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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 256 30 "Harmonic oscillator anima
tions" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 51 "
Martha Abell and James Braselton in their textbook " }{TEXT 277 35 "Di
fferential Equations with Maple V" }{TEXT -1 243 " (Academic Press) pr
ovide some lines of code that allow to show movies of the mass-spring \+
system by connecting straigh-line segments to depict a spring. The ide
a is to use the solution to the differential equation for the harmonic
 oscillator, " }{TEXT 257 1 "y" }{TEXT -1 1 "(" }{TEXT 258 1 "t" }
{TEXT -1 180 "), which provides the location of the moving endpoint of
 the spring, i.e., the attached mass as a function of time. We begin b
y providing the solution to the differential equation." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 218 "The parameters are \+
the mass and the spring constant, as well as the gravitational acceler
ation. We work implicitly in MKSA (SI) units, and pick values after th
e set-up of the equation and solution for the general case:" }}{PARA 
0 "" 0 "" {TEXT -1 1 "[" }{TEXT 260 1 "g" }{TEXT -1 25 " is measured i
n m/sec^2, " }{TEXT 259 1 "m" }{TEXT -1 8 " in kg, " }{TEXT 261 1 "k" 
}{TEXT -1 8 " in N/m]" }{TEXT 262 0 "" }{TEXT -1 0 "" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 23 "g:='g': m:='m': k:='k':" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 55 "The forces acting on the mass are gravity (in the form " 
}{TEXT 272 1 "F" }{TEXT -1 1 "_" }{TEXT 273 1 "y" }{TEXT -1 4 " = -" }
{TEXT 274 2 "mg" }{TEXT -1 149 ", i.e., position-independent at the su
rface of the Earth), and the spring force given by Hooke's law, which \+
counteracts the motion from equilibrium, " }{TEXT 271 1 "F" }{TEXT -1 
6 "_H =- " }{TEXT 270 1 "k" }{TEXT -1 1 "(" }{TEXT 269 1 "y" }{TEXT 
-1 1 "-" }{TEXT 268 1 "y" }{TEXT -1 7 "0). If " }{TEXT 267 1 "y" }
{TEXT -1 17 " is greater than " }{TEXT 266 1 "y" }{TEXT -1 41 "0, the \+
force accelerates in the negative " }{TEXT 265 1 "y" }{TEXT -1 27 " di
rection, and vice versa." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 972 "Note that one can make different kinds of springs
: the usual springs encountered occasionally are tightly wound, and ar
e to be stretched only (e.g., a slinky), but we know coil springs from
 car/motorcycle/bicycle suspensions that can be both stretched and com
pressed. With gravity present a tightly wound spring with a mass attac
hed acquires an equilibrium point from which the spring can be both co
mpressed and further stretched, i.e., the equilibrium point is no long
er given as the length of the spring. The equilibrium point is determi
ned as the point at which the gravitational attraction to the Earth's \+
surface is balanced by the spring force which results from expanding t
he spring from its original equilibrium point to suspend the mass. We \+
assume for simplicity that we have a stretch-only spring, which is ide
alized to be of zero length when totally compressed (natural state). T
he equilibrium position for the mass-spring system under gravity becom
es (we label " }{TEXT 275 1 "y" }{TEXT -1 112 "1 now as the height at \+
which the system balances, the original balancing position that enters
 Hooke's law being " }{TEXT 276 1 "y" }{TEXT -1 5 "0=0):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "y0:=0; y1:='y1':" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 22 "EQL:=-m*g-k*(y1-y0)=0;" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 18 "y1:=solve(EQL,y1);" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 52 "We measure the displacement of the mass starting at \+
" }{TEXT 263 1 "y" }{TEXT -1 107 "=0 (this being the equilibrium point
 without any mass attached) using the conventional direction (positive
 " }{TEXT 264 1 "y" }{TEXT -1 58 " is upwards). The gravitational acce
leration therefore is " }{TEXT 280 1 "F" }{TEXT -1 1 "_" }{TEXT 279 1 
"y" }{TEXT -1 2 "=-" }{TEXT 278 2 "mg" }{TEXT -1 32 ". The Newton equa
tion now reads:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "NE:=m*di
ff(y(t),t$2)=-m*g-k*(y(t)-y0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 21 "sol:=dsolve(NE,y(t));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 659 
"While every physics student definitely should be able to derive this \+
general answer to the differential equation by hand, we assume that s/
he can at least verify that the solution found by Maple is correct by \+
substitution. We have two integration constants to be determined by th
e initial conditions. One typical condition is to release the system f
rom rest at an arbitrary height. We could take the general solution, f
ind the velocity expression by differentiation, and determine _C1 and \+
_C2. Alternatively, we can specify the initial conditions, so that dso
lve comes up with the special solution of interest. We pick some arbit
rary height and zero velocity." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 24 "IC:=y(0)=-0.2,D(y)(0)=0;" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 26 "sol:=dsolve(\{NE,IC\},y(t));" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 33 "To plot we specify the constants:" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 19 "g:=10; m:=1; k:=10;" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 25 "Y_s:=unapply(rhs(sol),t):" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 72 "We graph a stretch of 10 seconds together with the e
quilibrium position:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "y1;
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "plot([Y_s(t),y1],t=0..1
0);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 229 "We recognize the simple h
armonic motion about the equilibrium position. To graph a cartoon side
 view of the mass-spring system we need a sequence of straight-line se
gments. We pick a radius for the spring and the number of turns:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "w:=0.2: nT:=10:" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 145 "Given a position for the mass y(t) we di
vide this length into 2 nT segments, and generate a list of points whi
ch alternate on the left and right." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 37 "DL:=proc(t) local i,L,d; global w,nT;" }}{PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 50 "d:=evalf(Y_s(t)/(2*nT+1)); L:=[[0.,0.],[w,d*0.
5]]:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 216 "for i from 1 to nT do: L:=
[op(L),[-w,d*(2*i-0.5)],[w,d*(2*i+0.5)]]; od: L:=[op(L),[0.,evalf(Y_s(
t))],[w/4,evalf(Y_s(t))],[w/4,evalf(Y_s(t))-w/8],[-w/4,evalf(Y_s(t))-w
/8],[-w/4,evalf(Y_s(t))],[0.,evalf(Y_s(t))]]; end:" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 36 "PL:=seq(plot(DL(j_t/10)),j_t=1..32):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "display(PL,insequence=true,a
xes=boxed);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 67 "There are many way
s how one can make this problem more interesting:" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 60 "1) introducing a friction
 term to make the problem realistic" }}{PARA 0 "" 0 "" {TEXT -1 44 "2)
 to consider a friven harmonic oscillator." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 28 "Let us start with problem 1:" }
}{PARA 0 "" 0 "" {TEXT -1 61 "we need to add a term which is proportio
nal to the velocity d" }{TEXT 282 1 "y" }{TEXT -1 2 "/d" }{TEXT 281 1 
"t" }{TEXT -1 265 ". Whenever the velocity gets appreciable, we want a
 force that reduces it. For positive velocity it has to be negative, f
or negative velocity (downward) it has to be positive, i.e., we want a
 friction constant times the negative of the velocity added to the for
ces:" }{TEXT 283 0 "" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 31 "b:='b': m:='m': g:='g': k:='k':" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 53 "NE:=m*diff(y(t),t$2)=-m*g-k*(y(t)-y0)-b*diff(
y(t),t);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 594 "Using mathematical j
argon we can say that this is still a non-homogeneous (due to the cons
tant gravity-related term), second-order ordinary differential equatio
n with constant coefficients. The method of solution is still the same
 [find the general solution to the homogeneous problem, (find and ) ad
d a special solution to the inhomogeneous equation to the general homo
geneous solution, and that is the general solution to the inhomogeneou
s problem with the boundary conditions still to be specified]. Maple s
till can do all the math work for us so we concentrate on understandin
g the physics:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "sol:=dsol
ve(NE,y(t));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 278 "The solution is \+
expressed in a 'weird' way if we expect oscillatory answers, but keep \+
in mind that the exponentials may turn out to have imaginary arguments
. In fact, the damped HO admits both oscillatory and non-oscillatory s
olutions (depending on the strength of the damping)." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 52 "Let us put in the same \+
initial conditions as before:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 26 "sol:=dsolve(\{NE,IC\},y(t));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 117 "Let us pick a small value for the damping coefficient in addit
ion to the previously chosen parameters [the units for " }{TEXT 284 1 
"b" }{TEXT -1 8 ": Ns/m]:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
27 "g:=10; m:=1; k:=10; b:=1/5;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 25 "Y_s:=unapply(rhs(sol),t):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 26 "plot([Y_s(t),y1],t=0..20);" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 35 "Voila, the solution is oscillatory." }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 37 "PL:=seq(plot(DL(j_t/10)),j_t=1..320):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "display(PL,insequence=true,a
xes=boxed);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 39 "We can think of re
al-life applications:" }}{PARA 0 "" 0 "" {TEXT -1 507 "The shock absor
bers in a car's (or other vehicle's) suspension system cooperate with \+
the coils (or other springs, such as multi-link suspension, or leaf sp
rings) to provide a damped oscillator. In fact, one is not interested \+
in the kind of motion displayed above, but would rather want to absorb
 the oscillatory energy in one-two swings. From the analytic form of t
he solution in the case with unspecified constants we can see when a c
haracter change occurs in the solution. This character change occurs a
s " }{TEXT 295 1 "b" }{TEXT -1 4 "^2-4" }{TEXT 294 2 "km" }{TEXT -1 
123 " turns zero, and the exponential has no longer complex-valued arg
uments [we add a tiny number to avoid a zero denominator]:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "g:=10; m:=1; k:=10; b:=sqrt(4*k)+10
^(-Digits);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "Y_s:=unapply
(rhs(sol),t):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "plot([Y_s(
t),y1],t=0..10);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "This case is \+
called the critically damped solution." }}{PARA 0 "" 0 "" {TEXT -1 
145 "There are interesting questions to be asked, such as: what happen
s if the damping is increased further? Is it desirable to do so in app
lications?" }}{PARA 0 "" 0 "" {TEXT -1 26 "Let us double the damping:
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "g:=10; m:=1; k:=10; b:=
2*sqrt(4*k)+10^(-Digits);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
25 "Y_s:=unapply(rhs(sol),t):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 26 "plot([Y_s(t),y1],t=0..10);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
161 "We see that it takes longer for the spring-mass system to reach e
quilibrium! One has introduced so much friction that it takes forever \+
to reach the final height." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 
"" 0 "" {TEXT -1 706 "There is an interesting representation that allo
ws to gain further insights. In a so-called phase-space diagram one pl
ots the velocity versus the position with time playing the role of the
 parameter. This curve [an abstract object called the phase-space traj
ectory] allows to read off how energy is dissipated in the system. For
 the case without friction it results in a closed curve. Solutions wit
h different initial conditions are characterized by similar-looking ph
ase-space trajectories whose size is a measure of the amount of total \+
energy which is conserved and transformed between kinetic and potentia
l energy forms. For the case with damping we obtain spirals, as energy
 is removed from the system." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 40 "Let us begin with undercritical damping." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "g:=10; m:=1; k:=10; b:=1/2;
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "Y_s:=unapply(rhs(sol),t
):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "V_s:=D(Y_s);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "plot([Y_s(t),V_s(t),t=0..20]
);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 358 "Due to the small amount of
 damping it may not be immediately obvious, but we wish to point out t
hat the reduction in mechanical energy (reduction of radius) occurs du
ring those phases when the magnitude of the velocity is largest. This \+
can be inferred from a careful inspection of the graph, or from a deta
iled plot of the total energy as a function of time:" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 50 "Etot:=t->1/2*m*V_s(t)^2+m*g*Y_s(t)+1/2*k*
Y_s(t)^2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "plot([Etot(t)-
Etot(0),V_s(t)],t=0..5);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 102 "We l
eave it as an exercise to look at the diagrams for the critically and \+
overcritically damped cases." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 195 "Now we turn to the case of a driven HO. We cho
ose a harmonic driving force with small amplitude to drive the HO away
 from the equilibrium. A circular frequency nu characterizes the drivi
ng force." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "b:='b': m:='m'
: g:='g': k:='k': nu:='nu':" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 70 "NE:=
m*diff(y(t),t$2)=-m*g-k*(y(t)-y0)-b*diff(y(t),t)+(1/10)*sin(nu*t);" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 66 "For the undamped case the interna
l frequency of the oscillator is:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 17 "omega:=sqrt(k/m);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 136 "We need to investigate what happens as nu is varied over a wid
e range, and particularly what happens as nu approaches omega (resonan
ce)." }}{PARA 0 "" 0 "" {TEXT -1 101 "For a small amplitude of the dri
ving force we can expect dramatic results only for nu close to omega.
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "g:=10; m:=1; k:=10; b:=
1/2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "omega;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "y1;" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 8 "nu:=3.1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 
"sol:=dsolve(\{NE,y(0)=y1,D(y)(0)=0\},y(t)):" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 49 "The result is long, and perhaps worth looking at." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "Y_s:=unapply(rhs(sol),t):" }
}{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "plot([Y_s(t),y1],t=0..20);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 373 "It is worthwhile to repeat the ca
lculation for different values of nu, and to observe the more complica
ted transient behaviour. Eventually the system responds with oscillati
ons at the driving frequency nu. This latter fact is clearly evident w
hen choices are made for nu quite different from omega. On the other h
and, we obtain appreciable oscillations only near resonance." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 96 "The phase-space
 diagram shows that the abstract phase-space trajectory approaches a l
imit cycle:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "V_s:=D(Y_s):
" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "plot([Y_s(t),V_s(t),t=0..20]);
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 394 "The emergence of a limit cyc
le tells us that there comes a point when as much energy is dumped int
o the mass-spring system by the driving force, as there is energy lost
 due to friction. We encourage the student to try out cases where the \+
friction is smaller, and therefore, larger-amplitude motion can be obs
erved. We also leave it as an exercise to explore different driving fo
rce frequencies." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 158 "To complete the understanding of the driven HO with damp
ing one has to figure out the dependence of the final amplitude on the
 frequency of the driving force." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 41 "b:='b': m:='m': g:='g': k:='k': nu:='nu':" }}{PARA 0 
"> " 0 "" {MPLTEXT 1 0 70 "NE:=m*diff(y(t),t$2)=-m*g-k*(y(t)-y0)-b*dif
f(y(t),t)+(1/10)*sin(nu*t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
3 "y1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "sol:=dsolve(\{NE,
y(0)=y1,D(y)(0)=0\},y(t)):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
25 "Y_s:=unapply(rhs(sol),t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 140 "Intuitively it is clear
 that for undercritical damping the internal oscillations die out on a
 time scale determined by the damping constant " }{TEXT 285 1 "b" }
{TEXT -1 187 ". What remains is the first term in the above expression
 whose time variation is just connected to the inhomogeneous term in t
he Newton equation, i.e., just to the periodic driving force." }}
{PARA 0 "" 0 "" {TEXT -1 79 "Let us assume that we are in the asymptot
ic regime, which is reached when exp(-" }{TEXT 286 2 "bt" }{TEXT -1 2 
"/2" }{TEXT 293 1 "m" }{TEXT -1 118 ") becomes negligible compared to \+
the first term. We simply pick out the first term in the solution. For
 our choice of " }{TEXT 287 1 "b" }{TEXT -1 43 "=1/2 (a strong, but un
dercritical damping) " }{TEXT 288 1 "t" }{TEXT -1 148 "_a=20 should be
 appropriate. We can verify from a graph, that we have reached the lim
it cycle, i.e., driven harmonic motion with constant amplitude." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "Y_as:=unapply(op(1,Y_s(t)),t
);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 118 "plot([subs(g=10,m=1,
k=10,b=1/2,nu=31/10,Y_as(t)),subs(g=10,m=1,k=10,y1)+1/20*sin(31/10*t)]
,t=20..40,color=[red,blue]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 110 "
The graph shows that the harmonic motion associated with the limit cyc
le is centered on the equilibrium point " }{TEXT 289 1 "y" }{TEXT -1 
52 "1, and is phase shifted compared to a simple sin(nu*" }{TEXT 290 
1 "t" }{TEXT -1 95 "). The phase shift is achieved by the combination \+
of sines and cosines present in the solution." }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 128 "Our interest is in extracting the amplitude as a func
tion of the parameters. First we have to subtract the equilibrium posi
tion:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "assume(nu,real);" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "A:=combine(Y_as(t)-y1,tri
g);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "A0:=simplify(subs(b=
0,A));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 108 "We note that in the ca
se without friction the asymptotic amplitude blows up on resonance [no
te that omega^2=" }{TEXT 292 1 "k" }{TEXT -1 1 "/" }{TEXT 291 1 "m" }
{TEXT -1 231 "]. This means that no limit cycle is reached, but that t
he amplitude grows linearly with time [the latter detail is obtained f
rom an analysis of the solution to Newton's equation, we leave it as a
n exercise to extract this result]." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 295 "For the case with friction the resonanc
e occurs at a shifted frequency, and the mentioned phase shift occurs.
 Let us look first at the asymptotic solution when nu is chosen to be \+
on  resonance in the frictionless (undamped HO) case. This answer shou
ld be reasonable for small friction constants " }{TEXT 296 1 "b" }
{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "simplify(a
lgsubs(nu=sqrt(k/m),A));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 172 "So l
et us assume that near resonance the amplitude is given by looking jus
t at the cosine term (as the two sine terms nearly cancel on resonance
 in the weakly damped case)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 38 "A1:=simplify(nu*b*cos(nu*t)/denom(A));" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 41 "For the amplitude of oscillation we have:" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "A1a:
=simplify(nu*b/denom(A));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 15 "Now \+
we specify " }{TEXT 298 1 "k" }{TEXT -1 5 " and " }{TEXT 297 1 "m" }
{TEXT -1 39 ", with the resonance at omega=sqrt(10)." }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 25 "A1as:=subs(m=1,k=10,A1a);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "P0:=plot([sqrt(10),t,t=0..1.5],colo
r=black):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "P1:=plot(subs(
b=1/5,A1as),nu=3..3.3,color=red):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 49 "P2:=plot(subs(b=1/15,A1as),nu=3..3.3,color=blue):" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "P3:=plot(subs(b=1/45,A1as),
nu=3..3.3,color=green):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "
display(P0,P1,P2,P3,axes=boxed,labels=[frequency,Amplitude]);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 122 "The graph shows that the damped o
scillator is capable of absorbing energy from the driving mechanism in
 the following way:" }}{PARA 0 "" 0 "" {TEXT -1 229 "1) very weak damp
ing: the oscillator can only absorb energy effectively if the mechanic
al energy is offered at a frequencies close to the natural frequency, \+
which is very close to the natural frequency of the undamped oscillato
r;" }}{PARA 0 "" 0 "" {TEXT -1 205 "2) for moderate damping the range \+
over which energy can be absorbed broadens somewhat; the resonance con
dition is shifted slightly towards lower frequencies when compared to \+
the undamped natural frequency." }}{PARA 0 "" 0 "" {TEXT -1 329 "3) fo
r stronger damping the resonance is broadened: the harmonic oscillator
 is able to absorb frequencies over a wider range equally, but the amp
litude of oscillation is reduced considerably. This is the result of t
he driving force not being able to be in phase with the internal oscil
latory motion even in the asymptotic regime." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 84 "For a more detailed analysis th
e reader may wish to consult the book by Ron Greene: " }{TEXT 299 30 "
Classical Mechanics with Maple" }{TEXT -1 20 " (Springer-Verlag). " }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "0 1 0" 0 }
{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }