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MATLAB代写-M5

时间：2021-04-04

Dynamics M5

Method of Multiple Scales: Parametric Oscillator

Andrea Cammarano

University of Glasgow

Room 427A, James Watt (South) Building

andrea.cammarano@glasgow.ac.uk

Andrea Cammarano (UG) Dynamics M5 1 / 18

Parametric Oscillator Equation

In the first lesson we have seen the equation of motion for a pendulum

with the constraint that moves in the vertical direction.

After a few calculation we obtained the following equation:

φ¨+

(

g

l

− d¨(t)

l

)

sinφ = 0, (1)

Andrea Cammarano (UG) Dynamics M5 2 / 18

Parametric Oscillator Equation

We highlighted, in the first lesson, that this equation belongs to a

particular set of equations called Matthieu-Hill Equations and that being

the state variable (or a function of the state variable) multiplied by the

forcing term, we generally call the forcing a parametric excitation. If we

assume that the movement of the constrain follows an harmonic law and

that all the energy losses in the system are captured by the viscous

damping, equation 1 becomes:

φ¨+ 2ζωnφ˙+

(

g

l

− d0Ω

2

l

cos(Ω t)

)

sinφ = 0. (2)

Andrea Cammarano (UG) Dynamics M5 3 / 18

Reducing the EOM to Ordered form

If we expand sinφ to its Taylor series truncated to the 2nd term

sinφ = φ− 1

3!

φ3 and we substitute in equation 3 we have

φ¨+ 2ζωnφ˙+

g

l

φ− g

6l

φ3 +

d0Ω

2

l

cos(Ω t)φ− d0Ω

2

6l

cos(Ω t)φ3 = 0. (3)

Note that in this form we have both linear parametric excitation terms

and nonlinear parametric excitation terms! The nonlinear parametric

excitation terms will be neglected here, since the principal resonance of

this equation occurs in a neighbourhood of 2ωn, where the nonlinear

parametric excitation terms are non resonant.

Andrea Cammarano (UG) Dynamics M5 4 / 18

Reducing the EOM to Ordered form

Ad mentioned this equation will be satudied for

Ω = 2ωn + εσ (4)

This resonance condition is similar to the condition introduced in the

previous example, but the central frequency is 2ωn rather than ωn. This

condition is called principal parametric resonance.

We now consider the following assumption

g

l

= ω2n (5)

g

6l

= εh˜ (6)

d0Ω

2

l

= εf˜ (7)

ζ = εζ˜ (8)

Andrea Cammarano (UG) Dynamics M5 5 / 18

Reducing the EOM to Ordered form

Substituting in equation 3 we have

φ¨+ 2εζ˜ωnφ˙+ ω

2

nφ− εh˜φ3 + εf˜ cos(Ω t)φ = 0. (9)

Now we introduce the MMS series for φ,

d

dt

,

d2

dt2

, truncating the series after

ε, we generate the zerosth and first order perturbation equations (we will

skip the details in this case, but you should have a good understanding of

the methodology at this point):

ε0 : D20φ0 + ω

2

nφ0 = 0 (10)

ε1 : D20φ1 + ω

2

nφ1 = −2D0D1φ0 − 2ζ˜ωnD0φ0 + h˜φ30 − f˜ cos(Ωt)φ0 (11)

Andrea Cammarano (UG) Dynamics M5 6 / 18

Applying the Perturbation Expansions

Note that there is no difference in the ε0 equation of the horizontal and

vertical case

V : D20φ0 + ω

2

nφ0 = 0

H : D20φ0 + ω

2

nφ0 = 0

whereas there are minimal difference in the ε1 equation

V : D20φ1 + ω

2

nφ1 = −2D0D1φ0 − 2ζ˜ωnD0φ0 + h˜φ30−f˜ cos(Ωt)φ0

H : D20φ1 + ω

2

nφ1 = −2D0D1φ0 − 2ζ˜ωnD0φ0 + h˜φ30 + f˜ cos(Ωt)

Consequently we could use most of the derivation used in the previous

example to obtain the solution for this case!

Andrea Cammarano (UG) Dynamics M5 7 / 18

Zeroth and First Order Perturbation Equation

Being the zeroth order perturbation equation identical to the previous

case, obviously the solution has the same form

φ0 = a(T1) cos(ωnT0 + α(T1))

= A(T1)e

iωnT0 + A¯(T1)e

−iωnT0 (12)

Substituting into the first order perturbation equation, as well as

applying differentiations with respect with T0, expanding terms and taking

out a common factor of eiωnT0 , we obtain:

D20φ1 + ω

2

nφ1 = e

iωnT0

{−2iωnD1A+ 2iωnD1A¯e−2iωnT0 (13)

− 2iζ˜ω2nA+ 2iζ˜ω2nA¯e−2iωnT0 + h˜A3ei2ωnT0 + h˜A¯3e−i4ωnT0

+ 3h˜A2A¯+ 3h˜A¯2Ae−2iωnT0−Af˜

2

eiΩT0 − A¯ f˜

2

ei(Ω−2ωn)T0

−Af˜

2

e−iΩT0 − A¯ f˜

2

e−i(Ω+2ωn)T0

}

Andrea Cammarano (UG) Dynamics M5 8 / 18

Deriving Secular Term Equations from First Order

Perturbation Equation

We use the resonance condition to get the secular terms equations

(including the complex conjugates).The resonance condition we are

expecting here is principal parametric resonance, i.e. Ω = 2ωn + εσ.

This is the condition for the forcing terms to become “almost” secular:

A¯

f˜

2

ei(Ω+2ωn)T0 → A¯ f˜

2

eiεσT0 → A¯ f˜

2

eiσT1 (14)

Therefore, the secular terms equation and its complex conjugate variant

are, respectively:

− 2iωnD1A− 2iζ˜ω2nA+ 3h˜A2A¯− A¯

f˜

2

eiσT1 = 0 (15)

2iωnD1A¯+ 2iζ˜ω

2

nA¯+ 3h˜AA¯

2 −Af˜

2

e−iσT1 = 0 (16)

Andrea Cammarano (UG) Dynamics M5 9 / 18

Deriving Steady-State Amplitude Equation from Secular

Terms Equations

Now, reusing the usual forms for the complex amplitude, A(T1) =

a

2

eiα

and A¯(T1) =

a

2

e−iα, and then differentiating with respect to T1 and

tidying up we obtain:

−iωna′ + ωnaα′ − iζ˜ω2na+

3

8

h˜a3 − af˜

4

ei(εσT0−2α) = 0 (17)

Separating into real and imaginary parts:

< : ωnaα′ + 3

8

h˜a3 − af˜

4

cos (σT1 − 2α) = 0 (18)

= : − ωna′ − ζ˜ω2na− a

f˜

4

sin (σT1 − 2α) = 0 (19)

Andrea Cammarano (UG) Dynamics M5 10 / 18

Deriving Steady-State Amplitude Equation from Secular

Terms Equations

In order to exploit the very slow varying dynamics (due to the amplitude

a and the phase α both being functions of time-scale T1), we transform to

an autonomous (time-independent) system via the substitution:

σT1 − 2α = Ψ (20)

Then, we impose the condition that a′ = Ψ′ = 0. After differentiating with

respect to T1, we get:

σ = 2α′ ⇔ α′ = σ

2

, (21)

which leads to

1

2

ωnaσ +

3

8

h˜a3 − af˜

4

cos Ψ = 0 (22)

ζ˜ω2na− a

f˜

4

sin Ψ = 0 (23)

Andrea Cammarano (UG) Dynamics M5 11 / 18

Deriving Steady-State Amplitude Equation from Secular

Terms Equations

By cancelling a throughout these equations, we find that they become:

f˜

4

cos Ψ =

1

2

ωnσ +

3

8

h˜a2 (24)

f˜

4

sin Ψ = −ζ˜ω2n (25)

Squaring both and adding them gives:

9

64

h˜2a4 +

3

8

h˜ωnσa

2 +

1

4

ω2nσ

2 + ζ˜2ω4n =

1

16

f˜2 (26)

From which we can deduce the steady-state amplitude equation:

a =

√√√√−4

3h˜

ωnσ ± 8

3h˜

√

f˜2

16

− ω4nζ˜2 (27)

Andrea Cammarano (UG) Dynamics M5 12 / 18

Deriving Steady-State Amplitude Equation from Secular

Terms Equations

We now return to physical parameters by multiplying numerators and

denominators on the right side by ε:

a =

√√√√ −4

3(εh˜)

ωn(εσ)± 8

3(εh˜)

√

(εf˜)2

16

− ω4n(εζ˜)2. (28)

We obtain

a =

√

−4

3h

ωn(Ω− 2ωn)± 8

3h

√

d20Ω

4

16l2

− ω4nζ2 (29)

Andrea Cammarano (UG) Dynamics M5 13 / 18

Numerical Example and Stability Analysis

We can obtain a plot of a against εσ. Remembering the resonance

condition defines εσ as follows: Ω = 2ωn + εσ or εσ = Ω− 2ωn+. In

other words, εσ is the detuning parameter which is a measure of the

difference between the excitation frequency and the principal

parametric resonance frequency, 2ωn. If εσ = 0, then Ω = 2ωn and we

have perfect parametric resonance. Anything else is detuned from this

condition. For the practical example a length l = 0.1 m is considered. The

constants in the EOM are as follow:

ζ =0.001

d0Ω

2

l

=0.5s−2√

g

l

=9.9045rad/s

g

6l

=16.35s−2

Andrea Cammarano (UG) Dynamics M5 14 / 18

Numerical Example and Stability Analysis

−0.5 −0.4 −0.3 −0.2 −0.1 0 0.1 0.2

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Andrea Cammarano (UG) Dynamics M5 15 / 18

Numerical Example and Stability Analysis

−0.5 −0.4 −0.3 −0.2 −0.1 0 0.1 0.2

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Andrea Cammarano (UG) Dynamics M5 16 / 18

Numerical Example and Stability Analysis

Andrea Cammarano (UG) Dynamics M5 17 / 18

Numerical Example and Stability Analysis

Andrea Cammarano (UG) Dynamics M5 18 / 18

学霸联盟

Method of Multiple Scales: Parametric Oscillator

Andrea Cammarano

University of Glasgow

Room 427A, James Watt (South) Building

andrea.cammarano@glasgow.ac.uk

Andrea Cammarano (UG) Dynamics M5 1 / 18

Parametric Oscillator Equation

In the first lesson we have seen the equation of motion for a pendulum

with the constraint that moves in the vertical direction.

After a few calculation we obtained the following equation:

φ¨+

(

g

l

− d¨(t)

l

)

sinφ = 0, (1)

Andrea Cammarano (UG) Dynamics M5 2 / 18

Parametric Oscillator Equation

We highlighted, in the first lesson, that this equation belongs to a

particular set of equations called Matthieu-Hill Equations and that being

the state variable (or a function of the state variable) multiplied by the

forcing term, we generally call the forcing a parametric excitation. If we

assume that the movement of the constrain follows an harmonic law and

that all the energy losses in the system are captured by the viscous

damping, equation 1 becomes:

φ¨+ 2ζωnφ˙+

(

g

l

− d0Ω

2

l

cos(Ω t)

)

sinφ = 0. (2)

Andrea Cammarano (UG) Dynamics M5 3 / 18

Reducing the EOM to Ordered form

If we expand sinφ to its Taylor series truncated to the 2nd term

sinφ = φ− 1

3!

φ3 and we substitute in equation 3 we have

φ¨+ 2ζωnφ˙+

g

l

φ− g

6l

φ3 +

d0Ω

2

l

cos(Ω t)φ− d0Ω

2

6l

cos(Ω t)φ3 = 0. (3)

Note that in this form we have both linear parametric excitation terms

and nonlinear parametric excitation terms! The nonlinear parametric

excitation terms will be neglected here, since the principal resonance of

this equation occurs in a neighbourhood of 2ωn, where the nonlinear

parametric excitation terms are non resonant.

Andrea Cammarano (UG) Dynamics M5 4 / 18

Reducing the EOM to Ordered form

Ad mentioned this equation will be satudied for

Ω = 2ωn + εσ (4)

This resonance condition is similar to the condition introduced in the

previous example, but the central frequency is 2ωn rather than ωn. This

condition is called principal parametric resonance.

We now consider the following assumption

g

l

= ω2n (5)

g

6l

= εh˜ (6)

d0Ω

2

l

= εf˜ (7)

ζ = εζ˜ (8)

Andrea Cammarano (UG) Dynamics M5 5 / 18

Reducing the EOM to Ordered form

Substituting in equation 3 we have

φ¨+ 2εζ˜ωnφ˙+ ω

2

nφ− εh˜φ3 + εf˜ cos(Ω t)φ = 0. (9)

Now we introduce the MMS series for φ,

d

dt

,

d2

dt2

, truncating the series after

ε, we generate the zerosth and first order perturbation equations (we will

skip the details in this case, but you should have a good understanding of

the methodology at this point):

ε0 : D20φ0 + ω

2

nφ0 = 0 (10)

ε1 : D20φ1 + ω

2

nφ1 = −2D0D1φ0 − 2ζ˜ωnD0φ0 + h˜φ30 − f˜ cos(Ωt)φ0 (11)

Andrea Cammarano (UG) Dynamics M5 6 / 18

Applying the Perturbation Expansions

Note that there is no difference in the ε0 equation of the horizontal and

vertical case

V : D20φ0 + ω

2

nφ0 = 0

H : D20φ0 + ω

2

nφ0 = 0

whereas there are minimal difference in the ε1 equation

V : D20φ1 + ω

2

nφ1 = −2D0D1φ0 − 2ζ˜ωnD0φ0 + h˜φ30−f˜ cos(Ωt)φ0

H : D20φ1 + ω

2

nφ1 = −2D0D1φ0 − 2ζ˜ωnD0φ0 + h˜φ30 + f˜ cos(Ωt)

Consequently we could use most of the derivation used in the previous

example to obtain the solution for this case!

Andrea Cammarano (UG) Dynamics M5 7 / 18

Zeroth and First Order Perturbation Equation

Being the zeroth order perturbation equation identical to the previous

case, obviously the solution has the same form

φ0 = a(T1) cos(ωnT0 + α(T1))

= A(T1)e

iωnT0 + A¯(T1)e

−iωnT0 (12)

Substituting into the first order perturbation equation, as well as

applying differentiations with respect with T0, expanding terms and taking

out a common factor of eiωnT0 , we obtain:

D20φ1 + ω

2

nφ1 = e

iωnT0

{−2iωnD1A+ 2iωnD1A¯e−2iωnT0 (13)

− 2iζ˜ω2nA+ 2iζ˜ω2nA¯e−2iωnT0 + h˜A3ei2ωnT0 + h˜A¯3e−i4ωnT0

+ 3h˜A2A¯+ 3h˜A¯2Ae−2iωnT0−Af˜

2

eiΩT0 − A¯ f˜

2

ei(Ω−2ωn)T0

−Af˜

2

e−iΩT0 − A¯ f˜

2

e−i(Ω+2ωn)T0

}

Andrea Cammarano (UG) Dynamics M5 8 / 18

Deriving Secular Term Equations from First Order

Perturbation Equation

We use the resonance condition to get the secular terms equations

(including the complex conjugates).The resonance condition we are

expecting here is principal parametric resonance, i.e. Ω = 2ωn + εσ.

This is the condition for the forcing terms to become “almost” secular:

A¯

f˜

2

ei(Ω+2ωn)T0 → A¯ f˜

2

eiεσT0 → A¯ f˜

2

eiσT1 (14)

Therefore, the secular terms equation and its complex conjugate variant

are, respectively:

− 2iωnD1A− 2iζ˜ω2nA+ 3h˜A2A¯− A¯

f˜

2

eiσT1 = 0 (15)

2iωnD1A¯+ 2iζ˜ω

2

nA¯+ 3h˜AA¯

2 −Af˜

2

e−iσT1 = 0 (16)

Andrea Cammarano (UG) Dynamics M5 9 / 18

Deriving Steady-State Amplitude Equation from Secular

Terms Equations

Now, reusing the usual forms for the complex amplitude, A(T1) =

a

2

eiα

and A¯(T1) =

a

2

e−iα, and then differentiating with respect to T1 and

tidying up we obtain:

−iωna′ + ωnaα′ − iζ˜ω2na+

3

8

h˜a3 − af˜

4

ei(εσT0−2α) = 0 (17)

Separating into real and imaginary parts:

< : ωnaα′ + 3

8

h˜a3 − af˜

4

cos (σT1 − 2α) = 0 (18)

= : − ωna′ − ζ˜ω2na− a

f˜

4

sin (σT1 − 2α) = 0 (19)

Andrea Cammarano (UG) Dynamics M5 10 / 18

Deriving Steady-State Amplitude Equation from Secular

Terms Equations

In order to exploit the very slow varying dynamics (due to the amplitude

a and the phase α both being functions of time-scale T1), we transform to

an autonomous (time-independent) system via the substitution:

σT1 − 2α = Ψ (20)

Then, we impose the condition that a′ = Ψ′ = 0. After differentiating with

respect to T1, we get:

σ = 2α′ ⇔ α′ = σ

2

, (21)

which leads to

1

2

ωnaσ +

3

8

h˜a3 − af˜

4

cos Ψ = 0 (22)

ζ˜ω2na− a

f˜

4

sin Ψ = 0 (23)

Andrea Cammarano (UG) Dynamics M5 11 / 18

Deriving Steady-State Amplitude Equation from Secular

Terms Equations

By cancelling a throughout these equations, we find that they become:

f˜

4

cos Ψ =

1

2

ωnσ +

3

8

h˜a2 (24)

f˜

4

sin Ψ = −ζ˜ω2n (25)

Squaring both and adding them gives:

9

64

h˜2a4 +

3

8

h˜ωnσa

2 +

1

4

ω2nσ

2 + ζ˜2ω4n =

1

16

f˜2 (26)

From which we can deduce the steady-state amplitude equation:

a =

√√√√−4

3h˜

ωnσ ± 8

3h˜

√

f˜2

16

− ω4nζ˜2 (27)

Andrea Cammarano (UG) Dynamics M5 12 / 18

Deriving Steady-State Amplitude Equation from Secular

Terms Equations

We now return to physical parameters by multiplying numerators and

denominators on the right side by ε:

a =

√√√√ −4

3(εh˜)

ωn(εσ)± 8

3(εh˜)

√

(εf˜)2

16

− ω4n(εζ˜)2. (28)

We obtain

a =

√

−4

3h

ωn(Ω− 2ωn)± 8

3h

√

d20Ω

4

16l2

− ω4nζ2 (29)

Andrea Cammarano (UG) Dynamics M5 13 / 18

Numerical Example and Stability Analysis

We can obtain a plot of a against εσ. Remembering the resonance

condition defines εσ as follows: Ω = 2ωn + εσ or εσ = Ω− 2ωn+. In

other words, εσ is the detuning parameter which is a measure of the

difference between the excitation frequency and the principal

parametric resonance frequency, 2ωn. If εσ = 0, then Ω = 2ωn and we

have perfect parametric resonance. Anything else is detuned from this

condition. For the practical example a length l = 0.1 m is considered. The

constants in the EOM are as follow:

ζ =0.001

d0Ω

2

l

=0.5s−2√

g

l

=9.9045rad/s

g

6l

=16.35s−2

Andrea Cammarano (UG) Dynamics M5 14 / 18

Numerical Example and Stability Analysis

−0.5 −0.4 −0.3 −0.2 −0.1 0 0.1 0.2

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Andrea Cammarano (UG) Dynamics M5 15 / 18

Numerical Example and Stability Analysis

−0.5 −0.4 −0.3 −0.2 −0.1 0 0.1 0.2

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Andrea Cammarano (UG) Dynamics M5 16 / 18

Numerical Example and Stability Analysis

Andrea Cammarano (UG) Dynamics M5 17 / 18

Numerical Example and Stability Analysis

Andrea Cammarano (UG) Dynamics M5 18 / 18

学霸联盟