{VERSION 3 0 "APPLE_PPC_MAC" "3.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 256 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 260 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 261 "" 0 1 0 0 0 0 0 1 1 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 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Heading 1" 0 3 1 {CSTYLE "" -1 -1 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 }1 0 0 0 8 4 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 256 1 {CSTYLE " " -1 -1 "" 1 18 0 0 0 0 0 1 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 26 "Vibrations of a Rigid Beam" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 55 "Copyright 2000 by James V. Herod. All \+ rights reserved." }}}{PARA 0 "" 0 "" {TEXT -1 4 " " }}{PARA 0 "" 0 "" {TEXT -1 30 " David L. Powers in his book, " }{TEXT 256 23 "Boundar y Value Problems" }{TEXT -1 205 ", Third Edition (Published by Harcour t Brace Jovanovich Publishers) gives a problem concerning the vibratio ns of a rigid beam. The text states that the displacement u(t, x) of a uniform thin beam satisfies" }}{PARA 0 "" 0 "" {TEXT -1 10 " \+ " }{XPPEDIT 18 0 "u[xxxx];" "6#&%\"uG6#%%xxxxG" }{TEXT -1 3 " = " } {XPPEDIT 18 0 "-1/(c^2);" "6#,$*&\"\"\"\"\"\"*$%\"cG\"\"#!\"\"F*" } {TEXT -1 2 " " }{XPPEDIT 18 0 "u[tt];" "6#&%\"uG6#%#ttG" }{TEXT -1 27 " , for 0 < x < L and t > 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 25 "The ends of the beam are " }{TEXT 260 16 "simply supported" }{TEXT -1 36 ", which produces boundary conditions " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 33 " \+ u(t, 0) = u(t, L) = 0 and " }{XPPEDIT 18 0 "u[xx];" "6#&%\"uG6#%#xxG " }{TEXT -1 9 "(t, 0) = " }{XPPEDIT 18 0 "u[xx];" "6#&%\"uG6#%#xxG" } {TEXT -1 10 "(t,L) = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 139 "A derivation of this equation can be found in many \+ texts on undergraduate partial differential equations. See for example Donald W. Trim's " }{TEXT 257 38 "Applied Partial Differential Equati ons" }{TEXT -1 42 " published by PWS KENT Publishing Company." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 334 " It \+ is natural to ask what is the difference between the transverse vibrat ions of a string and of a thin beam. An over simplified response would be that the beam offers resistance to bending. This resistance is res ponsible for changing the wave equation to the fourth order beam equat ion above. Hereafter we write this equation as" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 " " }{XPPEDIT 18 0 "u[tt];" "6#&%\"uG6#%#ttG" }{TEXT -1 5 " = " }{XPPEDIT 18 0 "-c^2; " "6#,$*$%\"cG\"\"#!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "u[xxxx];" "6 #&%\"uG6#%%xxxxG" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 109 "The constant c incorporates the rigidity and the linear density of the beam. As for the boundary conditions, \+ " }{TEXT 258 15 "simply fastened" }{TEXT -1 210 " is usually taken to \+ mean that the ends of the beam are held stationary, but the slopes at \+ the end points can move. One describes the remaining boundary conditio ns in terms of the bending moment of the beam. A " }{TEXT 259 15 "simp ly fastened" }{TEXT -1 50 " beam should have zero bending moments at t he end." }}{PARA 0 "" 0 "" {TEXT -1 61 " All that remains now is t o have the initial conditions: " }}{PARA 0 "" 0 "" {TEXT -1 33 " \+ u(0, x) = f(x) and " }{XPPEDIT 18 0 "u[t];" "6#&%\"uG6#%\"tG" }{TEXT -1 15 " (0, x) = g(x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {SECT 0 {PARA 3 "" 0 "" {TEXT -1 34 "Derivative of the general solutio n" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 71 " \+ We expect separation of variables to lead to solutions of the form" } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 33 " \+ u(t, x) = X(x) [ A cos(" }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" } {TEXT -1 12 " t) + B sin(" }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" } {TEXT -1 6 " t) ]." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 53 "That is, we expect vibrations in the time variable t." }} {PARA 0 "" 0 "" {TEXT -1 70 " In this problem, separation of varia bles will lead an equation to" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 20 " X''''/X = " }{XPPEDIT 18 0 "lam bda^2;" "6#*$%'lambdaG\"\"#" }{TEXT -1 10 " = - T ''/" }{XPPEDIT 18 0 "c^2;" "6#*$%\"cG\"\"#" }{TEXT -1 2 "T." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 19 "For the X function:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 24 " X '''' \+ - " }{XPPEDIT 18 0 "lambda^2;" "6#*$%'lambdaG\"\"#" }{TEXT -1 7 " X = \+ 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 46 "We \+ seek the general solution of this equation." }}{PARA 0 "" 0 "" {TEXT -1 2 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "restart; assume( lambda>0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "dsolve(diff(X (x),x,x,x,x)-lambda^2*X(x)=0,X(x),method=laplace);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 88 " We get combinations of sines, cosines, hyp erbolic sines, and hyperbolic cosines of " }{XPPEDIT 18 0 "sqrt(lambda )*x;" "6#*&-%%sqrtG6#%'lambdaG\"\"\"%\"xGF(" }{TEXT -1 26 ". For simpl icity, we write" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 14 " X(x) = C cos(" }{XPPEDIT 18 0 "sqrt(lambda)*x;" "6#*&-%% sqrtG6#%'lambdaG\"\"\"%\"xGF(" }{TEXT -1 11 " ) + D sin(" }{XPPEDIT 18 0 "sqrt(lambda)*x;" "6#*&-%%sqrtG6#%'lambdaG\"\"\"%\"xGF(" }{TEXT -1 5 " ) + " }}{PARA 0 "" 0 "" {TEXT -1 40 " \+ E cosh(" }{XPPEDIT 18 0 "sqrt(lambda)*x;" "6#*&-%%sqrtG6#%'lamb daG\"\"\"%\"xGF(" }{TEXT -1 12 " ) + F sinh(" }{XPPEDIT 18 0 "sqrt(lam bda)*x;" "6#*&-%%sqrtG6#%'lambdaG\"\"\"%\"xGF(" }{TEXT -1 3 " )." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 73 "As usual, to determine these constants, we apply the boundary conditions." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 64 "u(t, 0) = 0 implies that X(0) = 0, which implies that C + E = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "u[xx];" "6#&%\"uG6 #%#xxG" }{TEXT -1 67 "(t, 0) = 0 implies that X''(0) = 0, which implie s that - C + E = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 38 "From this, we conclude that C = E = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 72 "We see what are the im plications from the other end boundary conditions." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 49 "u(t, L) = 0 implies X(L) \+ = 0, which implies that " }}{PARA 0 "" 0 "" {TEXT -1 23 " \+ D sin(" }{XPPEDIT 18 0 "sqrt(lambda)*L;" "6#*&-%%sqrtG6#%'lambdaG \"\"\"%\"LGF(" }{TEXT -1 11 ") + F sinh(" }{XPPEDIT 18 0 "sqrt(lambda) *L;" "6#*&-%%sqrtG6#%'lambdaG\"\"\"%\"LGF(" }{TEXT -1 6 ") = 0." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "u[xx]; " "6#&%\"uG6#%#xxG" }{TEXT -1 55 "(t, L) = 0 implies that X ''(L) = 0, which implies that" }}{PARA 0 "" 0 "" {TEXT -1 24 " - \+ D sin(" }{XPPEDIT 18 0 "sqrt(lambda)*L;" "6#*&-%%sqrtG6#%'lambdaG\"\" \"%\"LGF(" }{TEXT -1 11 ") + F sinh(" }{XPPEDIT 18 0 "sqrt(lambda)*L; " "6#*&-%%sqrtG6#%'lambdaG\"\"\"%\"LGF(" }{TEXT -1 6 ") = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 24 "We conclude tha t F sinh(" }{XPPEDIT 18 0 "sqrt(lambda)*L;" "6#*&-%%sqrtG6#%'lambdaG\" \"\"%\"LGF(" }{TEXT -1 25 ") = 0, so that F = 0, and" }}{PARA 0 "" 0 " " {TEXT -1 16 " D sin(" }{XPPEDIT 18 0 "sqrt(lambda)*L;" "6#* &-%%sqrtG6#%'lambdaG\"\"\"%\"LGF(" }{TEXT -1 19 ") = 0, so that sin(" }{XPPEDIT 18 0 "sqrt(lambda)*L;" "6#*&-%%sqrtG6#%'lambdaG\"\"\"%\"LGF( " }{TEXT -1 8 " ) = 0. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 46 "We know everywhere the sine function is zero: " } {XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "(n*Pi/L)^2;" "6#*$*(%\"nG\"\"\"%#PiGF&%\"LG!\"\"\"\"#" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 95 "Thi s means that there is an infinity of solutions for the X equation and \+ they all have the form" }}{PARA 0 "" 0 "" {TEXT -1 26 " \+ X(x) = sin(" }{XPPEDIT 18 0 "n*Pi/L;" "6#*(%\"nG\"\"\"%#PiGF%%\"LG!\" \"" }{TEXT -1 4 " x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 140 "The equation in T is easier. First, there are no boun dary conditions on the T equation, and second, it is only second order . The equation is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 " " }{XPPEDIT 18 0 "(n*Pi/L)^4;" "6#*$*(%\"nG\"\" \"%#PiGF&%\"LG!\"\"\"\"%" }{TEXT -1 9 "= - T ''/" }{XPPEDIT 18 0 "c^2; " "6#*$%\"cG\"\"#" }{TEXT -1 2 "T." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 25 "Thus, T '' + " }{XPPEDIT 18 0 "c^2;" "6#*$%\"cG\"\"#" }{TEXT -1 1 " " }{XPPEDIT 18 0 "(n*Pi/L)^4; " "6#*$*(%\"nG\"\"\"%#PiGF&%\"LG!\"\"\"\"%" }{TEXT -1 7 " T = 0." }} {PARA 0 "" 0 "" {TEXT -1 23 "We solve this equation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "dsolve(diff(T(t),t,t)+c^2*(n*Pi/L)^4*T(t) =0,T(t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 " " 0 "" {TEXT -1 81 "We can now write down the general solution for the partial differential equation:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 13 " u(t, x) = " }{XPPEDIT 18 0 "sum((a[n]* cos(c*(n*Pi/L)^2*t)+b[n]*sin(c*(n*Pi/L)^2*t))*sin(n*Pi*x/L),n);" "6#-% $sumG6$*&,&*&&%\"aG6#%\"nG\"\"\"-%$cosG6#*(%\"cGF-*$*(F,F-%#PiGF-%\"LG !\"\"\"\"#F-%\"tGF-F-F-*&&%\"bG6#F,F--%$sinG6#*(F2F-*$*(F,F-F5F-F6F7\" \"#F-F9F-F-F-F--F?6#**F,F-F5F-%\"xGF-F6F7F-F," }{TEXT -1 2 " ." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 16 "Here is a check." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 106 "u:=(t,x)->sum((a[n]*cos(c*(n*Pi/L)^2*t)+\n b[n] *sin(c*(n*Pi/L)^2*t))*sin(n*Pi/L*x),\n n= 1..3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "simplify(diff(u(t,x),t,t)+\n \+ c^2*diff(u(t,x),x,x,x,x));" }}}{PARA 0 "" 0 "" {TEXT -1 33 "We check the boundary conditions." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "u(t,0);u(t,L);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "eval(sub s(x=0,diff(u(t,x),x,x)));\neval(subs(x=L,diff(u(t,x),x,x)));" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 22 "Details for an Example " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "restart: with(plots):" }}} {PARA 0 "" 0 "" {TEXT -1 162 "To see how this vibrating beam is differ ent from a vibrating string, it would be well to compare the two. We d isplace both these by one arch of the sine function." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "plot(sin(Pi*x),x=0..1);" }}}{PARA 0 "" 0 "" {TEXT -1 78 "To keep the two the same, we make c = 1 so that we are comparing solutions for" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 5 " " }{XPPEDIT 18 0 "u[tt];" "6#&%\"uG6#%#ttG" } {TEXT -1 3 " = " }{XPPEDIT 18 0 "u[xx];" "6#&%\"uG6#%#xxG" }{TEXT -1 9 " and " }{XPPEDIT 18 0 "u[tt];" "6#&%\"uG6#%#ttG" }{TEXT -1 3 " \+ + " }{XPPEDIT 18 0 "u[xxxx];" "6#&%\"uG6#%%xxxxG" }{TEXT -1 5 " = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 105 "Check \+ that the solution for the string equation with zero boundary condition s and no initial velocity is " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "s:=(t,x)->sin(Pi*x)*cos(Pi*t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "diff(s(t,x),t,t)-diff(s(t,x),x,x);" }}}{PARA 0 "" 0 " " {TEXT -1 62 "We graph this solution and also animate graph of the so lution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "plot3d(s(t,x),x=0 ..1,t=0..2,axes=normal, orientation=[-165,55]);" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 30 "animate(s(t,x),x=0..1,t=0..2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 103 "Check that the solution for the beam eq uation with zero boundary conditions and no initial velocity is " }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "b:=(t,x)->sin(Pi*x)*cos((Pi) ^2*t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "diff(b(t,x),t,t)+ diff(b(t,x),x,x,x,x);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 62 "We graph this solution and also animate graph of the solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "plot3d(b(t,x), x=0..1,t=0..2,axes=normal, orientation=[-165,55]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "animate(b(t,x),x=0..1,t=0..2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 161 " \+ Could you see the difference? The beam vibrated faster.\nThe strin g completes one cycle at t = 2, 4, 6, 8, ... . Watch. The following sh ould be three cycles." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "ani mate(s(t,x),x=0..1,t=0..6);" }}}{PARA 0 "" 0 "" {TEXT -1 38 "The beam \+ completes one cycle at t = 2/" }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 4 ", 4/" }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 5 " ,6/ " } {XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 40 " . The following should be three cycles." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "animate(b( t,x),x=0..1,t=0..6/Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}{PARA 0 "" 0 "" {TEXT -1 140 " This suggests that if a beam a nd a string are both struck, parameters for the two being equal, the b eam should vibrate at a high pitch." }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 261 11 "Assignment." }{TEXT -1 130 " Work out \+ the details for getting the separation of variables solution and the d 'Alembert's solution for the above string problem." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "10 4" 4 }{VIEWOPTS 1 1 0 1 1 1803 }