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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 44 "Amanda Smith, Katie Causby
, and Eric Calhoun" }}{PARA 0 "" 0 "" {TEXT -1 20 "Page 203, Problem #
2" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 17 "The \+
problem here:" }}{PARA 0 "" 0 "" {TEXT -1 99 "A mass of 8.0 lbs hangs \+
from a spring.  At equalibrium, the spring is stretched 6 in.  Pulling
 the " }}{PARA 0 "" 0 "" {TEXT -1 91 "mass 3 in below equalibrium, the
 spring bounces back with an upward velocity of 0.5 ft/sec." }}{PARA 
0 "" 0 "" {TEXT -1 79 "We are searching for the equation of motion, th
e natural frequency, amplitude, " }}{PARA 0 "" 0 "" {TEXT -1 63 "perio
d, and to plot a graph of the resulting harmonic motion.  " }}{PARA 0 
"" 0 "" {TEXT -1 90 "As we are neglecting any external forces, we solv
ed this as an example of the \"free case.\"" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 21 "with(plots): restart:" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 83 "Because we are working with units in US Customary units, \+
we must remember that the " }}{PARA 0 "" 0 "" {TEXT -1 56 "8.0 lbs is \+
the downward force, the mass times gravity.  " }}{PARA 0 "" 0 "" 
{TEXT -1 52 "Using gravity as 32 ft/sec^2, we solve for the mass:" }}
{PARA 0 "" 0 "" {TEXT -1 59 "          8.0 lbs = (32 ft/sec^2) * m,   \+
 or m = .25 slugs." }}{PARA 0 "" 0 "" {TEXT -1 99 "Thus, when we solve
 for k = [force(lbs)]/[equalibrium length(ft)], k in units lbs/ft or s
lug/sec^2." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "k:=8/.5;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"kG$\"+++++;!\")" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 61 "Using the equation for angular frequency, w = s
qrt ( k / m )." }}{PARA 0 "" 0 "" {TEXT -1 74 "The units here are sqrt
[(slug/sec^2)/(slug)] => sqrt[(1/sec^2)] or 1/sec, " }}{PARA 0 "" 0 "
" {TEXT -1 32 "the correct units for frequency." }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 15 "w:=sqrt(k/.25);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%\"wG$\"+++++!)!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 50 "Now
 we can easily solve for the natural frequency:" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "evalf(w/(2*
Pi));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+W&RKF\"!\"*" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 14 "Or the period:" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "evalf((2*
Pi)/w);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+N;)R&y!#5" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 71 "Notice the units for the period are 1/(1/
sec), or sec, which we expect." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 69 "Now we solve for the equation of motion u
sing our initial conditions." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 66 "eq1:=dsolve(\{diff(x(t),t$2)+w^2*x(t)=0,x(0)=.25,D(x)(0)=.5\},x(
t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq1G/-%\"xG6#%\"tG,&-%$sinG
6#,$F)$\"\")\"\"!$\"++++]i!#6-%$cosGF-$\"+++++D!#5" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 24 "y1:=unapply(rhs(eq1),t);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%#y1GR6#%\"tG6\"6$%)operatorG%&arrowGF(,&-%$sinG6#,$
9$$\"\")\"\"!$\"++++]i!#6-%$cosGF/$\"+++++D!#5F(F(F(" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 75 "We plot the motion over two complete cycles, un
til 4*Pi/w sec have elapsed." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 24 "plot(y1(t),t=0..4*Pi/w);" }}{PARA 13 "" 1 "" {INLPLOT "6%-%'CURV
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" 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 71 "The am
plitude is the sqrt(c1^2+c2^2) = sqrt((.0625)^2+(.25)^2) = .258. " }}}
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