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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 40 "Math 2250 Maple Project 2,
  October 2002" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 95 "NAME _____________________________________________  CLASS
TIME _____________   SECTION 2250- ___" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 99 "There are six (6) problems in this pr
oject. You are expected to answer the questions A, B, C , ... " }}
{PARA 0 "" 0 "" {TEXT -1 110 "associated with each problem.  The origi
nal worksheet \"project2-fall-2002.mws\" is a template for the solutio
n;" }}{PARA 0 "" 0 "" {TEXT -1 104 " you must fill in the code and all
 comments. Sample code can be copied with the mouse. Please use pencil
" }}{PARA 0 "" 0 "" {TEXT -1 85 " freely to annotate the worksheet and
 to clarify any code or figure presented herein." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 96 "The problem headers for t
he Fall 2002 revision of David Eyre's project (original was year 2000)
." }}{PARA 0 "" 0 "" {TEXT -1 494 " __________ 2.1. OVERDAMPED FREE OS
CILLATIONS.                     The first four problems study the\n __
________2.2. UNDERDAMPED FREE OSCILLATIONS.                   simplest
 linear model for x(t).\n __________2.3. UNDAMPED FORCED OSCILLATIONS \+
(c=0).\n __________2.4. DAMPED FORCED OSCILLATIONS (c>0).\n __________
2.5. LARGE SUSTAINED OSCILLATIONS.                       This is the s
econd, nonlinear model.\n __________2.6. MCKENNA NON-HOOKES LAW CABLE \+
MODEL.  This is the third, nonlinear model.\n" }}{PARA 0 "" 0 "" 
{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 44 "2.1.  PROBLEM (OVERDAMPE
D FREE OSCILLATIONS)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 76 "FREE OSCILLATIONS. Consider the general problem of free
 linear oscillations\n" }}{PARA 0 "" 0 "" {TEXT -1 63 "           m x'
' + c x' + k x=0,\n           x(0)=x0,  x'(0)=v0," }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 69 " where  m=1,  k=16 and  c
  is a non-negative constant.  The symbols  " }}{PARA 0 "" 0 "" {TEXT 
-1 76 " x0  and  v0  are the initial position and initial velocity,  r
espectively.\n" }}{PARA 0 "" 0 "" {TEXT -1 94 "    A. Suggest a value \+
for parameter c  > 0 so that the free oscillations are overdamped. Thi
s" }}{PARA 0 "" 0 "" {TEXT -1 108 "         value will be used in item
s B, C, D below. Check your answer by solving the characteristic equat
ion" }}{PARA 0 "" 0 "" {TEXT -1 68 "         using Maple's \"solve\" c
ommand (example: solve(r^2-1=0,r);)." }}{PARA 0 "" 0 "" {TEXT -1 4 "  \+
  " }}{PARA 0 "" 0 "" {TEXT -1 100 "    B. Use x(0)=1 and x'(0)= -2 fo
r the initial conditions and Maple's \"dsolve\" to find the explicit" 
}}{PARA 0 "" 0 "" {TEXT -1 95 "        real solution x(t). Plot the so
lution x(t) for t=0 to t=5 using Maple's \"plot\" command." }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 107 "    C. Suggest n
ew values for x(0) and x'(0) such that the solution is non-negative, i
ncreases near t=0 but" }}{PARA 0 "" 0 "" {TEXT -1 114 "        eventua
lly decreases with limit zero at t=infinity. Find the explicit solutio
n and plot it for t=0 to t=5." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 107 "    D. Suggest new values for x(0) and x
'(0) such that the solution changes sign exactly once on t>0, then " }
}{PARA 0 "" 0 "" {TEXT -1 96 "         decreases to zero at t=infinity
. Find the explicit solution and plot it for t=0 to t=5." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 121 "EXAMPLE(Wrong par
ameters! Change it!)\nde:=3*diff(x(t),t,t)+1.5*diff(x(t),t)+4*x(t)=0: \+
 # Define the differential equation" }}{PARA 0 "" 0 "" {TEXT -1 273 "i
c:=x(0)=1,D(x)(0)= -2:                                 # Define the in
itial conditions\ndsolve(\{de,ic\},x(t),method=laplace);              \+
# Symbolically solve for x(t)\nX:=unapply(rhs(%),t):                  \+
                  # Capture the symbolic answer as a function X(t)" }}
{PARA 0 "" 0 "" {TEXT -1 80 "plot(X(t),t=0..5);                       \+
                    # Plot the solution" }}{PARA 0 "> " 0 "" {MPLTEXT 
1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "#2.1-A" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "#2.1-B" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 6 "#2.1-C" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 6 "#2.1-D" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 44 "2.2. PROBLEM (UNDERDAMPED FREE OSCILLATIO
NS)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 68 "FR
EE OSCILLATIONS. Consider the problem of free linear oscillations\n" }
}{PARA 0 "" 0 "" {TEXT -1 61 "           m x'' + c x' + k x=0,\n      \+
     x(0)=0,  x'(0)=1," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 74 " where  m=2,  c=4 and  k  is a non-negative constant
. The underdamped case" }}{PARA 0 "" 0 "" {TEXT -1 75 " is studied her
e, for which the solution x(t) has at most one oscillation. " }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 72 "    A.  Find a \+
 parameter value k > 5  so that the solution x(t) changes" }}{PARA 0 "
" 0 "" {TEXT -1 77 "         sign infinitely many times and decays to \+
zero at t=infinity. Display" }}{PARA 0 "" 0 "" {TEXT -1 58 "         t
he value of  k  and the exact symbolic solution." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 80 "    B. Plot the exact sym
bolic solution x(t) from t=0 to t=5. Estimate from the " }}{PARA 0 "" 
0 "" {TEXT -1 52 "        graph the decimal value of the pseudoperiod.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 89 "    C
.  Use the solution formula for x(t) from 2.2.A to calculate the exact
 pseudoperiod." }}{PARA 0 "" 0 "" {TEXT -1 77 "          Verify that y
our answer is consistent with the estimate of 2.2.B.  " }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "#2.2-A  Define
  k, then solve and plot." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 
"#2.2-B" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "#2.2-C " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 49 "2.3. PROBLEM (UNDAMPED FORCED OSCILLATIONS (c=0))" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 57 "FORCED LI
NEAR OSCILLATIONS.  Consider the forced problem\n" }}{PARA 0 "" 0 "" 
{TEXT -1 103 "      x'' +  1.5625 x = 10 cos(wt),\n      x(0)=0,  x'(0
)=0,\n\n     where  w is a  non-negative constant." }}{PARA 0 "" 0 "" 
{TEXT -1 2 "  " }}{PARA 0 "" 0 "" {TEXT -1 76 "    A.  Choose  w=3.75,
  so that the forcing  frequency w is 3 times larger " }}{PARA 0 "" 0 
"" {TEXT -1 76 "          than the natural frequency w0=1.25. Solve fo
r x(t) using dsolve()." }}{PARA 0 "" 0 "" {TEXT -1 77 "          Plot \+
the solution x(t) on a suitable interval in order to show the " }}
{PARA 0 "" 0 "" {TEXT -1 48 "          global behavior of  the solutio
n x(t)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
108 "    B. The solution x(t) is the sum of two functions, one of peri
od  2Pi/w  and the other of period  2Pi/w0." }}{PARA 0 "" 0 "" {TEXT 
-1 72 "         Approximate the period of x(t) by examining the graph \+
in 2.2.A." }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT 
-1 95 "    C.  Calculate the period of x(t) from the solution formula \+
(see page 341 for how to do this" }}{PARA 0 "" 0 "" {TEXT -1 72 "     \+
    in general).  Verify that your answer is consistent with 2.3.B." }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 85 "    D. S
uggest a value for the forcing frequency  w  so that the oscillations \+
exhibit" }}{PARA 0 "" 0 "" {TEXT -1 55 "         resonance.  Show reso
nant behavior on a graph." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }
{TEXT -1 2 "  " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "#2.3-A " }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "#2.3-B  " }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 7 "#2.3-C " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 7 "#2.3-D " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 47 "2.4. PROBLEM (DAMPED FORCED O
SCILLATIONS (c>0))" }}{PARA 0 "" 0 "" {TEXT -1 31 "   Consider the for
ced problem\n" }}{PARA 0 "" 0 "" {TEXT -1 60 "      x'' + 2 x' + 21 x \+
= 10cos(t),\n      x(0)=0,  x'(0)=0,\n" }}{PARA 0 "" 0 "" {TEXT -1 59 
"    A. Solve for x(t) and plot the solution on t=0 to t=15." }}{PARA 
0 "" 0 "" {TEXT -1 5 "     " }}{PARA 0 "" 0 "" {TEXT -1 107 "    B.  E
xtract the steady-state solution xss(t) from the solution x(t) of 2.4.
A. Plot x(t) on t=0 to t=15." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 97 "    C. Consider the equation x'' + c x' + 21 x \+
= 5cos(wt), where  c=2, c=1 or c=1/2.  Compute the" }}{PARA 0 "" 0 "" 
{TEXT -1 107 "        amplitude function  C(w) of xss(t) [page 346] fo
r these three equations, then plot for w=0 to w=20 " }}{PARA 0 "" 0 "
" {TEXT -1 59 "        the three amplitude graphs on a single set of a
xes." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 103 "
    D. Consider the equation x'' + c x' + 21 x = 10 cos(wt). For each \+
case  c=2, c=1, c=1/2, print the " }}{PARA 0 "" 0 "" {TEXT -1 96 "    \+
     values  w*, C*  where C*=C(w*)=max \{C(w) : 0 <= w <= 20\}.  The \+
three data pairs should " }}{PARA 0 "" 0 "" {TEXT -1 89 "         show
 that C* becomes larger as c tends to zero.  SAVE YOUR MAPLE FILE FREQ
UENTLY" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
108 "         Maple Hint: Use Maple's mouse interface on the graphic o
f Part C. Specifically, click on a possible" }}{PARA 0 "" 0 "" {TEXT 
-1 104 "         maximum (horizontal tangent) in the graph to display \+
the values  w*, C* on the screen. Copy the" }}{PARA 0 "" 0 "" {TEXT 
-1 25 "         values on paper." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 66 "EXAMPLE(Beware! Wrong values!)\nF:=15: m:
=1: k:=25: c:='c': w:='w':" }}{PARA 0 "" 0 "" {TEXT -1 88 "C:=(w,c)->F
/sqrt((k-m*w*w)^2+(c*w)^2):\nplot(\{C(w,4),C(w,3),C(w,2)\},w=0..15,col
or=black);" }}{PARA 0 "" 0 "" {TEXT -1 47 "Cmax:=evalf(maximize(C(w,2)
,w=0..20,location));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "#2.4-A Solve and plot." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "#2.4-B Define and plot xss(t)." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "#2.4-C Plot C(w), three grap
hics on one set of axes" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "
#2.4-D Table of six data values for w*, C*" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 43 "2.5. PROBLE
M (LARGE SUSTAINED OSCILLATIONS)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 361 "NONLINEAR MODEL WITH GEOMETRY INCLUDED.
\nConsider the nonlinear, forced, damped oscillator equation for torsi
onal motion, with bridge geometry included,\n\n    x'' + 0.05 x' + 2.4
 sin(x)cos(x) = 0.06 cos (12 t/10) ,\n    x(0) = x0,  x'(0) = v0\n\nan
d its corresponding linearized equation\n\n   x'' + 0.05 x' + 2.4 x = \+
0.06 cos (12 t/10) ,  \n   x(0) = x0,  x'(0) = v0.\n" }}{PARA 0 "" 0 "
" {TEXT -1 84 "The spring-mass system parameters are m=1, c = 0.05,  k
 = 2.4,  w = 1.2 ,  F = 0.06." }}{PARA 0 "" 0 "" {TEXT -1 116 "Maple c
ode used to solve and plot the solutions appears below.\n\n   # WARNIN
G: set the parameters on the second line!" }}{PARA 0 "" 0 "" {TEXT -1 
58 "   m:=1:  F := 0.06:  w := 1.2: m:=1: c:= 0.05: k:= 2.4:  " }}
{PARA 0 "" 0 "" {TEXT -1 29 "   x0:=0: v0:=0: a:=0: b:=50:" }}{PARA 0 
"" 0 "" {TEXT -1 198 "   deNonLinear:= m*diff(x(t),t,t) + c*diff(x(t),
t) + k*sin(x(t))*cos(x(t)) = F*cos(w*t):\n   deLinear:= m*diff(x(t),t,
t) + c*diff(x(t),t) + k*x(t) = F*cos(w*t):\n   with(DEtools):  opts:=s
tepsize=0.1:" }}{PARA 0 "" 0 "" {TEXT -1 158 "   DEplot(deNonLinear,x(
t),t=a..b,[[x(0)=x0,D(x)(0)=v0]],opts,title='NonLinear');\n   DEplot(d
eLinear,x(t),t=a..b,[[x(0)=x0,D(x)(0)=v0]],opts,title='Linear');\n" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 103 "    A. L
et  x0=0,  v0=0.  Plot the solutions of the linear and nonlinear equat
ions from t=160 to t=260." }}{PARA 0 "" 0 "" {TEXT -1 79 "         The
se plots represent the steady state solutions of the two equations." }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 103 "   B. L
et  x0=1.2,  v0=0. Plot the solutions of the linear and nonlinear equa
tions from t=220 to t=320." }}{PARA 0 "" 0 "" {TEXT -1 108 "       The
se plots represent the steady state solutions of the two equation, wit
h new starting value x0=1.2." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 90 "    C.  Argue in a sentence why the two linear \+
plots have to be identical, based upon the " }}{PARA 0 "" 0 "" {TEXT 
-1 88 "          superposition formula x(t)=xh(t)+xss(t), even though \+
the homogeneous solution " }}{PARA 0 "" 0 "" {TEXT -1 103 "          x
h(t) is different for the two plots. Please include a discussion of th
e amplitude of  xh(t) " }}{PARA 0 "" 0 "" {TEXT -1 44 "          on th
e corresponding  t-interval. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 93 "    D.  Determine the ratio of the appare
nt amplitudes (a number > 1) for the nonlinear plots" }}{PARA 0 "" 0 "
" {TEXT -1 92 "         and explain why \"large sustained oscillations
\" is an appropriate description of the" }}{PARA 0 "" 0 "" {TEXT -1 
41 "         nonlinear steady-state behavior." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 6 "#2.5-A" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 6 "#2.5-B" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "#2
.5-C" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 55 "2.6.  PROBLEM. ( MCKENNA'S NON-HOOKE'S LAW CABLE \+
MODEL)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 68 
" MCKENNA'S NON-HOOKE'S LAW CABLE MODEL FOR THE TACOMA NARROWS BRIDGE
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 73 "The m
odel of McKenna studies the bridge with a nonlinear, forced, damped " 
}}{PARA 0 "" 0 "" {TEXT -1 78 "oscillator equation for torsional motio
n that accounts for the non-Hooke's law" }}{PARA 0 "" 0 "" {TEXT -1 
75 "cables coupled to the equations for vertical motion.  The equation
s in this" }}{PARA 0 "" 0 "" {TEXT -1 78 "case couple the torsional mo
tion with the vertical motion.  The equations are:" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 63 "    x'' + c x' - k G(x,y)
 = F sin wt,   x(0) = x0,  x'(0) = x1," }}{PARA 0 "" 0 "" {TEXT -1 63 
"    y'' + c y' + (k/3) H(x,y) = g ,     y(0) = y0,  y'(0) = y1," }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 81 "where x(t
) is the torsional motion and y(t) is the vertical motion. The functio
ns" }}{PARA 0 "" 0 "" {TEXT -1 74 "G(x,y) and H(x,y) are the models of
 the force generated by the cable when " }}{PARA 0 "" 0 "" {TEXT -1 
81 "it is contracted and stretched. Below is sample code for writing t
he differential" }}{PARA 0 "" 0 "" {TEXT -1 77 "equations and for plot
ting the solutions. It is ready to copy with the mouse." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 276 "with(DEtools):     \+
                                                     \nw := 1.3:  F :=
 0.05:  f(t) := F*sin(w*t):                              \nc := 0.01: \+
 k1 := 0.2:  k2 := 0.4:  g := 9.8:  L := 6:                  \nSTEP:=x
->piecewise(x<0,0,1):\nfp(t) := y(t)+(L*sin(x(t))):" }}{PARA 0 "" 0 "
" {TEXT -1 384 "fm(t) := y(t)-(L*sin(x(t))):\nSm(t) := STEP(fm(t))*fm(
t):\nSp(t) := STEP(fp(t))*fp(t):\nsys := \{             \n     diff(x(
t),t,t) + c*diff(x(t),t) - k1*cos(x(t))*(Sm(t)-Sp(t))=f(t),\n     diff
(y(t),t,t) + c*diff(y(t),t) + k2*(Sm(t)+Sp(t)) = g\}:\nic := [[x(0)=0,
 D(x)(0)=0, y(0)=27.25, D(y)(0)=0]]:\nvars:=[x(t),y(t)]:   \nopts:=ste
psize=0.1:  \nDEplot(sys,vars,t=0..300,ic,opts,scene=[t,x]);\n" }}
{PARA 0 "" 0 "" {TEXT -1 96 "The amazing thing that happens in this si
mulation is that the large vertical  oscillations take " }}{PARA 0 "" 
0 "" {TEXT -1 80 "all the tension out of the springs and they induce l
arge torsional oscillations." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 86 " A. TORSIONAL OSCILLATION PLOT. Get the sample \+
code above to produce the plot of x(t) " }}{PARA 0 "" 0 "" {TEXT -1 
39 "      [that's what scene=[t,x] means]. " }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 97 " B. Estimate the number of degr
ees the roadway oscillates based on the plot; recall that x in the" }}
{PARA 0 "" 0 "" {TEXT -1 95 "     plot is reported in radians. Comment
 on the agreement of this result with historical data." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 95 "     Hint: Average t
he five largest amplitudes in the plot to find an average maximum ampl
itude" }}{PARA 0 "" 0 "" {TEXT -1 73 "     for t=0 to t=300. Convert t
o degrees using Pi radians = 180 degrees." }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 81 " C. VERTICAL OSCILLATION PLOT. Mod
ify the DEplot code to scene=[t,y] and plot the" }}{PARA 0 "" 0 "" 
{TEXT -1 98 "      oscillation y(t) on t=0 to t=300. The plot is suppo
sed to show 30-foot vertical oscillations" }}{PARA 0 "" 0 "" {TEXT -1 
92 "      that dampen to 7-foot vertical oscillations after 300 second
s.Comment on the agreement" }}{PARA 0 "" 0 "" {TEXT -1 98 "      betwe
en these oscillation results and the historical data for Tacoma Narrow
s, especially the" }}{PARA 0 "" 0 "" {TEXT -1 66 "      visual data pr
esent in the film clip of the bridge disaster." }}{PARA 0 "" 0 "" 
{TEXT -1 6 "      " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "#2.6-A Torsional plot t-versus-x" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "#2.6-B  Roadway oscillatio
n estimate in degrees" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "#2
.6-C Vertical plot t-versus-y" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 0 "" }}}}{MARK "30 0 0" 31 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }
{PAGENUMBERS 0 1 2 33 1 1 }