exact solution for a linear forced oscillator: x + 2ε x + x = 2f cos(Ωt ) = f ei Ωt + cc
assume the particular solution has the form x = K ei Ωt + cc;
substitute it into EoM
( −Ω2 + i 2ε Ω − 1)K ei Ωt = f ei Ωt
⎛ 1 − Ω2 − i 2ε Ω ⎞
f
1 − Ω2 − i 2εΩ
i⎜
K=
f
⎟=
1 − Ω2 + i 2ε Ω ⎝ 1 − Ω2 − i 2ε Ω ⎠ 1 − Ω2 2 + 4ε 2Ω2
(
let 1 − Ω2 − i 2ε Ω =
1
a = ⎡ 1 − Ω2
⎢⎣
2
(
x particular = f
)
2
1 iφ 1
1
1
a e = a cos φ + i a sin φ → 1 − Ω2 = a cos φ;
2
2
2
2
+ 4ε 2 Ω2 ⎤
⎥⎦
1
2
and
ae (
i Ωt +φ )
(
)
2 ⎡ 1 − Ω2
⎣⎢
)
2
+ 4ε 2Ω2 ⎤
⎦⎥
1
2
tan φ =
− 2ε Ω =
1
a sin φ
2
sin φ −2ε Ω
=
cos φ 1 − Ω2
+ cc = f H cos ( Ωt + φ )
1
H=
(
⎡ 1 − Ω2
⎣⎢
)
2
+ 4ε 2Ω2 ⎤
⎦⎥
1
2
H is the amplication factor, which is the ratio the amplitude
of the response and to that of the excitation, and φ is the
phase angle between the response and the excitation:
H
Ф
Ω
Ω
We are interested in small damping (ε <<1) and Ω 1 so that H is very large.
approximation to the exact particular solution of the EoM valid when
1) ε <<1 and 2) Ω 1
introduce a 'detuning' factor, labeled σ : Ω = 1 + εσ
x particular =
fe(
i Ωt +φ )
(
2 ⎡ 1 − Ω2
⎣⎢
)
2
i Ωt +φ )
x particular
fe(
(
2ε 1 +σ
CONCLUSION :
1
2 2
)
+ 4ε 2 Ω2 ⎤
⎦⎥
1
2
+ cc
→
fe(
i Ωt +φ )
2 ⎡( 2εσ ) + 4ε 2 ⎤
⎣
⎦
2
1
2
Ω2 = 1 + 2εσ +
+ cc
+ cc
if we set up a familiar MMS expansion to approximate
the response, we must initially account for the amplitude of the response
being an order (in ε ) larger than the amplitude of the excitation.
MMS approximation to the exact solution
x + 2ε x + x = 2ε f cos(Ωt ) = ε f ei Ωt + cc
note: we have taken the difference in the order of ε into account
the approximate solution is expressed in the form
x (t ; ε )
x0 (T0 ,T1 ) + ε x1 (T0 ,T1 ) and substituted into EoM:
(
)
D02 x0 + x0 + ε D02 x1 + x1 + 2D0D1x0 + 2D0 x0 = ε f ei ΩT0 + cc
D02 x0 + x0 = 0
→
x0 = A (T1 ) eiT0 + cc
D02 x1 + x1 = − ( 2D0D1x0 + 2D0 x0 ) + f ei ΩT0 + cc = −2i ( D1A + A ) eiT0 + f ei ΩT0 + cc
if 2i ( D1A + A ) eiT0 remains, the particular solution for x1 will contain secular terms
if f ei Ωt remains the particular solution will contain a 'small-divisor' term; therefore
to eliminate all the troublesome (both secular and small-divisor) terms, we put
−2i ( D1A + A ) eiT0 + f ei ΩT0 = 0
→
− 2i ( D1A + A ) + f ei ( Ω−1)T0 = 0
as before, we put Ω − 1 = εσ and introduce polar coordinates: A =
1
iβ T
a(T1 )e ( 1 )
2
−i ( a′ + iaβ ′ + a ) + f ei (σεT0 − β ) = −i ( a′ + iaβ ′ + a ) + f ei (σT1 − β ) = −i ( a′ + iaβ ′ + a ) + f ei γ = 0
a′ + a = f sin γ
aβ ′ = −f cos γ
γ ′ = σ − β′
steady state: a and γ are constants; then σ = β ′
a = f sin γ
aσ = −f cos γ
→
tan γ = −
sin γ
1
=−
cos γ
σ
and
a=
f
1
2 2
(1 + σ )
again the approximation to the exact solution of the exact EoM is the same as
the exact solution of the approximation to EoM
nonlinear EoMs
free vibration
hinge
moment about O − mgl sinθ = IOθ
O
l
θ
θ+
cg
()
→ θ ( t ) + sinθ ( t ) = 0
mg
t = ωt
1
sinθ = θ − θ 3 +
6
1
→ θ +θ − θ3 +
6
let θ = θmax x ( t )
x ( t;ε )
→
x+x−
x0 (T0 ,T1 ) + ε x1 (T0 ,T1 )
()
mgl
sinθ = θ t + ω 2 sinθ t = 0
IO
2
θmax
6
x +
3
=0
x + x −εx = 0
3
ε=
(
2
θmax
6
)
→ D02 x0 + x0 + ε D02 x1 + x1 + 2D0D1x0 − x03 = 0
D02 x0 + x0 = 0
and D02 x1 + x1 = −2D0D1x0 + x03
free vibration
D02 x0 + x0 = 0
hinge
D02 x1 + x1 = −2D0D1x0 + x03
O
x0 = A eiT0 + cc
l
θ
cg
(
D02 x1 + x1 = −2iD1A eiT0 + cc + A eiT0 + A e− iT0
)
3
mg
(
D02 x1 + x1 = −2iD1A eiT0 + cc + A eiT0 + A e − iT0
−2iD1A + 3 A2 A = 0
a′ = 0
→
→
A=
a = a constant,
⎛
3
)
3
1 iβ
ae
2
→
aβ ′ +
3 3
a =0
8
= −2iD1A eiT0 + A3 ei 3T0 + 3 A2 A eiT0 + cc
− i (a′ + iaβ ′) +
→
3 3
a =0
8
3
8
β ′ = − a2
→
3
8
β = − a 2T1 + φ
⎞
1 i ⎜⎝T0 − 8 a T1 +φ ⎟⎠
3
⎛
⎞
⎛ 3
⎞
x0 = A e + cc = a e
+ cc = a cos ⎜T0 − a 2T1 + φ ⎟ = a cos ⎜ t − a 2ε t + φ ⎟
2
8
⎝
⎠
⎝ 8
⎠
2
iT0
θ
2
2
⎡⎛ θmax
⎛ 3 a 2θmax
⎞
t + φ ⎟ = θmax cos ⎢⎜ 1 −
θmax x0 = aθmax cos ⎜ t −
8
6
16
⎝
⎠
⎣⎢⎝
⎤
⎞
t
φ
+
⎥ where we put a = 1
⎟
⎠
⎦⎥
θ
2
⎡⎛ θmax
θmax cos ⎢⎜ 1 −
16
⎢⎣⎝
⎤
⎞
t
φ
+
⎥
⎟
⎥⎦
⎠
is the same as the approximation to the
exact solution of the exact EoM
forced vibration
x + 2ε x + x + α x 3 = 2ε f cos(Ωt ) = ε f ei Ωt + cc
x
x0 (T0 ,T1 ) + ε x1 (T0 ,T1 )
D02 x0 + x0 = 0
→
(
)
D02 x0 + x0 + ε D02 x1 + x1 + 2D0D1x0 + 2D0 x0 + α x03 = ε f ei ΩT0 + cc
→
x0 = A (T1 ) eiT0 + cc
D02 x1 + x1 = −2D0D1x0 − 2D0 x0 − x03 + f ei ΩT0 + cc
(
= −2i ( A′ + A ) eiT0 + cc − α A eiT0 + A e− iT0
(
= −2i ( A′ + A ) eiT0 − α A3 ei 3T0 + 3 A2 A eiT0
) +f e
)+f e
3
i ΩT0
i ΩT0
+ cc
+ cc
elimnate troblesome terms
−2i ( A′ + A ) − 3α A2 A + f ei ( Ω−1)T0 = 0
→
A=
1 iβ
a e & Ω − 1 = εσ
2
γ = σ T1 − β
→
next page
from the previous page:
3
3
−i ( a′ + iaβ ′ + a ) ei β − α a3 ei β + f eiσT1 = 0 → − i ( a′ + iaβ ′ + a ) − α a3 + f eiγ = 0
8
8
3
3
f
a′ + a = f sin γ & aβ ′ − α a3 = −f cos γ → γ ′ = σ − α a 2 + cos γ
8
8
a
γ = a constant
steady state:
a = a constant
a = f sin γ
3
aσ − α a3 = −f cos γ
8
&
2
⎡
⎤
3
⎛
⎞
a 2 ⎢1 + ⎜ σ − α a 2 ⎟ ⎥ = f 2
8
⎠ ⎥⎦
⎢⎣ ⎝
3 2
f2
→ σ = αa ±
−1
8
a2
→
γ ′ = σ − β′ = 0
comparison of the responses of a linear EOM and a Duffing equation
for the nonlinear EOM, there can be more than one response for the same
excitation and correspondingly a jump phenomenon can exist
can we verify that this is true?
x + 2 μ x + x + α x3 = k cos ( Ωt )
jumps and hysteresis in the resonant responses obtained
from a Duffing’s equation (see beneath the graphs)
frequency increasing
frequency decreasing
jump
down
unstable
branch
unstable
jump up
jump
down
jump up
x + 2 μ x + x + α x3 = k cos ( Ωt )
excitation increasing
excitation decreasing
note the amplitude of the response barely changes at first, but suddenly
there is a big change: the amplitude of the response changes by
approximately a factor of three – there’s a jump
all results obtained by using ode45 in MATLAB
x + 2 μ x + x + α x3 = k cos ( Ωt )
when the amplitude of the response starts to decrease the amplitude of
the response remains quite large
x + 2 μ x + x + α x3 = k cos ( Ωt )
different responses at the same frequency and amplitude of the
excitation – holy moly, two responses at the same excitation
what going on?
x + 2 μ x + x + α x3 = k cos ( Ωt )
basins of attraction
jump up
excitation increasing
excitation decreasing
jump
down
γ
a′ = −a + f sin γ
3
aβ ′ − α a3 = −f cos γ
8
γ = σ T1 − β → β ′ = σ − γ ′
3
8
f
a
γ ′ = σ − α a 2 + cos γ
a
how do we determine stability?