analytical mechanics ( am ) - kvischolten/aam/aam-oscill.pdf · november 9, 2014 advanced...

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Analytical Mechanics ( AM ) Olaf Scholten KVI, kamer v3.008; tel. nr. 363-3552; email: [email protected] Web page: http://www.kvi.nl/˜scholten Book: Classical Dynamics of Particles and Systems, Stephen T. Thornton & Jerry B. Marion ISBN10: 0-495-55610-6 — ISBN13: 978-0-495-55610-7 Content: - Numerical methods for solving physics problems is stressed. The subjects that are being addressed include: - Damped and driven oscillators; non-linear effects (higher harmonics, hysteresis and chaos); Fourier transforms and Greens - Variational calculus; Lagrange multipliers; Hamilton and Lagrange formalism, Lagrange density for continuous systems - Central potentials, Kepler problem, - Rotating rigid body, non-inertial systems, inertial tensors, Euler angles. Generalizations to problems in Quantum Mechanics, Relativistic Mechanics, Field Theory, ......

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Analytical Mechanics( AM )

Olaf ScholtenKVI, kamer v3.008; tel. nr. 363-3552; email: [email protected] page: http://www.kvi.nl/˜scholten

Book:

Classical Dynamics of Particles and Systems,Stephen T. Thornton & Jerry B. MarionISBN10: 0-495-55610-6 — ISBN13: 978-0-495-55610-7

Content:

- Numerical methods for solving physics problems is stressed.The subjects that are being addressed include:- Damped and driven oscillators; non-linear effects (higher harmonics, hysteresisand chaos); Fourier transforms and Greens- Variational calculus; Lagrange multipliers; Hamilton and Lagrange formalism,Lagrange density for continuous systems- Central potentials, Kepler problem,- Rotating rigid body, non-inertial systems, inertial tensors, Euler angles.Generalizations to problems in Quantum Mechanics, Relativistic Mechanics, FieldTheory, ......

November 9, 2014 Advanced Analytical Mechanics (1a)-2 - Prelim.1 -

Preliminaries

website: //www.kvi.nl/∼scholten/AAM/college.htm

Homeworks are an integral part ofthe course!Language English?

Read through slides BEFORE lectureI may give flash exams (pop-quiz) during class

Assumed prior knowledge:– Mechanics-1 & 2,– Complex analysis,– Introduction to Programming and Numerical Methods

November 9, 2014 Advanced Analytical Mechanics (1a)-3 - Prelim.2 -

Mechanics-2

from website: Mechanics-2 Ocasys:

Upon completion of Mechanics-2, the stu-dent:

- is able to describe the three-dimensional motion of a rigid body as a translationof and a rotation about the center of mass;- knows the definition of the moment of inertia tensor and is able to compute theprincipal values of this tensor for sets of discrete particles and for simple contin-uous bodies;- is able to apply conservation of angular momentum for bodies rotating in twoand in three dimensions (including the effect of an external angular impulse);- is able to derive the complete equations of motion, i.e. involving translationand rotation, of rigid bodies subject to external forces and torques in two dimen-sions;- can explain precession of spinning objects and compute the precession fre-quency;- is able to identify the degrees of freedom of a mechanical system and to for-mulate its Lagrangian when subjected to conservative forces;- can apply constraint potentials in order to incorporate constraints in the La-grangian;- is able to show that the conservation laws of energy, momentum and angularmomentum are contained in the Lagrangian;- is able to derive the equations of motion for coupled linear oscillators, anddetermine eigenfrequency and eigenmodes for systems up to two degrees offreedom.

November 9, 2014 Advanced Analytical Mechanics (1a)-4 - Oscil.1 -

Harmonic Oscillatorsparticle, mass m, subject to force F (x)In general: F (x) = F0 + xF1 +

12x

2F2 + · · ·Equilibrium point: F (x0) = 0 choose here x0 = 0Good approximation near equilibrium: F (x) = xF1 = −x kEquation of motion (eom) from F = ma

−k x = mx

Solution:

x(t) = A sin(ωt− δ)with ω2 = k/m and A and δ determined by initial conditionsx(t = 0) and x(t = 0).Alternative:

x(t) = A+eiωt +A−e

−iωt

ω: Angular frequency (hoekfrequentie) [radians/second]ν: Frequency = ω/2π oscillations per secondτ : Oscillation time = 1/ν

Example: Sphere (radius R, mass M) attached by point on its surface, makingsmall oscillationsEquation of motion: N = IθTorque: N = R× F = −RMgθmoment of inertia: I = 2

5MR2 +MR2

7

5MR2θ = −RMgθ

ω2 = 57g/R

[prblm 3.1,3,4,5,6,7,8,9]

November 9, 2014 Advanced Analytical Mechanics (1a)-5 - Oscil.2 -

2D Harmonic Oscillator

Motion in 2 directions around equilibriumFx = −k x and Fy = −k yCoupled equations of motion

mx+ k x = 0

my + k y = 0

Solution, ω2 = k/m: (2 integration constants each)

x(t) = A cos(ωt− α)y(t) = B cos(ωt− β)

special cases: α = β and α− β = π/2

More general Fx = −kxx and Fy = −kyyCoupled equations of motion

mx+ kxx = 0

my + kyy = 0

Solution:

x(t) = A cos(ωxt− α)y(t) = B cos(ωyt− β)

Lissajous figures

[Fig. 3.3]

November 9, 2014 Advanced Analytical Mechanics (1a)-6 - Oscil.3 -

Phase Diagram

One dimensional oscillator

x(t) = A cos(ωt− δ)

Here we consider position as function of timeInitial conditions: value of x and x at a particular time

Consider x and x two coordinates (in phase space) specifying position and ve-locity (motion) of particle at a particular time.Curve given by functional p(x, x) = 0

Question: at which points (x, x) in phase space can the particle be found?Answer (directly from the solution):

x2/A2 + x2/(ω2A2) = 1

Moves with time in clock-wise direction on circle

Manipulating the eq. of motion x+ ω2x = 0gives d

dt x = −ω2x thus dxdx = d

dt x×dtdx = −ω2 x

xgiving x dx = −ω2x dx, or x2 = −ω2x2 + C

Poincare section:Point in phase diagram at times tnω = n 2π

To Do:- write computer program to numerically solve eom of oscillator - make plots ofx(t) and x(x)- check different initial conditions

November 9, 2014 Advanced Analytical Mechanics (1a)-7 - Oscil.4 -

Computer simulations-0

Solving x = −ω2x numerically:x(t) = [x(t+ δ/2)− x(t− δ/2)]/δ and thus

x(t) =([x(t+ δ)− x(t)]/δ − [x(t)− x(t− δ)]/δ

)/δ or

x(t) =(x(t+ δ)− 2x(t) + x(t− δ)

)/δ2 and thus

x(t+ δ)− 2x(t) + x(t− δ) = −δ2ω2x(t)

giving

x(t+ δ)︸ ︷︷ ︸New point

= (2− δ2ω2)x(t)− x(t− δ)︸ ︷︷ ︸previous points

Initial conditions (take grid, t = n δ):x(0) = aand x(0) ≈ x(δ/2) = [x(δ)− x(0)]/δ = a givingx(δ) = a+ a δ

November 9, 2014 Advanced Analytical Mechanics (1a)-8 - Oscil.5 -

Computer simulations-0b

Solving x = −ω2x numerically:

x(t+ δ)︸ ︷︷ ︸New point

= (2− δ2ω2)x(t)− x(t− δ)︸ ︷︷ ︸previous points

Initial conditions (take grid, t = n δ):x(0) = a; x(0) = a

Use http://www.onlinecompiler.net/fortran.html or GFORTRAN or whatever

program ho! Solving a Harmonic oscillator

integer ireal x(0:100),del,omgaprint *,"Hello World, give me delta and omega"read *, del,omgawrite(9,*) del,omgax(0)=1. ; x(1)=1.do i=1,99

x(i+1)=(2 - del**2 * omga**2) * x(i) - x(i-1)write(9,*) i+1 , x(i+1)

enddoprint *,"end ho",del,omgaend program ho

November 9, 2014 Advanced Analytical Mechanics (1a)-9 - Oscil.6 -

Computer simulations-1

Simple harmonic oscillator

x+ 2βx+ ω2rx = 0

parameters ωr = 2, β = 0.

-3 -2 -1 0 1 2 3-4

-2

0

2

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Phase

x

(dx/

dt)/

-3 -2 -1 0 1 2 3-4

-2

0

2

..........................

Poincare

-2-10123

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time [0.01 sec]

ampl

itude

NL-pendulum

Sun Mar 29 200922:02:50

Initial conditions at t = 0: x = 2 and x = 0.

Q: take Initial conditions at t = 0: x = 0 and x = 2ω

November 9, 2014 Advanced Analytical Mechanics (1a)-10 - Oscil.7 -

Damped Oscillator

[chapter 3.5]Eq. of motion (eom):

x+ 2βx+ ω2rx = 0

trial solution: x(t) = Aeγt givesγ2 + 2βγ + ω2

r = 0Solutions: γ = −β ±

√β2 − ω2

r

when ω2r ≥ β2 define ω1 =

√ω2r − β2

Solution now: γ = −β ± i ω1

x(t) = e−βt[Aeiω1t +B e−iω1t

]Numerically:x(t) = [x(t+ δ)− x(t− δ)]/(2 δ) and thus

x(t+ δ)− 2x(t) + x(t− δ) =−δ2ω2x(t)− δ β [x(t+ δ)− x(t− δ)]

giving

x(t+ δ) =(2− δ2ω2)x(t) + (−1 + δ β)x(t− δ)

1 + δ β

To Do:[prblm 3.2,10,11,12,13,14,16]

November 9, 2014 Advanced Analytical Mechanics (1a)-11 - Oscil.8 -

Computer simulations-2a

Damped harmonic oscillator

x+ 2βx+ ω2rx = 0

parameters ωr = 2, β = 0.05

-3 -2 -1 0 1 2 3-4

-2

0

2

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Phase

x

(dx/

dt)/

-3 -2 -1 0 1 2 3-4

-2

0

2

..........................

Poincare

-2-10123

0 1000 2000 3000 4000

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4000 5000 6000 7000 8000-3-2-10123

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time [0.01 sec]

ampl

itude

NL-pendulum

Sun Mar 29 200922:05:04

Initial conditions at t = 0: x = 2 and x = 0.

Q: Take different initial conditions[prblm 3.17,18,21,22,23]

November 9, 2014 Advanced Analytical Mechanics (1a)-12 - Oscil.9 -

Computer simulations-2b

Damped harmonic oscillator

x+ 2βx+ ω2rx = 0

parameters ωr = 2, β = 2.00

-3 -2 -1 0 1 2 3-4

-2

0

2

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Phase

x

(dx/

dt)/

-3 -2 -1 0 1 2 3-4

-2

0

2

..........................

Poincare

-2-10123

0 1000 2000 3000 4000

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time [0.01 sec]

ampl

itude

NL-pendulum

Sun Mar 29 200922:16:45

Initial conditions at t = 0: x = 2 and x = 0.

Q: Change friction force

November 9, 2014 Advanced Analytical Mechanics (1a)-13 - Oscil.10 -

Driven & Damped Oscillator

[chapter 3.6]Driving force F (t) = mA0 cos(ωt)Eq. of motion (eom):

x+ 2βx+ ω2rx = A0 cos(ωt)

Same as real part of

x+ 2βx+ ω2rx = A0e

iωt

trial solution: x(t) = Aeiωt (why same frequency?) gives−Aω2 + 2i Aβω +Aω2

r = A0

Special solution:

A =A0

(ω2r − ω2) + 2iβω

= |A|e−iδ

with tan δ = −Im(A)/Re(A) = 2βω/(ω2r − ω2)

|A| = |A0|/√(ω2r − ω2)2 + 4β2ω2 Special solution now:

x(t) = |A|<(e−iδeiωt

)= |A| cos(ωt− δ)

Full sol. = Special sol. + sol. homogeneous eq.To Do:- check importance real & imaginary parts

[prblm 3.24,25]

November 9, 2014 Advanced Analytical Mechanics (1a)-14 - Oscil.11 -

Computer simulations-3

Damped & Driven harmonic oscillator

x+ 2βx+ ω2rx = A0 cos(ωdt)

parameters ωr = 2, β = 0.1, A0 = 2, ωd = 2.3

-3 -2 -1 0 1 2 3-4

-2

0

2

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Phase

x

(dx/

dt)/

-3 -2 -1 0 1 2 3-4

-2

0

2

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Poincare

-2-10123

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time [0.01 sec]

ampl

itude

NL-pendulum

Sun Mar 29 200922:33:11

Initial conditions at t = 0: x = 2 and x = 0.

Q: Take different initial conditions;Change driving frequency;Change friction force.

November 9, 2014 Advanced Analytical Mechanics (1a)-15 - Oscil.12 -

Non-linear Oscillator

[chapter 4.2]

Spring constant k, unextended length l0mass m then

Fx = −2k(s− l0) sin θ

with s =√l2 + x2 and sin θ = x/s

Problem 4.1: for d = l − l0 we have

Fx = −2k dxl− k l0

x

l

3+ · · ·

LL

SS

x

eom including driving force F0 cosωt:

mx+ 2k dx

l+ k l0

x

l

3− F0 cosωt = 0

Substitute y = x/l, ω20 = 2k d

m l , ε = k l0ml , g = F0

ml giving

y = −ω20y − εy3 + g cosωt

November 9, 2014 Advanced Analytical Mechanics (1a)-16 - Oscil.13 -

x = −ω20x− εx3 + g cosωt

Interested in the effect of the x3 term on the motion.Assume that this correction is small and that we can write

x(ε; t) = x(0)(t) + εx(1)(t) + ε2x(2)(t) + · · ·

explain! Substitute in eom.

Collect terms proportional to ε0

x(0) = −ω20x

(0) + g cosωt

gives x(0)(t) = g/(ω20 − ω2) cosωt ≡ A(0) cosωt

Collect terms proportional to ε1

x(1) = −ω20x

(1) − [x(0)]3

use cos3 ωt = 34 cosωt+

14 cos 3ωt

gives x(1)(t) = A(1) cosωt+B(1) cos 3ωtwith A(1) = − 3

4 [A(0)]3/(ω2

0 − ω2)

and B(1) = − 14 [A

(0)]3/(ω20 − 9ω2)

Conclusion:Due to x3 term solution has frequencies 3ω and higher.

[prblm 4.1,2,3,4,5,6,7,8]

November 9, 2014 Advanced Analytical Mechanics (1a)-17 - Oscil.14 -

Computer simulations-4

Damped & Driven Anharmonic Oscillator

x+ 2βx+ ω2rx+ ω2

rC x3 = A0 cos(ωdt)

parameters ωr = 2, β = 0.1, A0 = 2, ωd = 0.8, C = 1.3

-3 -2 -1 0 1 2 3-4

-2

0

2

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.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

Phase

x

(dx/

dt)/

-3 -2 -1 0 1 2 3-4

-2

0

2

.. .........

Poincare

-2-10123

0 1000 2000 3000 4000

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time [0.01 sec]

ampl

itude

NL-pendulumThu Apr 29 2010

08:59:03

Initial conditions at t = 0: x = 2 and x = 0.

Q: what happens if sign C is reversedand if sign ω2

r is reversed?

November 9, 2014 Advanced Analytical Mechanics (1a)-18 - Oscil.15 -

Hysteresis

[4.5]Simple estimate by assuming 2 linear regimes with different spring constants

[prblm 4.9,10]

Computer simulations-5

Oscillator with memory effect

x+ 2βx+ ω2rx+ ω2

rC x3 = A0 cos(ωdt)

parameters ωr = 2, β = 0.1, A0 = 1.8, ωd = 0.8, C = 1.1

-2-10123

0 1 2 3 4 5 6

.. . . . . . . . . . . . . . . . . . . .. . . . . . .

. .

. . . . . . . . . . . ............................

...............

Tune upTune down

0 1 2 3 4 5 60

1

2

3

4

.. . . . . . . . .. . . . . .

. . . . . .. . . . . .

. .

. . . . . . . . . . . ............................

............... x3

oscil

Frequency

Max

Am

plPh

ase

pendulum

Tue Mar 31 200923:19:23

Initial conditions at slowly tune up or down frequency

Q: try different non-linearities and strength friction force

November 9, 2014 Advanced Analytical Mechanics (1a)-19 - Oscil.16 -

Chaos

Computer simulations-6

Driven and damped non-linear Oscillator inverted linear part

x+ 2βx+ ω2rx+ ω2

rC x3 = A0 cos(ωdt)

parameters ω2r = −4, β = 0.1, A0 = 4., ωd = 4.3, C = −1.1

-3 -2 -1 0 1 2 3-4

-2

0

2

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Phase

x

(dx/

dt)/

-3 -2 -1 0 1 2 3-4

-2

0

2

.... ...

...

.. ...

.. ...

.. ... .. ... .. .

..

..

.. ..

.. .. .. ... .. ...

Poincare

-2-10123

0 1000 2000 3000 4000

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4000 5000 6000 7000 8000-3-2-10123

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time [0.01 sec]

ampl

itude

NL-pendulumWed Apr 01 2009

23:00:02

Q: try different non-linearities and strength friction force

November 9, 2014 Advanced Analytical Mechanics (1a)-20 - Oscil.17 -

In the coming lectures:

Fourier transform

Solve damped & driven oscillator for a general periodic force[3.8][prblm 3.27,28,29,39,40]

Greens function

Solve damped & driven oscillator for an impulsive force[3.9][prblm 3.31,32,33,34,35,38]

other: [prblm 3.42,43,44,45]

November 9, 2014 Advanced Analytical Mechanics (1a)-21 - Oscil.18 -

Useful

sinα =eiα − e−iα

2i

cosα =eiα + e−iα

2

sinα+ β = cosα sinβ + sinα cosβ

cosα+ β = cosα cosβ − sinα sinβ

sinα ≈ α

cosα ≈ 1− 12α

2