{VERSION 4 0 "IBM INTEL NT" "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 Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 261 "" 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 1 }{PSTYLE "Normal " -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }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 }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 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {PARA 256 "" 0 "" {TEXT -1 37 "Module 43: Vibrations of a Rigi d Beam" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 34 " David L. Powers in his book, " } {TEXT 256 23 "Boundary Value Problems" }{TEXT -1 205 ", Third Edition \+ (Published by Harcourt Brace Jovanovich Publishers) gives a problem co ncerning the vibrations of a rigid beam. The text states that the disp lacement 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#,$*&\"\"\"F%*$%\"cG \"\"#!\"\"F)" }{TEXT -1 2 " " }{XPPEDIT 18 0 "u[tt];" "6#&%\"uG6#%#tt G" }{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 boundar y 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#&%\"u G6#%#xxG" }{TEXT -1 10 "(t,0) = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 139 "A derivation of this equation can be fou nd in many texts on undergraduate partial differential equations. See \+ for example Donald W. Trim's " }{TEXT 257 38 "Applied Partial Differen tial Equations" }{TEXT -1 42 " published by PWS KENT Publishing Compan y." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 334 " \+ It is natural to ask what is the difference between the transverse \+ vibrations of a string and of a thin beam. An over simplified response would be that the beam offers resistance to bending. This resistance \+ is responsible for changing the wave equation to the fourth order beam equation 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 rigi dity and the linear density of the beam. As for the boundary condition s, " }{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 condi tions in terms of the bending moment of the beam. A " }{TEXT 259 15 "s imply fastened" }{TEXT -1 50 " beam should have zero bending moments a t the end." }}{PARA 0 "" 0 "" {TEXT -1 61 " All that remains now i s to have the initial conditions: " }}{PARA 0 "" 0 "" {TEXT -1 33 " \+ u(0, x) = f(x) and " }{XPPEDIT 18 0 "u[t];" "6#&%\"uG6#%\"t G" }{TEXT -1 15 " (0, x) = g(x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {SECT 1 {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-F6F7F8 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 "0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }