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{SECT 0 {PARA 18 "" 0 "" {TEXT -1 8 "Math 374" }}{PARA 18 "" 0 "" 
{TEXT -1 15 "Harmonic Motion" }}{PARA 19 "" 0 "" {TEXT -1 26 "Fall 199
9      Jay Treiman" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 18 "Load the pa
ckages." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "with(DEtools):" }
}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 41 "The Differential Equation With
out Damping" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 65 "The differential equation for a mass on a spring without fricti
on" }}{PARA 0 "" 0 "" {TEXT -1 11 "is given by" }}{PARA 0 "" 0 "" 
{TEXT -1 5 "     " }{XPPEDIT 18 0 "m*diff(x(t),`$`(t,2)) = -k*x(t)" "6
#/*&%\"mG\"\"\"-%%diffG6$-%\"xG6#%\"tG-%\"$G6$F-\"\"#F&,$*&%\"kGF&-F+6
#F-F&!\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 69 "The solutio
n to this differential equation is simple harmonic motion." }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "deq1
 := m*diff(x(t),t$2) = -k*x(t);\ndsolve(deq1,x(t));" }}}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 67 "If one puts in values f
or the constants and  initial conditions one" }}{PARA 0 "" 0 "" {TEXT 
-1 42 "can see the graph of the general solution." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 121 "deq1a := su
bs(\{m=1,k=1\},deq1);\nDEplot(deq1a,x(t),t=-1..20,[[D(x)(0)=1,x(0)=1]]
,stepsize=0.1,\n           linecolor=[red]);" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 17 "Includ
ing Damping" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 58 "One can include a simple damping term that is proportional" }}
{PARA 0 "" 0 "" {TEXT -1 66 "to velocity. This could be friction on a \+
table.  The differential " }}{PARA 0 "" 0 "" {TEXT -1 16 "equation is \+
now:" }}{PARA 0 "" 0 "" {TEXT -1 8 "        " }{XPPEDIT 18 0 "m*diff(x
(t),`$`(t,2)) = -k*x(t)-K*diff(x(t),t);" "6#/*&%\"mG\"\"\"-%%diffG6$-%
\"xG6#%\"tG-%\"$G6$F-\"\"#F&,&*&%\"kG\"\"\"-%\"xG6#%\"tGF5!\"\"*&%\"KG
F5-%%diffG6$-F76#F9F9F5F:" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 49 "The type of solution changes with \+
the constant K." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 49 "Here are a couple plots with different values of " }
{XPPEDIT 18 0 "K;" "6#%\"KG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "deq2 := m*diff(x(t
),t$2) = -k*x(t)-K*diff(x(t),t);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 126 "deq2a := subs(\{m=1,k=1,K=.25\},deq2);\nDEplot(deq2a
,x(t),t=0..30,[[D(x)(0)=1,x(0)=1]],stepsize=0.1,\n           linecolor
=[red]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 125 "deq2b := subs(
\{m=1,k=1,K=.5\},deq2);\nDEplot(deq2b,x(t),t=0..20,[[D(x)(0)=1,x(0)=1]
],stepsize=0.1,\n           linecolor=[red]);" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 124 "deq2c := subs(\{m=1,k=1,K=1\},deq2);\nDEplot(de
q2c,x(t),t=0..20,[[D(x)(0)=1,x(0)=1]],stepsize=0.1,\n           lineco
lor=[red]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 124 "deq2d := su
bs(\{m=1,k=1,K=2\},deq2);\nDEplot(deq2d,x(t),t=0..20,[[D(x)(0)=1,x(0)=
1]],stepsize=0.1,\n           linecolor=[red]);" }}}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 24 "Animating the cha
nge in " }{XPPEDIT 18 0 "K;" "6#%\"KG" }{TEXT -1 1 "." }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 47 "What follows shows how
 the solution changes as " }{XPPEDIT 18 0 "K;" "6#%\"KG" }{TEXT -1 9 "
 changes." }}{PARA 0 "" 0 "" {TEXT -1 46 "It generates a list of the p
lots generated by " }{HYPERLNK 17 "DEplot" 2 "DEplot" "" }{TEXT -1 7 "
.  Then" }}{PARA 0 "" 0 "" {TEXT -1 26 "is is displayed using the " }
{HYPERLNK 17 "display" 2 "plots,display" "" }{TEXT -1 9 " command." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
12 "with(plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 306 "plot_l
ist := []:\nfor i from 0 to 25 do\n    K1 := evalf(.1*i);\n    deq2_ :
= eval(subs(\{m=1,k=1,K=K1\},deq2));\n    plot_list := [op(plot_list),
\n                 eval(DEplot(deq2_,x(t),t=0..20,[[D(x)(0)=1,x(0)=1]]
,\n                 stepsize=0.25,linecolor=[red]))];\n    od:\ndispla
y(plot_list,insequence=true); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 0 "" }}}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 9 "Exercises" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 62 "Explain \+
what is happening to the solutions of the differential" }}{PARA 0 "" 
0 "" {TEXT -1 62 "equation in terms of the roots of the characteristic
 equation." }}{PARA 0 "" 0 "" {TEXT -1 62 "(Hint: When are the roots o
f the characteristic equation real " }}{PARA 0 "" 0 "" {TEXT -1 21 "(i
maginary, complex)?" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 41 "What happens if one adds a forcing term, " }
{XPPEDIT 18 0 "f(t) = cos(t/5);" "6#/-%\"fG6#%\"tG-%$cosG6#*&F'\"\"\"
\"\"&!\"\"" }{TEXT -1 1 "," }{TEXT -1 7 " to the" }}{PARA 0 "" 0 "" 
{TEXT -1 70 "differential equation?  Plot the solution to the differen
tial equation" }}{PARA 0 "" 0 "" {TEXT -1 26 "and chack what happens a
s " }{XPPEDIT 18 0 "K;" "6#%\"KG" }{TEXT -1 9 " changes." }}}}}{MARK "
3 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 }