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{SECT 0 {PARA 256 "" 0 "" {TEXT 256 20 "Damped Resonance and" }}{PARA 
256 "" 0 "" {TEXT 257 27 "Damped Amplitude Modulation" }}{PARA 257 "" 
0 "" {TEXT 258 14 "Example Report" }}{PARA 258 "" 0 "" {TEXT 259 17 "G
ilbert Weinstein" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "
" 0 "" {TEXT -1 12 "Introduction" }}{PARA 0 "" 0 "" {TEXT -1 92 "We co
nsider a damped forced oscillator near resonance with the following eq
uation of motion:" }}{PARA 259 "" 0 "" {XPPEDIT 18 0 "diff(diff(x(t),t
),t)+2*sigma*diff(x(t),t)+(1+sigma^2)*x(t) = A*cos((1+delta)*t)" "6#/,
(-%%diffG6$-F&6$-%\"xG6#%\"tGF-F-\"\"\"*(\"\"#F.%&sigmaGF.-F&6$-F+6#F-
F-F.F.*&,&F.F.*$F1F0F.F.-F+6#F-F.F.*&%\"AGF.-%$cosG6#*&,&F.F.%&deltaGF
.F.F-F.F." }{TEXT -1 1 "," }}{PARA 0 "" 0 "" {TEXT -1 6 "where " }
{XPPEDIT 18 0 "sigma" "6#%&sigmaG" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "de
lta" "6#%&deltaG" }{TEXT -1 6 ", and " }{TEXT 260 1 "A" }{TEXT -1 114 
", are constant parameters.  To find the natural resonance of the osci
llator, we solve the characteristic equation:" }}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 50 "eq := lambda^2 + 2*sigma*lambda + (1+sigma^2) = 0;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "r:=solve(eq,lambda);" }}}{PARA 0 "
" 0 "" {TEXT -1 90 "The natural frequency of the oscillator is the ima
ginary part of the characteristic roots." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 26 "assume(sigma>0); Im(r[1]);" }}}{PARA 0 "" 0 "" {TEXT 
-1 25 "The natural frequency is " }{XPPEDIT 18 0 "omega=1" "6#/%&omega
G\"\"\"" }{TEXT -1 53 ".  Thus, the oscillator is in resonance exactly
 when " }{XPPEDIT 18 0 "delta=0" "6#/%&deltaG\"\"!" }{TEXT -1 1 "." }}
}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 11 "Definitions" }}{PARA 0 "" 0 "" 
{TEXT -1 60 "We define a few procedures to facilitate our investigatio
ns." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 85 "deq := diff(x(t),t,t) + 2*sigma*dif
f(x(t),t) + (1+sigma^2)*x(t) = A*cos((1+delta)*t);" }}}{PARA 0 "" 0 "
" {TEXT -1 62 "We will need initial conditions for the differential eq
uation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "ic := x(0)=0, D(x
)(0)=1;" }}}{PARA 0 "" 0 "" {TEXT -1 13 "The function " }{TEXT 0 5 "ca
se " }{TEXT -1 49 "is used to set the parameters to given values in " 
}{TEXT 0 3 "deq" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 63 "case := (alpha,beta,a) -> subs(sigma=alpha,delta=beta,A=a,deq)
;" }}}{PARA 0 "" 0 "" {TEXT -1 14 "The procedure " }{TEXT 0 3 "sol" }
{TEXT -1 39 " returns the solution as an expression." }}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 44 "sol := proc(d) rhs(dsolve(\{d,ic\},x(t)))
 end;" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 32 "Undamped Oscillator in
 Resonance" }}{PARA 0 "" 0 "" {TEXT -1 51 "We begin with the undamped \+
oscillator in resonance." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "
deq1 := case(0,0,1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "p:=
sol(deq1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "plot(p,t=0..5
0*Pi);" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT 263 16 "Damped Resonance" }}
{PARA 0 "" 0 "" {TEXT -1 69 "We now 'turn on' damping.  We plot the so
lution over 50 cycles, with " }{XPPEDIT 18 0 "sigma" "6#%&sigmaG" }
{TEXT -1 43 " equal to 1/5, 1/10, 1/20, 1/40, and 1/100." }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "deq1:=case(1/5,0,1): p:=sol(deq1): \+
plot(sol(deq1),t=0..50*Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
63 "deq1:=case(1/10,0,1): p:=sol(deq1): plot(sol(deq1),t=0..50*Pi);" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "deq1:=case(1/20,0,1): p:=s
ol(deq1): plot(sol(deq1),t=0..50*Pi);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 63 "deq1:=case(1/40,0,1): p:=sol(deq1): plot(sol(deq1),t=
0..50*Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "deq1:=case(1/
100,0,1): p:=sol(deq1): plot(sol(deq1),t=0..50*Pi);" }}}{PARA 0 "" 0 "
" {TEXT -1 107 "It seems that the damping causes the amplitude to satu
rate@a certain ceiling.  As the damping parameter " }{XPPEDIT 18 0 
"sigma" "6#%&sigmaG" }{TEXT -1 259 " becomes smaller, both the saturat
ion level and the saturation time increase.  If the damping is very sm
all, and one looks only@a relatively small number of cycles, it wil
l seem as if the amplitude increases linearly.  For example, in the ca
se above, with " }{XPPEDIT 18 0 "sigma" "6#%&sigmaG" }{TEXT -1 321 " =
 1/100, the saturation may not be apparent in the first 50 cycles.  In
 this sense, the undamped oscillator in resonance is the limit of the \+
damped ones.  This is however, a singular limit, for even if the satur
ation is not initially visible, it will eventually become evident, no \+
matter how small the damping parameter." }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 35 "We will now increase the amplitude \+
" }{TEXT 261 1 "A" }{TEXT -1 86 " of the forcing term, with fixed damp
ing.  We plot the solution with A set to 2 and 4." }}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 63 "deq1:=case(1/20,0,2): p:=sol(deq1): plot(sol(
deq1),t=0..50*Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "deq1:
=case(1/20,0,4): p:=sol(deq1): plot(sol(deq1),t=0..50*Pi);" }}}{PARA 
0 "" 0 "" {TEXT -1 94 "This has the effect of pushing the saturation l
evel up without increasing the saturation time." }}}{SECT 1 {PARA 3 "
" 0 "" {TEXT -1 28 "Time-Averaged Kinetic Energy" }}{PARA 0 "" 0 "" 
{TEXT -1 120 "In order to test our conjectures over a very larger numb
er of cycles, we use the kinetic energy averaged over one cycle:" }}
{PARA 264 "" 0 "" {XPPEDIT 18 0 "k(s) = (2*Pi)^(-1)*int(diff(x(t),t)^2
,t=s-Pi..s+Pi)" "6#/-%\"kG6#%\"sG*&)*&\"\"#\"\"\"%#PiGF,,$F,!\"\"F,-%$
intG6$*$-%%diffG6$-%\"xG6#%\"tGF:F+/F:;,&F'F,F-F/,&F'F,F-F,F," }{TEXT 
-1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 30 "If our conjecture is correct, \+
" }{XPPEDIT 18 0 "k(s)" "6#-%\"kG6#%\"sG" }{TEXT -1 24 " should have a
 limit as " }{XPPEDIT 18 0 "s->infinity" "6#f*6#%\"sG7\"6$%)operatorG%
&arrowG6\"%)infinityGF*F*F*" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 13 "The function " }{TEXT 0 1 "k" }
{TEXT -1 94 " computes the kinetic energy averaged over one cycle of a
 solution passed as an expression in " }{TEXT 262 1 "t" }{TEXT -1 1 ".
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "k := (s,p) -> int(diff(p
,t)^2,t=s-Pi..s+Pi)/(2*Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
58 "deq1:=case(0,0,1): p:=sol(deq1): plot(k(s,p),s=0..500*Pi);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 101 "deq1:=case(1/5,0,1): p:=sol
(deq1): kp5 := k(s,p): plot(kp5,s=0..50*Pi); evalf(limit(kp5,s=infinit
y));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 106 "deq1:=case(1/10,0,
1): p:=sol(deq1): kp10 := k(s,p): plot(kp10,s=0..100*Pi); evalf(limit(
kp10,s=infinity));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 106 "deq1
:=case(1/20,0,1): p:=sol(deq1): kp20 := k(s,p): plot(kp20,s=0..200*Pi)
; evalf(limit(kp20,s=infinity));" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 100 "deq1:=case(1/40,0,1): p:=sol(deq1): kp := k(s,p): pl
ot(kp,s=0..400*Pi); evalf(limit(kp,s=infinity));" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 102 "deq1:=case(1/100,0,1): p:=sol(deq1): kp := k
(s,p): plot(kp,s=0..1000*Pi); evalf(limit(kp,s=infinity));" }}}{PARA 
0 "" 0 "" {TEXT -1 121 "It seems our first conjecture is supported by \+
these findings.  We now compare saturation levels and saturation times
 for " }{XPPEDIT 18 0 "sigma" "6#%&sigmaG" }{TEXT -1 23 " = 1/5, 1/10,
 and 1/20:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "plot(\{kp5,kp1
0,kp20\}, s=0..50*Pi);" }}}{PARA 0 "" 0 "" {TEXT -1 92 "Finally, we co
mpare saturation levels and saturation times for different forcing amp
litudes:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 131 "deq1:=case(1/20
,0,2): p:=sol(deq1): kp2:=k(s,p): deq1:=case(1/20,0,4): p:=sol(deq1): \+
kp4:=k(s,p): plot(\{kp20,kp2,kp4\}, s=0..50*Pi);" }}}}{SECT 1 {PARA 3 
"" 0 "" {TEXT -1 0 "" }{TEXT 265 29 "Undamped Amplitude Modulation" }}
{PARA 0 "" 0 "" {TEXT -1 26 "We now consider undamped (" }{XPPEDIT 18 
0 "sigma=0" "6#/%&sigmaG\"\"!" }{TEXT -1 24 ") amplitude modulation (
" }{XPPEDIT 18 0 "delta<>0" "6#0%&deltaG\"\"!" }{TEXT -1 45 ").  We pl
ot the solution over 50 cycles with " }{XPPEDIT 18 0 "delta" "6#%&delt
aG" }{TEXT -1 32 " set to 1/10, 1/20, 1/50, 1/100." }}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 63 "deq1:=case(0,1/10,1): p:=sol(deq1): plot(sol
(deq1),t=0..50*Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "deq1
:=case(0,1/20,1): p:=sol(deq1): plot(sol(deq1),t=0..50*Pi);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "deq1:=case(0,1/50,1): p:=sol
(deq1): plot(sol(deq1),t=0..50*Pi);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 64 "deq1:=case(0,1/100,1): p:=sol(deq1): plot(sol(deq1),t
=0..50*Pi);" }}}{PARA 0 "" 0 "" {TEXT -1 162 "The frequency of the amp
litude modulation decreases while the amplitude of the amplitude modul
ation increases.  Thus, when the oscillator is very near resonance (" 
}{XPPEDIT 18 0 "delta" "6#%&deltaG" }{TEXT -1 327 " very small), and o
ne only looks@a relatively small number of cycles, it seems as if t
he amplitude increases linearly.  In this last example, with delta=1/1
00, it may be difficult to notice the sinusoidal nature of the amplitu
de modulation over the first 80 cycles or so.  Thus, amplitude modulat
ion turns into resonance as " }{XPPEDIT 18 0 "delta->0" "6#f*6#%&delta
G7\"6$%)operatorG%&arrowG6\"\"\"!F*F*F*" }{TEXT -1 1 "." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 70 "deq1:=case(0,1/10,1): p:=sol(deq1): kp := k(s
,p): plot(kp,s=0..50*Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
71 "deq1:=case(0,1/20,1): p:=sol(deq1): kp := k(s,p): plot(kp,s=0..100
*Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "deq1:=case(0,1/50,
1): p:=sol(deq1): kp := k(s,p): plot(kp,s=0..200*Pi);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "deq1:=case(0,1/100,1): p:=sol(deq1)
: kp := k(s,p): plot(kp,s=0..500*Pi);" }}}{PARA 0 "" 0 "" {TEXT -1 37 
"This seems to support our conjecture." }}}{SECT 1 {PARA 3 "" 0 "" 
{TEXT 266 27 "Damped Amplitude Modulation" }}{PARA 0 "" 0 "" {TEXT -1 
81 "Finally, we consider the full model.  We will only consider two si
tuations, when " }{XPPEDIT 18 0 "sigma->0" "6#f*6#%&sigmaG7\"6$%)opera
torG%&arrowG6\"\"\"!F*F*F*" }{TEXT -1 6 " with " }{XPPEDIT 18 0 "delta
" "6#%&deltaG" }{TEXT -1 39 " small and fixed, and vice versa, when " 
}{XPPEDIT 18 0 "delta->0" "6#f*6#%&deltaG7\"6$%)operatorG%&arrowG6\"\"
\"!F*F*F*" }{TEXT -1 6 " with " }{XPPEDIT 18 0 "sigma" "6#%&sigmaG" }
{TEXT -1 17 " small and fixed." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 20 "We begin with fixed " }{XPPEDIT 18 0 "del
ta" "6#%&deltaG" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 58 "deq1:=case(1/50,1/10,1): p:=sol(deq1): plot(p,t=0..50*Pi);" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "deq1:=case(1/100,1/10,1): p
:=sol(deq1): plot(p,t=0..100*Pi);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 60 "deq1:=case(1/200,1/10,1): p:=sol(deq1): plot(p,t=0..2
00*Pi);" }}}{PARA 0 "" 0 "" {TEXT -1 290 "On the short range, this loo
ks just like resonance, with the amplitude shooting up almost linearly
.  On the long range, there seems to be some amplitude modulation whic
h is eventually damped into a sinusoidal wave with several components.
  We use our average kinetic energy on this problem:" }}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 73 "deq1:=case(1/200,1/10,1): p:=sol(deq1): k
p:=k(s,p): plot(kp,s=0..200*Pi);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 73 "deq1:=case(1/400,1/10,1): p:=sol(deq1): kp:=k(s,p): p
lot(kp,s=0..400*Pi);" }}}{PARA 0 "" 0 "" {TEXT -1 115 "This seems to c
onfirm our conjecture, although a finer structure is faintly discernib
le insider the main wave form." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 25 "We now turn to the case: " }{XPPEDIT 18 
0 "sigma" "6#%&sigmaG" }{TEXT -1 7 " fixed." }}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 59 "deq1:=case(1/20,1/50,1): p:=sol(deq1): plot(p,t=0..
100*Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "deq1:=case(1/20
,1/100,1): p:=sol(deq1): plot(p,t=0..100*Pi);" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 60 "deq1:=case(1/20,1/200,1): p:=sol(deq1): plot(p,t
=0..100*Pi);" }}}{PARA 0 "" 0 "" {TEXT -1 233 "This looks very much li
ke damped resonance with the saturation level rising slowly as delta->
0.  It also seems that there is another wave component with period abo
ut 50 inside the main form.  We use our averaged kinetic energy again.
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "deq1:=case(1/20,1/50,1):
 p:=sol(deq1): kp:=k(s,p): plot(kp,s=0..100*Pi);" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 73 "deq1:=case(1/20,1/100,1): p:=sol(deq1): kp:=k
(s,p): plot(kp,s=0..100*Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 73 "deq1:=case(1/20,1/200,1): p:=sol(deq1): kp:=k(s,p): plot(kp,s=0.
.100*Pi);" }}}}{EXCHG }{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "11" 0 }
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