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ionnelle

ecˆplacement
x par rapport
`placement
la
´eau
longation
de
able, etc. proportionnelle
une barre,
contribution
au d´ea
x p
Nous
allonse
icinom´
introduire e
unenologique.
premi`ere force ph´enom´enologique.
C’est-`
a-dire que
nous n’allons pas
etroduire
force
ph´
C’est-`
a
-dire
que
premi`
e
re
force
e
nom´
nologiq
chercher
les une
causes physiques
r´ese
elles de
cette
force mais
donner
a prioriph´
son
expression.
Nouse
consid´
erons
le
au

e
placement
rencontre
dans
tous
les
lin´eairement
proportionnelle
au

e
placement
se
rencon
L’oscillateur harmonique: première approche

e
force
mais
donner
a
priori
son
expression.
’´
e
quilibre
:
allongement
de
ressort,
torsion
de
ar rapportr´
une position
d’´equilibre
: allongement
dea rep
ysiques
ea`elles
de cette
force mais
donner
Il existe une force phénoménologique qui est linéairement proportionnelle au déplacement et se
rencontre
dans beaucoup
de phénomènes
de déformation
faible par
rapport à une d´
position
orce
qui
s’exerce
entre
deux
points
e
pend
l
e d’équilibre.
approximation, la force qui s’exerce entre
!
!
ologique.
C’est-`
a
-dire
que
nous
n’allons
pas
!
!
premi`
e
re
force
ph´
e
nom´
e
nologique.
C’est-`
a
-dire
que
n
nt
OP
et
OP
la
position
de
deux
objets
p
1
2
entre ces points. Soient OP1 et OP2 la posi
erdea cette
prioriforce
son expression.
consid´
mais donnerNous
a priori
sonerons
expression. N
pos.
La
force
qui
s’exerce
entre
eux

e
pend
d
ces deux points au repos. La force qui s’exer
un cas o`
u, en premi`ere approximation, la force qui s’exerce entre deux points d´epend lin´eairement de la
!
!
variation de distance entre ces points. Soient OP1 et OP2 la position de deux objets ponctuels et L0 la

distance qui s´eparerait ces deux points au repos. La force qui s’exerce entre eux d´epend de l’´elongation par
rapport a` cette distance au repos. Ph´enom´enologiquement, pour d´ecrire une situation stable, on veut que

la force ram`ene toujours les points vers leur distance d’´equilibre L 0 . Cela veut dire que si la distance entre
!
P1 et P2 , c’est-`
a-dire la longueur du vecteur P1 P2 , est plus grande que la longueur L0 , alors la force est
attractive, si la distance est plus petite, la force doit ˆetre r´epulsive.

entre
d´epend entre
lin´eairement
de lad´epend lin
ion, ladeux
forcepoints
qui s’exerce
deux points
´ecenologiquement,
pour

e
crire
une
situation
st
!
!
au
repos.
eOP
nom´
eponctuels
nologiquement,
pour po
d
position
deOP
deux
et L
ts.
Soient
et objets
position de
deux
0 la objets
1 Ph´
2 la

distance
d’´
e
quilibre
L
.
Cela
veut
dire
que
s
s’exerce
entre
eux

e
pend
de
l’´
e
longation
par
0
nts
au
repos.
La
force
qui
s’exerce
entre
eux

e
pend
de
urs lesOn peut
points
vers leur distance d’´equilibre L 0
donc mod´eliser la force par une expression qui explicite son caract`ere lin´eaire :
!
!
our

e
crire
une
situation
stable,
on
veut
que
est
plus
grand
que
Ph´
enom´
e
nologiquement,
pour

e
crire
une
situation
sta
ra P
P
,
est
plus
grande
que
la
longueur
L
,




1
2
0
est
plus
petit
que
!
!
!
!
longueur!F du
Pk2 ,kP P!est
= k kvecteur
P P k L 1
et P
F 1=
k L 1plus grand
re L 0leur
.o`uposition
Cela
veut
dire
que
si
la
distance
entre
d’équilibre
vers
distance
d’´
e
quilibre
L
.
Cela
veut
dire
que
si
position
d’équilibre
0
k
(en
N/m)
est
une
constante
ph´
e
nom´
e
nologique
qui

e
termine
le
rapport
entre
l’intensit´
e
de
la
force
et
orce
doit
ˆ
e
tre

e
pulsive.
ce
est
plus
la
force
doit
ˆ
e
tre

e
pulsive.
la d´
eformation. petite,
!
grande
quePlaPlongueur
, alors que
la force
est
u vecteur
, est plusL0grande
la longueur
L , a
5-19

1/2

1 2

0

1/2

2/1

1 2

0

2/1

distance qui s´eparerait ces deux points au repos. La force qui s’exerce entre eux d´epend de l’´elongation par
rapport a` cette distance au repos. Ph´enom´enologiquement, pour d´ecrire une situation stable, on veut que

L’oscillateur harmonique: première approche
la force ram`ene toujours les points vers leur distance d’´equilibre L 0 . Cela veut dire que si la distance entre
!
P1 et P2 , c’est-`
a-dire la longueur du vecteur P1 P2 , est plus grande que la longueur L0 , alors la force est
attractive, si la distance est plus petite, la force doit ˆetre r´epulsive.

On peut donc
mod´eune
liser la force
par une expression
qui explicite
son caract`eson
re lin´eaire
:
mod´eliser la force
par
expression
qui
explicite
caract`
ere lin´eaire :

On modélise la force en explicitant
linéaire:
⇣ son caractère

!
!
!
!
5-19

!
F 1/2 =

F 1/2 =


!
kla d´eformation.
kP1 P2 k

k

kP1 P2 k



1 1/2

L0

!
1 1/2

et

F 2/1 =


!
k kP1 P2 k

!
F 2/1 =

L0



!
1 2/1


!
k kP1 P2 k

o`
u k (en N/m) est une constante ph´enom´enologique qui d´etermine le rapport entre l’intensit´e de la force et

L0

et

L0

Si les vitesses initiales sont sur la droite qui joint les deux points entre eux, tout se passe selon cette

ligne et on simplifie grandement le probl`eme en y alignant un des axes de r´ef´erence. Le probl`eme devient



!
1 2/1

alors unidimensionnel. Prenons l’axe x selon la droite. Si X indique la position du premier point et X la
st une constante
ph´enom´enologique qui d´etermine le rapport entre l’intensit´e
position du deuxi`eme point (X > X ), la composante de la force dans la direction x s’´ecrit alors :
1

2

2

1

où k (en
détermine
5-20 N/m) est une constante
F1/2 x = phénoménologique
k (X2 X1 L0 ) et F2/1qui
(X2 X1 L0le
) rapport entre l’intensité de la
x =k
force et la déformation.

ses initiales Les
sont
sur
la droite
qui
joint les deux points entre eux, tout se pas
´equations
du mouvement
s’´ecrivent
:

m a e=me
k (Xen Xy alignant
L ) et m a =un
k (X des
X axes
L )
plifie grandement le probl`
de r´ef´erence. Le pro
5-21

2 2

2

1

0

1 1

2

1

0

Forces/ L’oscillateur harmonique

72

!
F 1/2 =

k

!
kP1 P2 k

!
1 1/2

L0

et

!
F 2/1 =

!
kP1 P2 k

k

chercher les causes physiques r´eelles de cette force mais donner a priori son expression. Nous consid´erons

un cas
o`
u, en premi`
approximation,
s’exerce entre
deux
points e
d´re
epend
lin´
eairement
ut donc mod´eliser
la⇣force
parereune
expression
quiquiexplicite
son
caract`
eaire
: de la
⌘ la force
⇣lin´

L0

!
1 2/1

! harmonique:
!
!
!
!
L’oscillateur
première
approche
k kP1 P2 k L0 1 1/2 et F 2/1 = k kP1 P2 k L0 1 2/1

!
F 1/2 =

!
!
variation de distance entre ces points. Soient OP1 et OP2 la position de deux objets ponctuels et L0 la

distance qui s´eparerait ces deux points au repos. La force qui s’exerce entre eux d´epend de l’´elongation par



rapport ph´
a` cette
distance au
repos. Ph´enom´enologiquement,
d´ecrire une situation
stable, on veut que
une constante
enologique
qui d´epour
termine
le rapport
entre l’intens
⇣ enom´




! la force ram`ene toujours
! les points!
! entre
vers leur distance!
d’´equilibre L 0 . Cela veut !
dire que si la distance
F 1/2 = k kP1 P2 k L0 1 1/2 et! F 2/1 = k kP1 P2 k L0 1 2/1

P et P , c’est-`
ae
-dire
la longueur
du vecteur P Pqui
, est plus
grande que la longueur
L , alors la force
est
est une constante
ph´
nom´
enologique
d´etermine
le rapport
entre
l’intensit´e
1

2

1 2

0

attractive, si la distance est plus petite, la force doit ˆetre r´epulsive.

N/m) est une
constante
enologique
qui d´etermine
le rapport
entreentre
l’intensit´
e de la
force se
et
s.eninitiales
sont
sur ph´
laenom´
droite
qui joint
les deux
points
eux,
tout

rmation.

sses
initiales sontlesur
la droite
quiy joint
les deux
points
entre
eux,
tout seLe
pasp
fie
grandement
probl`
e
me
en
alignant
un
des
axes
de

e

e
rence.
les vitessesxinitiales sont sur la droite qui joint les deux points entre eux, tout se passe selon cette
mplifie
grandement
le leprobl`
eeme
endroite.
alignant
unaxes
desdeaxes

ef´eprobl`
rence.
pro
t on simplifie
grandement
me la
en
yy
alignant
un X
des
r´ef´erence.
Le
edu
me Le
devient
nnel.
Prenons
l’axe
xprobl`
selon
Si
indique
ladeposition
premi

1
sionnel.
Prenons
l’axe
droite.
X1 indique
la du
position
nidimensionnel.
Prenons
l’axexxselon
selon laladroite.
Si XSi
la position
premier du
pointpremier
et X2 la p
1 indique
`eme point (X2 > X1 ), la composante de la force dans la direction x s’´ecri
n du
deuxi`
eme point
la composante
composante de ladeforce
dans ladans
direction
x s’´ecrit alors
1 ),la
uxi`
eme
point
(X (X>2 >
XX),
la force
la direction
x : s’´ecrit a
2

1

On peut donc mod´eliser la force par une expression qui explicite son caract`ere lin´eaire :

F1/2 xk= (X
k (X2 X
X1
F1/2 x =
2
1

F1/2 x =
5-19

k (X
! 2

F 1/2 =

k

LL
0 ) ) et
0

X
L⌘0!)
⇣ 1
!
kP1 P2 k

L0

1 1/2

F2/1 xF= k (X2= k
X1 (X
L0 ) X
et
2
1
2/1 x

et ! F2/1⇣x =
k (X

!
! 2

et

F 2/1 =

k

kP1 P2 k

L0

1 2/1

X1

L0 )

L0 )

uations du mouvement s’´ecrivent :
o`
u kmouvement
(en N/m) est une s’écrivent:
constante ph´enom´enologique qui d´etermine le rapport entre l’intensit´e de la force et
Les équations du
mouvement
s’´
e
crivent
:
la d´eformation.
du mouvement
s’´ecrivent :
m2Sia2les=vitesses
k initiales
(X2 sont
X1sur laLdroite
etjointmles1 adeux
k (X
Xtout
0 ) qui
1 =points
2 eux,
1 seLpasse
0 ) selon cette
entre
ligne et on simplifie grandement le probl`eme en y alignant un des axes de r´ef´erence. Le probl`eme devient

Forces/
alors unidimensionnel.
Prenons
l’axe x selon
la droite.
Si X m
indiqueala position
du
premierL’oscillateur
point X
et X la harmonique
m2m
a22a=
k
(X
X
L
)
et
=
k
(X
2
1
0
1
1
2
k (X2 X1 L0 ) et m1 a1 = k (X2 X1 1 L0L
) 0)
2 =
1

2

72

position du deuxi`eme point (X2 > X1 ), la composante de la force dans la direction x s’´ecrit alors :

5-20

F1/2 x =

k (X2

Les ´equations du mouvement s’´ecrivent :

X1

L0 )

et F2/1 x = k (X2

X1

L0 )

Forces/
L’oscillat
Forces/
L’oscillateur

def

L’´elongation
e = X2 première
X1 L0 approche
est l’´el´ement p
L’oscillateur
harmonique:
mouvement s’´ecrivent :
d’´evolution :
m 2 a2 =

k (X2

L0 )

X1

et m1 a2
1 = k (X2

X1

L0 )

d
k
e(t)
=
(X
2
Forces/ L’oscilla
2
def
L’élongation est
l’élément déterminant,
dt
m
ation e = X2 X1 L0 est l’´el´ement physiquement significatif.2
On ob

il
est
intéressant
d’exprimer
les
X2 X1 L0 est l’´el´ement physiquement significatif. On obtient ais´e1men
ion :
équations avec cette nouvelle
variable
k son ´equation
= X
X
L est l’´el´ement physiquement significatif. On obtient =
ais´ement
m2
d2
k
k
e(t)
=
(X
X
L
)
(X
X
L
2
1
0
2
1
0

d2
k
k
2
dtk (X2 Xm
m1 L 0 )
2 L0k)
d
e(t)
=
(X
X
1
2
1
m
+
2


2 e(t) =
1
(X
X
L
)
(X
X
L
)
dtdt
m
m
1
m 2✓
m
1
1
=
k



(X2 X1 L0 )
+
1
1 1 =1 k
m2
mX22 Lm
=
+ + (X (X
X)11 L0 )
= kk
def

2

1

0

2

2

2

2

1

0

1

2

=

=

donc

=

1

m2 m1
donc

d2 e
=
ke 2

ke

0

0

m2m2 m1 m1 ✓

✓✓
◆ ◆ m1 + m 2
m1 + =
m2 k
m
k
e
1 + m2
k m2 m1
em2 m1

d2 e donc
µ 2 2 = ke
dd te
µ

µ

1

2

donc

e

2

d e
µ
=
d t2

ke



◆ m
m
1
m1 + m 2 2

= kharmonique:
e première approche
L’oscillateur
m2 m1
o`
u µ = (m1 m2 /(m1 + mµ2 )
donc
donc
La
solution
de
cette
2
dd2 e e
o`
u µ = donn´
(m1 m
/(m
µµ d t2 2= =k e k e
ee 2par
: 1 + m2 )) es
dt

La
m1 + m2 )) est appel´e la “masse r´eduite” du
syst`esolution
me.
Masse
réduite

de système
cette ´equa
du

+demcette
appel´
e la “masse
r´eeduite”
du
syst`
e
me.
quation
du mouvement
est donn´
simple
:e e(t)
est
une
fonction

e
riodiqu
5-23
e(t)
2 )) ´eest
par :

La solution de cette équation différentielle est une fonction périodique du temps donnée
ette
´
e
quation
du
mouvement
est
simple
:
e(t)
est
u
par:
s

e(t) = e(t0 ) cos[!(t

e˙ 0
sin[!(t
t5-23
0 )] +
!

Physiquement, cela
signifi
k

t0 )]

avec ! =

e(t) = e

µ petite e
que la masse est

l’´eavec
quation
!⌧
= 2⇡.
la signifie que l’´elongation oscille autour de z´ero
une fr´
equence
d’autant

e˙ 0
Physiquement,
cela
signifie
qu
grande. La p´eriode d’oscillation ⌧ est d´ete

petite
et que
la
” k est oscille
Ici aussi
on
aurait
pu
t0fréquence
)]
avec
e(t)
=
e(t
)“rigidit´
cos[!(t
t0 )]autour
+ du pointsin[!(t
Physiquement
cela
que e
l’élongation
d’équilibre
avec une
0signifie

que la rigidité
k (des ressorts)
grande.
⇡. d’autant plus grande que la masse est petite et que
la!masse
estestpetite

et
qu
autre temps peut aussi ˆetr
aurait pu construire un temps en utilisant la vitesse initiale e˙ 0 et l’amplitude
l’´equation
= 2⇡. initiales et
des !⌧
conditions

gnifie
l’´elongation
oscille autour
de l’avantage
z´ero avec
aussi ˆetreque
construit
par analyse dimensionnelle.
Il pr´esente
d’ˆetreun
i
p
aurait

Ici temps
aussi caract´
on eristique.
pu con

neous
second-order
linear
differential
e
fficients:
L’oscillateur harmonique: équation différentielle

:erhaps
β and γthe
are first
real (i.e.
non-complex)
constants.
Without
lo
mathematician
to
derive
the
genera
dx
dx

linéaire
du
second
ordre
2
2
+
β
+ γx =If0 the coefficie
ed
that
the
coefficient
of
d
x/dt
is
unity.
2
footsteps.
As Euler second-order
suggested,
we
assumed
tha
hehis
case
of homogeneous
linear
equation
dt
dt
Interlude
mathématique....
yponential
multiply all
the
coefficients
by
1/α
to
get
an
equation
of
(60)
2 loss of
that
the simple
first-order
eq
where
β andform
γ différentielle
are real
(i.e.satisfy
non-complex)
constants.
Without
Soit
l’équation
linéaire
homogène
du second
ordre
d
x
2
2
2
thatsatisfy
the first
coefficient
of d x/dt
is unity.
If
the
coefficient
is
àassumed
coefficients
constants:
was
perhaps
the
mathematician
to derive
the
general
constant,
eq.
(60),
then
dx
dx
+
2


dt

simply
multiply
allhave
the
coefficients by 1/α to get an equation of
the fo
2
of
generality,
we
2
ollow his footsteps.
d x As Euler
dx suggested, we assumed that
2
isEuler
αexponential
(αwas
!= perhaps
0), weform
canfirst
the
mathematician
tosimple
derive the
general solutio
+
β
+
γx
=
0
me
that
satisfy
the
first-order
equ
2
λt
2
λt
dt d (e
d(e
)
)
Let’s follow
his
footsteps.
As dt
Euler
suggested,
we assumed
that eq. (
eλform
of
eq.
(60).
λt
is
a
constant,
satisfy
eq.
(60),
then
+
β
+
γ(e
)
=
0
the samesont
exponential
form
that
satisfy
the
simple
first-order
equation
des réels constants
et nous
supposons
que
le
coefficient
est
1
devant
la
2
dt
dt
dérivée
seconde
sans
perte
de
généralité
si
ce
coefficient
est non nul (il suffit
de of ge
ution
of
this
equation.
real
(i.e.
constants.
loss
where
λ is non-complex)
a constant, satisfy eq.
(60), then Without
diviser l’équation par ce coefficient pour se ramener à cette équation).2
2
λt
2 of 2 d2 (eλt )
q.
(60)
has
solution
coefficient of d x/dt is unity. Ifd(ethe) coefficient
is α
λt
λt
2λtλt + β
+
γ(e
)
=
0
2
λt
λt
d(e
)
d
(e
)
on
(6).
If
y
=
exp(λt)
2+ γe
λ dt
eto2 2get
++βλe
+
=
the coefficients by 1/α
equation
of020the form
βan dt
γ(eλt ) =

dt

d

re β and γ are real (i.e. non-complex
d
γ
are
real
(i.e.
non-complex)
c
med that the coefficient of d x/dt
hat
the coefficient
of dtd x/dt byis1/u
ply multiply
all the
dt coefficients
L'exponentielle ne
s’annule jamais et merci
M. Euler pour avoir eu
cette idée géniale...

i.e. first mathematician
the
general
solution
ethe
exp(λt)
is never zero, to2 2derive
λtλt
λt
λt
λt + γe
λt
λ
e
+
βλe
0
λ e + βλe + γe by
= 0=1/α
tiply
all
the
coefficients
t
As
Euler suggested,
we assumed
that eq. (60
ertsteps.
was
perhaps
the
first
mathematicia
(61)
and, because
exp(λt)
is never
zero,2
ecause
exp(λt)
is
never
zero,
λ + βλ
+ γ = 0equation (6
ial form that satisfy the simple
first-order

λt
λt λ + βλ + γ = 0
2
λ eharmonique:
+ βλe +équation
γe = différentielle
0
L’oscillateur
2
λ e + βλe + linéaire
γe = 0 λ2λdu
(62)
second
ordre
+
βλ
+
γ
=
0
2To
+
βλ
+
γ
=
0
is
never
zero,
find
λ,
we
just
need
to
solve
the
quadratic
algebric
e


>
0.
Then
there
are
two
real
roots
o
Interlude
mathématique....
ever zero,

2
λt
dt
use
exp(λt)
is
never
zero,
use exp(λt) is never zero,
dt2

2 λt

roots:

λt

λt

!
solve
the
quadratic
algebric
equation
(63).
We
thus
fi
,we
wejust
justneed
to
solve
the
quadratic
algebric
equation
(63).
We
thu
2
λ need
+ βλto
+
γ
=
0
(63)
2
# λ + βλ + 1γ "
−β
± β − 4γ#
!
!
=
0
!
λ
=
!
1,2
2
2
2−
2
−β
±
β

β the−quadratic
4γ algebric
andequation
λ2 −β
=
−β
β 2− 4γ
± Weβthus
−find
4γ −
o solve
(63).
two
2

λ1,2
λ1,2==

22 22

!

2 − 4γ
−β
±
β
Let us
first consider
the
case
β


>
λ1,2 =
(64)0. Then there are
22
2
tstconsider
the
case
β
consider the case β −−4γ4γ>>0.0.Then
Thenthere
thereare
aretwo
tworeal
realroots
rootsofofeq.eq.(6

eed to solve the quadratic algebric equation (63).
"
#
"
!
!
ase β − 4γ > 0.
Then
there
are
two
real
roots
of
eq.
(63):
1 exp(m
1
##
"
#
"
"
#
!! −β
!
nential11#"solutions,
t)
and
exp(m
t).
I
2
!
2
1
1
2
1
±
β


λ
=
−β
+
β


and
λ
=

2
2−
1
2
"
#
2
2
=
−β
+
β


and
λ
=
−β

β

!λλ
!
11 =
−β
+ 1=
β 2
− 4γ and 2λ2 =
−β − β − 4γ 2
λ
2
2
1,2
−β
+
β


and
λ
=
−β

β


(65)
2
2
general solution
can be2 written as a linear
2
2

2

2

2

exponential
solutions,
exp(m
and
exp(m
InInthe
p
exponential
solutions,
exp(m
exp(m
the previous
1 t)
2 t).
etwo
twodifferential
differential
exponential
t)
and
exp(m
t).
the
1 t) and solutions,
2 t). Inexp(m
There
are
two
differential
exponential
solutions,
exp(m
t)
1
2
1
2
the
case
β


>
0.
there
are
two
real
root
the general
solution
can
be
written
asThen
a linear
combination
of as
was
shown
that
the
general
solution
can
be
written
a
linear
combin
was
shown
that
the
general
solution
can
be
written
as
a
linear
comb
λ
t
λ
t
1 that the 2general solution can be
section it was 1shown
wr
2
ions:
utions:
La solution
est une combinaison
λ t
λ t
x(t)
=
C
e
+
C
e
(66)
1
2
these
solutions:
λ1λt 1 t
λ2λ
t 2t "
"
#
!
!
linéaire
avec C1
x(t)
CC
1 e1 e ++
2 e2 e 1 de deux exponentielles
x(t)==CC
1
λ1 t
λ2 t

x(t) = C e
1

+C e

2

C2 des=
constantes
stant. These constants 2can be determined from the et
initial
conx(t)
C1 e
2

22 e− 4γ
+
C
−β
+
β


and
λ
=
−β

β
These
constants
can
be
determined
from
and
areconstant.
constant.These
Theseconstants
constantscan
canbe2
bedetermined
determinedfrom
fromthe
theinit
in
2andCC2 2are

onential solutions,
exp(m1 t) and exp(m2 t).
roots:
L’oscillateur harmonique: équation différentielle −β ±
general solution
can
be
written
as
a
linear
λ
1,2 =
linéaire du second ordre
Interlude mathématique....

x(t) =

2
λ
t
λ
t
1 first consider
2 the case β − 4γ > 0. T
Let
us
C1 e + C2 e

fonction du signe
on aura:
. EnThese
constants
λλλ1 >
>0
0
>0
0 ,,, λ
λ22 >
>
0
λ
1
1
2

<0 0
> >0 0, λ, λ<
λλ11λ>
1 0 , λ22 <
2 0

λλλ111<<<000,,,λλλ222<<<000

X
X

X
X
X

Time

X
X
X

There are two differential exponential solutio
λ2 , different
aretheobserved
(fig
section behaviours
it was shown that
general solutio
these solutions:
λ
x(t)
=
C
e
1
Time
Time
Time

x

nd

"
#
!
1
2 − 4γ
λ
=
−β
+
β
can 1be 2determined froman

Time
Time

Temps

Time
Time

where
C
and
C
are
constant.
These
consta
1
2
Figure
4:
The
behaviour
of
x(t)
depends
on
the
sign
of
the
λ
and
λ
(eq.
66).
1
2
Figure 4: The behaviour of x(t) depends on the sign of the λ1 1 and λ
2 2 (eq. 66).
ditions.

222 − 4γ < 0 the two roots take the form
hen
β
hen β − 4γ < 0 the two roots take the form

ever zero,

L’oscillateur
harmonique:
2 Time équation différentielle
Time take the
β − 4γFigure
< 0 the4:
two
roots
λ + βλ +When
γ=0
(63)
The
behaviou
our
of
x(t)
depends
on
the
sign
of
the
λ
and
λ
(eq.
linéaire
du
second
ordre
rm
1
2
iour of x(t) depends on the sign of the λ1 and λ2 (eq.
2

λ = p + iq and λ =onpthe
−sign
iq of th

1equation
The behaviour
of x(t)
o Interlude
solve themathématique....
quadratic Figure
algebric 4:
(63). We thus
find depends
two2

λ = p + iq

1
β 2 − 4γ
2
When2 β 2 − 4γ < 0 the two roots take(64)
the form

−β ±

!

λ1,2 take
=take the
roots
form
When
β


<
0
the
two
r
o
roots
the
form
! (67)
nd λ2 = p −with
iq
1
2
2
case β − 4γ > 0. Then there
two real roots
of eq. (63):
p =are −β/2
and
q
=


β
λ1 = p + iq2and
λ2 =
p − iq an
p
=
−β/2
#
"
#
!
!
1
21!
2 −p
=
p
+
iq
and
λ
=

iq
λ
=
p
+
iq
and
λ
−β + λ
β


and
λ
=
−β

β

(65)
2
2
1
2
1 with 2

2
β

qng= solutions
4γ − are
2

(68)

1!
2
p
=
−β/2
and
q
=


β
The
corresponding
solutions are2
exponential solutions,
exp(m
1 t) and exp(m2 t). In the previous

!
1
with
1
t the general solution can be written as a linear combination
of
2
The corresponding solutions are
pp==−β/2
−β/2 and
and qq =
=
x(t) = C1 eλ1 t + C2 eλ2 t

22

4γ − β

(66)(p+iq)t

pt iqt

pt iqt
x1 (t) =
e
= (p−iq)t
e (p−iq)t
e a
(p+iq)t
pt
iqt
x1These
(t) constants
= e can be determined
= xe1(t) e=frome theand
xe 2 (t)
= e conand x=
=e
=
stant.
initial
2 (t)e
(p+iq)t

ons
are
ns are

It is often
useful
to
limit
the
set
of
solutions
The
corresponding
solutions
dtoxlimit
(t)
=
e
=
e
e
(69)
(see(see
Appendix):
2
the
set
of
solutions
to
the
real
solution
Appendix):
(p+iq)t
pt iqt
iqt
(p−iq)t
pt
−iqt
(p+iq)t
pt
(p−iq)t
pt
−iqt
e
=
e
e
and
x
(t)
=
e
=
e
e
= e e and x22(t)pt
=e e
pt

It is(p−iq)t
often useful
limit
the
pt −iqt
λ1 and λ2 , different
behaviours
areto
observed
(fig.set
4).of solutions to the real solutions

2
2 + 2iwν (w 2 − ν 2 − 2iwν)
2
w

ν
0
0
p
−β/2
and
q
=


β
Figure
4:
The
behaviour
of
x(t)
depends
on
the
sign
ofbet
L’oscillateur
harmonique:
équation
différentielle
λ1 = p + iq and
−Figure
iq
orm
2 λ22 = p 2
4:
The
w0 − ν − 2iwν
take
form
oroots
roots!
take the
the
form
=
λ
=
p
+
iq
linéaire
du
second
ordre
1
2
2
2
2
1
(w

ν
)

(2iwν)
22
0
! the form2
When
β − 4γ < 0 the two roots take
Interlude
mathématique....
1
qng
=


β
(68)
2
2
2
When
β


<
0
the
nd λsolutions
(67)
are
2 = p − iq with
p
=
−β/2
and
q
=


β
w

ν

2iwν
0
2
=
2
λλ11==pp++iq
and
λ
=
p

iq
2
2
2ν 2
iq and λ22 (w0 − λν1 )=2 p++4w
iq and
λ2 =
p − iq an
p
=
−β/2
!
1
2
2
esponding
solutions
are
2
w

ν
2w
q=
4γ −with
β
0(68)
=

i
!(p−iq)t
(p+iq)t
pt
iqt
2
2
2
1
2
2
2
2
2 )2
!
2
(w

ν
)
+
4w
ν
(w

ν
1
1
x1 (t) = e
= e e 0 and
xwith
(t)q == e4γ0 − β
p = −β/2
2and

=A

2
The
corresponding
solutions
are
pp==−β/2
q
=


−β/2 and
and
q
=
β
(p+iq)t
pt iqt
(p−iq)t2
pt −iqt
x
(t)
=
e
=
e
e
and
x
(t)
=
e
=
e
e
1
2
2
2
According the Moivre formula (see Appendix), an exponentia

The(p−iq)t
corresponding pt
solutions
−iqt are

dtoxlimit
(t) =the
=
e
e
(69)
2always
beerewritten
as:
set
of
solutions
to
the
real
solution
(p+iq)t
pt
iqt
(p−iq)t
pt
−iqt
n
useful
to
limit
the
set
of
solutions
to
the
real
solutions.
By
the
ons
are
The
corresponding
sol
ns are


nd x2 (t) = e

=e e

endix):

xe (t)
=
e + i sin(φ)
= e (p−iq)t
e
(69)
=
cos(φ)
pt iqt
= e e and x2 (t) = e

1
(p+iq)t

x1 (t) = e

formule
d’Euler, encore merci
Leonhart :-)
Applied
to
eq.
(253),
this
leads
to:
oto the
real
solutions.
By
the
Moivre’s
theorem
the real solutions.
thetoMoivre’s
theorem
x (t
It isIt
often
useful
limit theto
setlimit
of solutions
to the
solution
is By
often
useful
the set
ofreal
solution

1
(p+iq)t
pt iqt
iqt
(p−iq)t
pt
pt
pt −iqt
(p+iq)t
pt
(p−iq)t
pt
−iqt
e x1 (t)
=
e
e
and
x
(t)
=
e
=
e
e
e
(cos(qt)
+
i
sin(qt))
and
x
(t)
=
e
(cos(qt)

i
sin
2
1
=
e
e
and
x
(t)
=
e
e
(see(see
Appendix):
pt
pt
2
2
2
Appendix):

A x1 (t)
w0 −=
ν e (cos
) = e (cos(qt) + i sin(qt))
and
It is often useful to lim
cos(φ) =

2
2 )2 + 4w 2 νpt
2
B
(w

ν
xi1 (t)
= e (cos(qt) +(70)
i sin(qt))
and" x1 (t)1= e (cos
nd
x
(t)
=
e
(cos(qt)

sin(qt))
0
1
(see
Appendix):
he
setofofsolutions
solutions
to the
the
real
solutions.
By
Moivre’s
equation
(63)
isptlinear,
the real
linearsolutions.
combination
xthe
=
x
(t)
+
pt
1
e
set
to
By
the
Moivre
1
2
A
−2wν
x
(t)
=
e
(cos(qt)
+
i
sin(qt))
nd
x
(t)
=
e
(cos(qt)

i
sin(qt))
(70)
1
i
1
sin(φ)
=
x (t) are also
solutions.
Those
solutions
are
" p
2
"
2
2
2
2
2 1
Because
linear
combination
x
1 equation 1(63) is linear, B
"
"the
(w − ν ) + 4w
ν =1 e=
x1 (t)

pt

pt

2
ptpt −iqt
ptlinear combination x"
(p−iq)t
pt
pt
Because
equation
(63)
is
linear,
the
x
(t)
=
e
(cos(qt)
+
i
sin(qt))
and
x
(t)
=
e
(cos(qt) −−
i sin(qt))
x)2=
(t)e= e(cos(qt)
= e+ ei sin(qt)) and x1(69)
1
1 = e (cos(qt)
1
(t)
i
sin(qt))
i
i
Because
equation (63)
is line
x
(t)

x
(t)
are
also
solutions.
Those
solutions
are
2
1
2 the
2solutions
i
i By the
pt
seful
to
limit
the
set
of
to
the
real
solutions.
Moivre
useful
to
limit
set
of
solutions
to
the
real
solutions.
By
the
M
x
(t)

x
(t)
are
also
soluti
1
1
2
1
pt
pt
"
2
2 x (t) + x (t) an
ecause
(63)
linear,
the linear
combination
x
=
he
realequation
solutions.
Byis the
Moivre’s
theorem
1 "
2 pt
1
1
2
2
"
pt
1
1 (sin
1
dix):
"
ndix):
i
x
(t)
=
e
(cos(qt))
and
x
(t)
=
e
Interlude
mathématique....
1
2
(63)
linear,
linear
combination
x2=
(
xn2 (t)
− 2 xis
are also the
solutions.
Those
solutions
are x1 = 2 x1 (t)
1 (t)
1 + x2"1 (t)

L’oscillateur
harmonique:
équation
différentielle
(see
solutions. By the Moivre’s theorem Appendix):
linéaire
du
second
ordre
t substitution)
cos(qt)
+
i
sin(qt))
and
x
(t)
(cos(qt) + i sin(qt)) and x (t)==e e (cos(qt)
(cos(qt)−−i

x (t) = e
rposition
principle,
the
general
real
solution
of
eq.
(60
are
also
solutions.
Those
solutions
are
pt
pt pt (this can
ptpt pt
"
pt
"
be
verified
by
direct
substitution)
xe1x(t)
==
e e(cos(qt)
+ i+
andand
x1(70)
(t)
=
− i sin(qt))
=xtion
(cos(qt)

i
xsin(qt))
(t)
=sin(qt))
ei sin(qt))
(cos(qt))
x2x(t)
=e
e=(cos(qt)
(sin(qt))
pt
(t)
(cos(qt)
and
(t)
e
(cos(qt)

i
sin(
(this
can
be
verified
by
direct
1
1
1
e these
(cos(qt)solutions:
− i sin(qt))
(70)
11
of
" "
1 (t) =
is linear,
the
linear
x
=
x
(t
Thus,
according
tocombination
the
superposition
principle,
the
general
1
3)
is
linear,
the
linear
combination
x
=
x
(t
"
pt
"
pt
Thus,
according
to
the
superp
1
1
Because
equation
(63
2
1
2
his can be
direct
substitution)
x1verified
(t) =beebywritten
(cos(qt))
and
x
(t)
=
e
(sin(qt))
2
1
1
"
as
linear
combination
of
these
solutions:
be
written
as
linear
combinat
1 + x2 (t)
i
" i x1 (t)
uation
(63)
is
linear,
the
linear
combination
x
=
1 Those
1the
" pt
"combination
1
1
pt
"
"
1
solutions.
solutions
are
2
2 eq.
quation
(63)
is
linear,
linear
x
=
x
(t)
+
mbination
x
=
x
(t)
+
x
(t)
and
x
=
x
(t)

x
(t)
are
al
lso
Those
solutions
are
1
1
2
on according
xsolutions.
=
x
(t)
+
x
(t)
and
x
=
1
2
1
(t)
=
C
e
cos(qt)
+
C
e
sin(qt)
hus,
to
the
superposition
principle,
the
general
real
solution
of
(
1
2
2
2
2
1
2
1
2
2
2
1
2
2
2
(t)
are
also
solutions.
Those
solutions
are
pt
pt
1
x(
ons
are
x
(t)
are
also
solutions.
Those
solutions
are
x(t)
=
C
e
cos(qt)
+
C
e
sin(qt
e
written
as
linear
combination
of
these
solutions:
fied
by
direct
substitution)
1
2
1
e
"" which
pt
"
pt
" ptbe pt
" " the
pt initial
"
pt
pt
ptwhere
C
and C
1
2 are constant
pt
nts
can
determined
from
condition
x
(t)
=
e
(cos(qt))
and
x
(t)
=
e
(sin(qt))
x(t)
=
C
e
cos(qt)
+
C
e
sin(qt)
"
pt
"
pt
to
the
superposition
principle,
the
general
real
solution
of
e
xx12(t)
=
e 1xewhere
(cos(qt))
and
x
(t)
=
e
(sin(qt))
C
and
C
constants
can
be
determined
fro
1are
2which
xpt1=(t)
=
(cos(qt))
and
x
(t)
=
e
(sin(qt))
2
(t)
e (sin(qt))
(71)
1
2
2 2x2 (t)The
=behaviour
e (sin(qt))
1 (t) = e (cos(qt)) and
of x(t) depend
=ear
e combination
(sin(qt))The behaviour
(71)
of
these
solutions:
of
x(t)
depends
on
the
p
and
q
(fig.
5).
here
C
and
C
are
constants
which
can
be
determined
from
the
initial
conditio
(On
peut
vérifier
par
substitution)
ds
on
the
p
and
q
(fig.
5).
1
e verified
by 2direct substitution)
Note
that
eq.
(72)
can
be rewb
(this
can
be
verified
be
verified
by
direct
substitution)
Note
that
eq.
beqrewritten
he
behaviour
of
x(t)
depends
the can
p and
(fig.pt5). as (see Appendix):
by
direct
substitution)
pton(72)
direct
substitution)
=Appendix):
C1 e ofcos(qt)
+ can
C
sin(qt)real solution of eq
rding
to the
superposition
principle,
the
the general
real
eq. (60)
written
asx(t)
(seesolution
2 e general
Thus, according to th

ording
to
the
superposition
the
general
pt real solution
ote that eq.
(72) can
be rewritten asprinciple,
(see Appendix):
!cos(qt + φ)
x(t)
=
Ke
as
linear combination of these solutions:
ons:
2
2
C
+
C
φc
sa
with
K
=
he
superposition
principle,
the
general
real
sol
1
2 and
eneral
real
solution
of
eq.
(60)
can
be
written
as
linear
superposition
principle,
the
general
real
solu
as
linear
combination
of
these
solutions:
pt
!can be determined
pt
are constants which
from the initial con

2

x(t)
φ)
2 +=φ)
2Ke cos(qt +
x(t)
=
Ke
cos(qt
pt
pt
pt
C2 and(72)
sin(φ) = C1 /K and co
K=
= C Ce1 +cos(qt)
e sin(qt) with
x(t)
+φ Csatisfying
e sin(qt)

+
C2 !
ombination

1 solutions:
2
of
these
solutions:
mbination
of
these
pt q (fig. 5).
pt
fthx(t)
depends
on
the
p
and
x(t)
= C1 e sin(φ)
cos(qt)
C2and
e cos(φ)
sin(qt)
φ satisfying
= C+/K
= C /K.
K = C + C and
2
1

2
2

1

etermined
from
the initial
satisfying
sin(φ)
= Cconditions.
/K and cos(φ) = C /K.
t

2

The
behaviour
of
x(t)
depends
on
the
p
and
q
(fig.
5).
Note
that
eq.
(72)
can
be
rewritten
as (see Appendix):
on the p and q (fig. 5).

ds L’oscillateur harmonique: Note
that eq. (72)
can be rewritten as (se
équation
différentielle
Note that eq. (72) can be rewritten as (see Appendix):
pt

du second ptordre x(t) = Ke cos(qt
x(t) = K
written as (seelinéaire
Appendix):
Interlude mathématique....

with
K ==
x(t)

! x(t) = Ke cos(qt
+ φ)
!
C22 and
sin(φ)
= C1 /K
with K = C12 +with
C12 + C22 and
φ satisfying
sin(a
K =φ satisfying

pt
2
and φ satisfying
Ke+ C2cos(qt
+ φ) sin(φ) = C1/K and cos(φ) = C2/K.

!

C12

p > 0, q large
p > 0, q large

p > 0, q small
p > 0, q small

15

X

Time
Time

X

p > 0, q large

X

X

X

petit
p > 0, q small
X

X

X

15
satisfying sin(φ)
=
C
/K
and
cos(φ)
=
C
/K.
1
2
p > 0, q large
grand
p > 0, q small

15Time
Time

X

Time

Time

X

p < 0, q large
p < 0, q large

X

X

X

X
X

X

Time

Time

p < 0, q large
p < 0, q large
grand

p < 0, q small
petit
p < 0, q small
p < 0, q small
p < 0, q small

Time

Time

Simple
harmonic
oscillator
L’oscillateur
différentielle
dx 22 0
dt
d
x
+
w
x
=
0
harmonique:
équation
2
0
2
+
w
x
=
0
dt
2
0
2

2
d
x
dt
uation
represents
a
simple
harmonic
oscillator
(H
d
x
2
linéaire
du
second
ordre
!
+
w
x
=
0
2
0
lve the characteristic
equation:
2
+
w
x
=
0
dt
der
linear
differential
equation
is
d
to
solve
the
characteristic
equation:
ueInterlude
described
above.
First,
we
can
rewrite
the
equati
2
0
mathématique....
oui
mais
et
notre
oscillateur
harmonique
dans
tout
2
λ
=
±
−w
=
±iw
0
0
dt

hus
to solve the
characteristic
equation:
ça need
!
2
2
2

λ +w
=w02 = 0 2
λ20 +
0
dx

d x
olutions
are
2
istic equation:
2
+
w
x
=
0
2
2
=
−w
x
0
!
2+ w = 0
0!
λ
2
dt
0
dtλ = ±λ =−w
2
2
± =
−w±iw
=0±iw
0

0

0

équation caractéristique
s need
to solve the
iw0 tcharacteristic!equation:
−iw0 t

imple
harmonic
oscillator
(HO).
It
can
be
so
x
(t)
=
C
e
and
x
(t)
=
C
e
1
1
2
2
lar solutions
are 2 are
particular
2
2 solutions
λ
=
±
−w
=
±iw
0
0
λ
+
w
=
0
irst, we can 0rewrite the equation as follows:
2
2
t = solutions:
any
linear
combination
of
these
two
iw t iw t
−iw+t−iw
λ
w
0
x
(t)
=
C
e
and
x
(t)
=
C
e
0
ce, two xparticular
solutions
are
(t)
=
C
e
and
x
(t)
=
C
e
1 1
1
2
2
1
2
2
0

0

0

0

2

solution
combination
of
these
two
solutions:
danyxlinear
n is any islinear
combination
of
these
two
solutions:
2
!
iw0 t−iw t
−iw0 t
+
w
x
=
0
iw
t
0 x1 (t)0 = C1 e
0 x2
and
(t)
=
C
e
2
2
2
2
x(t) = C1 e iw t+
C
e
dt
2
λ
=
±
−w
=
±iw
0
0
0
−iw
t
0
0
0
iwC
−iw
t
x(t)
=
+
C
0 t1e
0e
2
x(t) = C e
+C e

!
= ± −w = ±iw

1
2 combination of these two sol
general
solution
is
any
linear
aracteristic equation:

x
x1 (t) = C1 e iwand
x2 (t) = C2 e iw
t
−iw
t
(see
Appendix):
t

x1 (t) = C1 eto the
and xreal
=
C
e
21(t)
2
x
(t)
=
C
e
and
x
(t)
=
C
e
convenient
to
limit
the
solutions
solutions.
This
1
2
2
L’oscillateur
harmonique:
équation
différentielle
The general solution is iw
anyt linear combination
of theseThe
two
solutions:
general
solution is any
−iw
x
0 linear combination
0
TheAppendix):
general
solution
is
any
of of
these
two
solutions:
The
general
solution
is
any
linear
combination
of these tw
The
general
solution
is
any
linear
combination
these
two
solutions:
(see
C
e
+
C
e
any
linear
combination
of
these
two
solutions
1
1
"
ordre
x1 (t) = linéaire du second =
C1 cos(w0 t)
iw0 t
−iw0 x iw t
−iw t
2
C
e
+
C
e
x(t)
=
C
e
e
iw tiw+
−iw
t C
−iwtça
t=! C1 eiw t + C2 e−iw t
1
2
1 mais et notre 1oscillateur
Interlude mathématique....
oui
harmonique
dans
tout
x(t)
"
x(t) x(t)
= C=1 e=
C
C1 eC+ cos(w
+ 2Ce2 e It ist)
x1 (t) =
1
0 usually more convenien
by noticing
that (see
Appe
2
It
is
usually
more
convenient
to
limit
the
solutions
to
the
re
ItItisisusually
more
convenient
to
limit
the
solutions
to
the
real
solutions.
Th
iw
t
−iw
x
iw
t
−iw
t
0
0
It
is
usually
more
convenient
to
limit
the
solutions
to
the
real
solutions.
This
0
0
usually
more
convenient
to
limit
the
solutions
to
the
real
solutions.
Thi
C
e

C
e
2
2
by
noticing
that
(see
Appendix):
x(t)
=
C
e
+
C
e
" noticing
1
2
by
noticing
that
(see
Appendix):
by
that
(see
Appendix):
x(
x2 (t)
=(see Appendix):
=
C
sin(w
t)
2
0
by noticing
that
iw
t
−iw
x
iw0 t
−iw0 x
C1 e
+ C1 e
"
2
iw
t
−iw
x
C
e

C
e
−iw
x = and
= C1 co
2C1 eiw
C1tte + C+1 eC−iw
1 ex1 (t)
" avec une peu2de
iw
x
"
Oui mais en fait
2 0 t)
""
C
e
+
C
x
(t)
=
=
C
cos(w
1
1 eC2 sin(w
x
(t)
=
=
t)
1
1
0
x
(t)
=
=
C
cos(w
t)
2
1cos(w00t)
1
trigonométrie....
x
(t)
=
=
C
2
x(
1
1 2
and
enient
to
limit
the
solutions
to
the
real
solution
2
s. The general
real
solution
of
eq.
(85)
can
thus
be
written
2
iw t
−iw
x
C
e

C
e
2
2
and
"
x
(t)
=
= Cgen
iw
t
−iw
x
2 si
and
2
are
also
solutions.
The
C
e

C
e
and
2t
2
2
"
iw
−iw
x =
Appendix):
iw
t
−iw
x
x
(t)
=
C
sin(w
. The general real solution
of
eq.
(85)
can
thus
be
written
a
−C
C22ee
2
0 t)
2 C
22e
C
e

""
2 general
xx22(t)
=
=real
C22sin(w
sin(w00t)t)
are
also solutions. The
solution
of eq. (85) ca
(t)
=
=
C
"
i
2C
iw
t
−iw
x
x
(t)
=
C
cos(w
t)
+
t)
are also solutions.
general0 real solution
of eq. 0(85)
can thus be written
0 The
0 2 sin(w
1
C
e
+
C
e
1
1
"
"
further
are
also
solutions.
The
general
real solution of eq.
eq. x(85)
(85)
can
thus
be
writte
(t) This
=can
C1expression
cos(w
+can
C
"
are
also
solutions.
The
general
thus
be
written
0 t)
2 sin(w0
x1 (t) =x (t) = C1 cos(w0t)
=
C
cos(w
t)
+
C
sin(w
t)
1
0
2
0
"
can further be rearranged
in
Appendix):
x
(t)(see
= C1can
cos(w
C2rearranged
sin(w0 t) in (see Appendi
0 t) +be
further
2This expression
""
xx (t)
=(see
C1 cos(w
+
C22Appendix):
sin(w00t)t) "
(t)
C
sin(w
Avec further
encore un
petit
effort
de trigonométrie...
00t)
This
expression
can further
rearranged
in +
(see
an
be
rearranged
inbe
Appendix):
with
A = C + C and φ
"
0

0 0

0
0

0
0 0

0

0

0

0

0

0
0

"
1

0

0 0
0

"
2

0

0

0

0

0
0

00

2
1

2
2

x (t) = A cos(w0 t + φ)

−iw
" 0t
0x
The solution is thus period
"
Thisexpression
expression can
caniw
further
be
rearranged
in
(see
Appendix):
This
further
be
(see
Appendix):
x
(t)
=
A
cos(w
t
+
φ)
2
" 20
2
2

C e


C
e
with
A
= xC(t)+=CAand
φ satisfying
sin(φ) by
=C
and co
c
"
determined
the/A
initial
cos(w
t + φ)
x2 (t) = "x" (t) = The
=
C
sin(w
t)
2
0
the
frequency
solution0ist thus
periodic. The amplitude of
A the
andoscilla
the
A
cos(w
+
φ)
2
A = C + C sin(φ)
and
φ satisfying
sin(φ)
=condition,
C
/A
andi.e.cos(φ)
= C/A.
/A.
2determined
x"=
(t)by
=the
A
cos(w
t
+
φ)
+ C2 andwith
φ satisfying
C
/A
and
cos(φ)
=
C
initial
the
value
of
1 cos(w0 t + φ)
2 x at t
1

2
1

2
2

"

2

0

0 1

1

2

2 general real solution of eq. (85) ca
are also solutions. The

L’oscillateur
harmonique:
équation
différentielle
are also" solutions.
The general real
solution of eq.
(85) can thus be written
x (t) = C1 cos(w
t)
+
C
sin(w
t)
"
0
2
0
This
expression
can
further
x
(t)
=
C
cos(w
t)
+
C
sin(w
t)
x
(t)
=
C
cos(w
t)
+
C
sin(w
1
0
2
0
linéaire du second ordre
"

1

0

2

0

"
x
(t) = C1can
cos(w
C2rearranged
sin(w0 t) in (see Appendi
0 t) +be
This
expression
further
rearranged in (see Appendix):

can
further
becanouifurther
This
expression
be oscillateur
rearranged
in (see
Interlude
mathématique....
mais et notre
harmonique
dansAppendix):
tout ça !

"
This expression can further be rearranged in (see Appendix):
with
A = C12 + C22 and φ
"
x (t) = A cos(w t + φ)
0

The solution is thus period
2cos(w0 t + φ)
"
A
" 2=
with A x
= (t)
C
+
C
and
φ satisfying
sin(φ) by
=C
and co
c
1 /A
determined
the
initial
x (t)
A
cos(w
t
+
φ)
1 = 2
0
0
"
theamplitude
frequency of
the
oscilla
"
The
solution
is
thus
periodic.
The
A
and
the
2
2
2
2
+ C2φand
φ satisfying
sin(φ)
cos(φ)
= C=
/A.
= withCA1 =
+ CC
satisfying
==condition,
C11/A
/Aand
and
cos(φ)
2of
2 /A
2 1and
determined
bysin(φ)
the
initial
i.e.
the
value
xCat
t
""

x (t) = A cos(w t + φ)

with2 A

+ C2 andThe
φ solution
satisfying
sin(φ)
=
C
/A
and
cos(φ)
=
C
/A.
1
2
the
frequency
of
the
oscillations
and
the
period
is
T =
is thus periodic. The amplitude A and the phase φ of thus
the osc

La solution
est donc
bien périodique
avec une
amplitude
A et une fréquence
phase φ of th
The
solution
is thus
periodic.
The
amplitude
A and wthe
phase
o et une
determined by the initial condition, i.e. the value of x at time t = 0, x0 . Par
determined
by
theThe
initial
condition,
i.e.
theperiod
value
x Tat=time
=
0,6).xo0
thus
periodic.
amplitude
A the
and
theis of
phase
φ18of0t (fig.
the
the
frequency
of
the
oscillations
and
thus
2π/w
A = AA
A < Aref
w0 > ww
=and
Arefw
w0ref
= w0ref
A < ArefA < A
> w0ref
Aand
==Aw
and
w0 = w0ref
w0 > w0ref
ref
0
0ref
0
0
ref
ref
theinitial
frequency
of the oscillations
the of
period
thus Tt =
= 2π/w
he
condition,
i.e. the and
value
x atistime
0, x00. (fig.
Pa

X

X

X

18

X

X

X

x

X

X

18 T = 2π/w (fig. 6).
the oscillations and the period is thus
0

18
TimeTime Time

TimeTime Time

Temps

Time Time Time

sin(x
+ y) =
sin(x) cos(y) + cos(x) sin(y)
+ y) = cos(x) cos(y)
− sin(x)
sin(y)
(239)
équation
différentielle
dixL’oscillateur harmonique:
cos(x + y) = cos(x) cos(y) − sin(x) sin(y)

+sin(x)
y) = cos(y)
sin(x)
cos(y)
+
cos(x)
sin(y)
sin(xsin(x
+ y) =
+
cos(x)
sin(y)
du+ cos(x)
second
sin(x + y)linéaire
= sin(x) cos(y)
sin(y) ordre
ometry
complex
numbers
+
y)
=y)cos(y)
cos(x)
−sin(x)
sin(x)
sin(y)
cos(xcos(x
+ y)and
=
cos(x)
−cos(y)
sin(x)
cos(x
+
= cos(x)
cos(y) −sin(y)
sin(y)

(23
(23

sin(x (238)
+ y) = sin(x) cos(y) + cos(
(238)
cos(x + y) = cos(x) cos(y) − sin(

(239)
(239)

oui=mais
et la trigonométrie
dans tout ça(240)
!
AInterlude
sin(x)mathématique....
+ B cos(x)
K sin(x
+ φ)
recall some formula with are used in this summary. The demonstrations
A sin(x) ++
B cos(x)
= K sin
A
sin(x)
+
B
cos(x)
=
K
sin(x
φ)

sis
aresuch
out ofthat
the scope
of
the
present and
summary.
cos
φ
=
A/K
sin
φ
=
B/K,
<
and φ −π
is such
thatφ
cos<
φ =π.
A/K a
A sin(x) + B cos(x) = K sin(x +with
φ) K = A + Band
(240)

A sin(x)
+
B cos(x)
=cos(x)
K sin(x
+Kφ)sin(x + φ)
(240)

2
2
A
sin(x)
+
B
=
(24
relations
Indeed,
h= KA2=
A
+
B
and
φ
is
such
that
cos
φ
=
A/K
and
sin
φ
=
B
2
+ B and φ is such that cos φ = A/K and sin φ = B/K, and −π < φ < π.
2

2

nd2 φ is such that cos φ = A/K and sin φ = B/K, and −π < φ < π.

A
A sin(x) + B cos(x) = K sin(x) +
K
!
A
= K
sin(x
K
(236)
= K(cos φ sin(
= K sin(x + φ)

B
eed,and φ is such that cos φ = A/K and sin φ = B/K, and −π < φ <

A sin(x) A
B B
A sin(x)=
+ tan(x)
B cos(x)
= K sin(x)
+ K cos(x)
cos(x)
+ B cos(x)
K = sin(x)
+
K
B
A cos(x)K
B K K
K
!
"A
! sin(x)
" sin(x) + K cos(x)
sin(x)
+
B
cos(x)
(x) + B cos(x) =A K
+
cos(x)
AK
B= K
B
= K A sin(x)
+ cos(x)
K
K
A
B
K
K
tions:
!
"
K
K
A sin(x) + B cos(x)
K sin(x)
+
K
cos(x)
!
= K =A sin(x)
+
cos(x)
B
K
A"
B
functions
=
K(cos
φ
sin(x)
sin φK
cos(x))
K+Hyperbolic
= K K sin(x)
! + cos(x)
=
K
sin(x)
+
cos(x)
K
K
2
2
A
B
By
definition,
the
hyperbolic
sine,
cosine
and
tangent
(241)
cos (x) + sin=(x)K=sin(x
1 + φ)
(237)
K
K
=
K
sin(x)
+
cos(x)
== K(cos
φ
sin(x)
+
sin
φ
cos(x))
K(cos φ sin(x)K+ sin φ cos(x))
e +e
K
cosh(x)
=
= K(cos φ sin(x)
+ sin
2 φ cos
bolic functions == K
+
φ)
(241)
K sin(x
sin(x
+
φ)
(241)
= K(cos φ sin(x) + sin φ cos(x)) sinh(x) = e − e
ition,sin(x
the hyperbolic
sine,cos(y)
cosine+and
tangent
are: = K sin(x +
φ)
2
+ y) = sin(x)
cos(x)
sin(y)
(238)

ns

= Kx sin(x
+ φ)
−x

cos(x + y) = cos(x) cos(y) −
e sin(x)
+ e sin(y)
cosh(x) =

2are:
erbolic
sine,
cosine
and
tangent
perbolic functions x −x

x

−x

x

−x

(24

sinh(x)
ex
tanh(x) =
= x
(239)
cosh(x)
e

(242)

L’oscillateur harmonique: équation différentielle
complex number
and trigonometric
functions:
linéaire
du second ordre
ix
Interlude mathématique....
maistrigonometric
trigonométrie
dans tout ça ! (2
complex
number
and
functions:
e
= oui
cos(x)
+etilasin(x)

as:

s:

−ix

e

ix

e

−ix

e

= cos(x) − i sin(x)

= cos(x) + i sin(x)
= cos(x) − i sin(x)

eix + e−ix
cos(x) =
2
ixix
−ix
ee −+e e−ix
sin(x) =
cos(x)
=
2i2
ix

e −e
sin(x) =
2i

−ix

(2

(2

(2


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