analytical mechanics ( am ) - kvischolten/aam/aam-oscill.pdf · november 9, 2014 advanced...
TRANSCRIPT
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
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-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 200922:02:50
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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-2
0
2
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Poincare
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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
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Poincare
-2-10123
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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]