\"3w ***\\i5`h(=FC$\"3%*****\\7iI_PF-7$$\"3WLLL3En$4#FC$\"3Anmm;_M(=%F-7$$ \"3qmm;/RE&G#FC$\"3aLLL3y_qXF-7$$\"3\")*****\\K]4]#FC$\"3v******\\1!>+ &F-7$$\"3$******\\PAvr#FC$\"3H+++]Z/NaF-7$$\"3)******\\nHi#HFC$\"3'*** ****\\$fC&eF-7$$\"3jmm\"z*ev:JFC$\"3ELL$ez6:B'F-7$$\"3?LLL347TLFC$\"37 mmm;=C#o'F-7$$\"3,LLLLY.KNFC$\"3ulmmm#pS1(F-7$$\"3w***\\7o7Tv$FC$\"3y* ***\\i`A3vF-7$$\"3'GLLLQ*o]RFC$\"3$ommmwy8!zF-7$$\"3A++D\"=lj;%FC$\"3b ,+]i.tK$)F-7$$\"31++vV&RY2aF C$\"3I,+]ih2&=*F-7$$\"39mm;zXu9cFC$\"3unmmT3^q()F-7$$\"3l******\\y))Ge FC$\"3r++++VAU$)F-7$$\"3'*)***\\i_QQgFC$\"3\"H++]ZHK#zF-7$$\"3@***\\7y %3TiFC$\"3&=++vVIy^(F-7$$\"35****\\P![hY'FC$\"3y,++DRqnqF-7$$\"3kKLL$Q x$omFC$\"3?MLLL_CjmF-7$$\"3!)*****\\P+V)oFC$\"3%)******\\#*RJiF-7$$\"3 ?mm\"zpe*zqFC$\"3pom;/E3SeF-7$$\"3%)*****\\#\\'QH(FC$\"3*3+++:qAT&F-7$ $\"3GKLe9S8&\\(FC$\"3bOL$3(>t4]F-7$$\"3R***\\i?=bq(FC$\"37++](ej*)e%F- 7$$\"3\"HLL$3s?6zFC$\"3uMLL$e&exTF-7$$\"3a***\\7`Wl7)FC$\"3Q++]P4\"pu$ F-7$$\"3#pmmm'*RRL)FC$\"3;mmmm+7KLF-7$$\"3Qmm;a<.Y&)FC$\"3Nomm\"\\Oz!H F-7$$\"3=LLe9tOc()FC$\"37LL$3Pls[#F-7$$\"3u******\\Qk\\*)FC$\"3k,+++Br +@F-7$$\"3CLL$3dg6<*FC$\"3_LLLe)ywl\"F-7$$\"3ImmmmxGp$*FC$\"3ImmmmWUh7 F-7$$\"3A++D\"oK0e*FC$\"3E&****\\PY$*Q)F07$$\"3A++v=5s#y*FC$\"3_%)*** \\izbM%F07$$\"\"\"F)F(-%'COLOURG6&%$RGBG$\"#5!\"\"F(F(-%+AXESLABELSG6$ Q\"x6\"Q!Fj[l-%%VIEWG6$;F(F][l%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 49 "A Fourier representation for a finite basis size:" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "Nb:=9;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#NbG\"\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "b:=[seq(IP(f0,fBn(j)),j=1..Nb)];" } }{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"bG7+,$**\"\"#\"\"\"\"\"&!\"\"%#Pi G!\"#F(#F)F(F)\"\"!,$**F(F)\"#XF+F,F-F(F.F+F/,$**F(F)\"$D\"F+F,F-F(F.F )F/,$**F(F)\"$X#F+F,F-F(F.F+F/,$**F(F)\"$0%F+F,F-F(F.F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "f0_ap:=add(b[j]*fBn(j),j=1..Nb);" } }{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&f0_apG,,*&#\"\"%\"\"&\"\"\"*&%#PiG !\"#-%$sinG6#*&F,F*%\"xGF*F*F*F**&#F(\"#XF**&F,F--F/6#,$*(\"\"$F*F,F*F 2F*F*F*F*!\"\"*&#F(\"$D\"F**&F,F--F/6#,$*(F)F*F,F*F2F*F*F*F*F**&#F(\"$ X#F**&F,F--F/6#,$*(\"\"(F*F,F*F2F*F*F*F*F<*&#F(\"$0%F**&F,F--F/6#,$*( \"\"*F*F,F*F2F*F*F*F*F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 " plot([f0,f0_ap],x=0..1,color=[red,blue]);" }}{PARA 13 "" 1 "" {GLPLOT2D 947 161 161 {PLOTDATA 2 "6&-%'CURVESG6$7U7$$\"\"!F)F(7$$\"3e mmm;arz@!#>$\"3)HLLL$3VfV!#?7$$\"3[LL$e9ui2%F-$\"3Inmm\"H[D:)F07$$\"3n mmm\"z_\"4iF-$\"3TLLLe0$=C\"F-7$$\"3[mmmT&phN)F-$\"3ILLL3RBr;F-7$$\"3C LLe*=)H\\5!#=$\"3imm;zjf)4#F-7$$\"3gmm\"z/3uC\"FC$\"3MLL$e4;[\\#F-7$$ \"3%)***\\7LRDX\"FC$\"3$)****\\i'y]!HF-7$$\"3]mm\"zR'ok;FC$\"3:LL$ezs$ HLF-7$$\"3w***\\i5`h(=FC$\"3%*****\\7iI_PF-7$$\"3WLLL3En$4#FC$\"3Anmm; 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What observation do you make about the Fourier co efficients as compared to the present case?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 258 11 "Exercise 2:" }}{PARA 0 "" 0 " " {TEXT -1 109 "Change the intial condition to represent a case where \+ the string is in equilibrium, but is hit with a hammer." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 259 11 "Exercise 3:" }} {PARA 0 "" 0 "" {TEXT -1 113 "Graph the time evolution of the first th ree modes. Find the relationship between the frequencies for these mod es." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "0 0 0" 13 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }