Periodically forced damped pendulum

Carmen Hoek, Thomas Dijk, Jelmer Levenga

February 10, 2019

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Contents1 History 3

2 Small angle approximation 32.1 introduction to the problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.2 Solving for small angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.3 evaluating the small-angle equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

3 Forced damped pendulum 8

4 Numerical solutions 114.1 Euler’s method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

4.1.1 The first derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114.1.2 Code for the numerical approximation . . . . . . . . . . . . . . . . . . . . . 114.1.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

5 Bibliography 15

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1 HistoryThe pendulum has been known for quite a long time, the first recorded use of a pendulum wasin the first century AD, in China. The pendulum was used as a seismometer. In the case ofan earthquake a pendulum would hit a lever which would cause a small ball to fall down, whichindicated the direction of the earthquake. [4]The first time the pendulum was described was by Galileo Galilei around 1582 [5]. He discoveredthat for a pendulum the period of a swing is independent of the mass of at the end as well as theamplitude the mass was raised to, but proportional to the square root of the length of the stringonly. [6]. This is the property of pendulums that allow them to be used in clocks as timekeepers,the period of a pendulum is set, as soon as long as the length of the rod it is attached to is known.Newton used that fact to prove that the force of gravity acting on any object is proportional to itsmass [6]. In 1747 Daniel Bernoulli updated the previously known model and calculated that forvery small angles, the period of the pendulum is not constant. [7]

2 Small angle approximation

2.1 introduction to the problemBefore tackling the full problem, looking at a simplified version will greatly help in the understand-ing of this problem, that simplest version is the undamped and not driven pendulum: a mass Mon a massless string of length l, which is attached to a frictionless pivot. The angle the pendulummakes with the vertical axis will be called θ. A diagram of the problem is given in Figure 1

Figure 1: A diagram of the undamped undriven pendulum

The pendulum is not driven, so no force, besides gravity and the tension in the string, actson it once let go. To find the motion of the mass, these two forces should be equated. The forceof gravity is given by F = −mg where g the acceleration due to gravity. The force of gravitycan be decomposed into two parts, a part tangential to the path of the ball and a componentperpendicular to it. Only the tangential part is capable of accelerating the ball. The tangentialpart is given by

−mg sin θ

The acceleration of a mass on a curved plane is given by the angular acceleration times the lengthof the rod.

a = lθ

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Using F = ma, the force associated with that acceleration is given by

F = mlθ.

Then equating the two gives−mg sin θ = mlθ.

θ +g

lsin θ = 0

and per convention gl is replaced by ω2:

θ + ω2 sin θ = 0. (1)

A possible damping force is given by friction, the magnitude of such a force scales with the (angu-lar) velocity and opposes the motion: Fdamping = −bθ, where b is a constant.

The driving force is periodic so that is given by: Fdriving = Fd cosωdt. Here Fd is a constant and gives the amplitude of the force and ωd is the frequency of the drivingforce, which in general is different from ω.

The full equation to be solved is now given as:

Mθ + bθ +Mω2 sin θ = Fd cosωdt.

dividing out the mass to clean up the equation slightly gives

θ + 2βθ + ω2 sin θ = F cosωdt.

where 2β = bM and F = Fd

M .

2.2 Solving for small anglesSolving this problem analytically is only possible for small angles, because at small angles sin θ ≈ θand the equation becomes

θ + 2βθ + ω2θ = F cosωdt. (2)

This can be rewritten in a slightly more usable form using the Euler identity:

θ + 2βθ + ω2θ =F

2(eiωdt + e−iωdt). (3)

This is a linear differential equation, thus the general solution will be the sum of the homogeneousand particular solution.

x+ 2bx+ ax = 0 (4)

The solution can be found by making an Ansatz, that

xh(t) = Aeαt.

substituting that into equation 4 gives

Aα2 + 2bα+ a = 0.

This is a quadratic equation in α with two roots: α1 and α2

α = b±√b2 − a

These solutions can be added using the superposition principle:

xh(t) = Aeα1t +Beα2t.

This simplifies toxh(t) = e−bt(Ae

√b2−at +Be−

√b2−at). (5)

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and if we defineΩ =

√b2 − at,

xh(t) = e−bt(AeΩt +Be−Ωt). (6)

Where A and B are constants, given by the initial conditions. Substituting this back into equation4 solves the equation so the Ansatz made was a good one. For the pendulum the homogeneoussolution is given by

θh(t) = e−βt(Ae√β2−ω2t +Be−

√β2−ω2t). (7)

The particular solution still has to be found. To solve an equation of the form

x+ 2bx+ ax = C0eiω0t. (8)

where a,b and C0 are constants, an Ansatz is made:

x(t) = Aeiω0t (9)

Substituting that into equation 8 this becomes:

−ω20eiω0t + 2biω0Ae

iω0t + aAeiω0t = C0eiω0t.

− ω20 + 2biω0A+ aA = C0 (10)

Solving for A gives:

A =C0

−ω20 + 2biω0 + a

.

x(t) =C0

−ω20 + 2biω0 + a

eiω0t. (11)

Substituting this into equation 8 solves this equation so the initial Ansatz has been a good one.Applying this to equation 3 is fairly straightforward, but care should be taken, because there aretwo complex exponentials. Luckily the superposition principle tells us that the full solution willbe one term that corresponds to eiω0t and one that corresponds to e−iω0t:

θp(t) =F

2(−ω2d + 2iβωd + ω2)

eiωdt − F

2(−ω2d − 2iβωd + ω2)

e−iωdt. (12)

While this is the analytic solution to this particular solution, it is not very useful at this point,because it is still in complex form. It can however be rewritten to give

θp(t) =F (ω2 − ω2

d)

(ω2 − ω2d)2 + 4β2ω2

d

cosωdt+2Fβωd

(ω2 − ω2d)2 + 4β2ω2

d

sinωdt. (13)

We define a quantity R such that

R =√

(ω2 − ω2d)2 + (2βωd)2.

Substituting that into equation 13 gives

θp(t) =F

R(ω2 − ω2

d

Rcosωdt+

2βωdR

sinωdt) (14)

From the trigonometric identity

A cosx+B sinx =√A2 +B2 cosx− φ

Wheretanφ =

B

A,

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Which can be applied to the equation as

θp(t) =F

Rcos (ωdt− φ). (15)

Here φ can be found from the amplitudes of the sine and cosine terms from equation 14 :

cosφ =ω2 − ω2

d

R

sinφ =2βωdR

(16)

Though the previous identity already gave an expression for φ, the sine and cosine are still ratheruseful in finding φ in certain extreme situations. combining these gives

tanφ =2βωdω2 − ω2

d

. (17)

Now that both the homogeneous solution and the particular solution are known, the full solutionis given by:

θ(t) =F

Rcos (ωdt− φ) + e−βt(AeΩt +Be−Ωt). (18)

[2] [3]

2.3 evaluating the small-angle equationFrom this equation it is immediately obvious that over a long enough time, the homogeneoussolution dies off and all that is left is the particular solution. At that time, the function becomes

θ(t) =F

Rcos (ωdt− φ).

The amplitude, from hereon called C, of this equations scales with 1R so it scales as

C =1

R=

1√(ω − ωd)2 + (2βωd)2

(19)

It is then possible to say something about φ, and the amplitude in three extreme cases, with smalldamping:

1 In the situation where βωd << ω2 − ω2d, or for simplicity, ωd ≈ 0. we find from equation 16

thatsinφ = 0, φ = 0

In the amplitude the ωd term drops out so the amplitude scales as:

C ∝ 1

ω2.

2 In the limit that ωd → ω from equation 17 we find

tanφ ≈ ±∞, φ =π

2

Where φ is restricted to positive values only. In the amplitude we find that

limωd→ω

C =∞

This is called resonance and dramatically increases the amplitude of the system.

3 When βωd << ω2d − ω2,or for simplicity, ωd ≈ ∞ it can be seen from equation 17 that

tanφ ≈ 0, φ = π

The amplitude then becomes

C ∝ 1

ω2d

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A plot of equation 18 is given in Figure 10 and the corresponding phase portrait is given in Figure4with the same initial conditions. For the plot the following initial conditions were used:l = 1g = 9.81ωd = 1.5β = 0.1F = 2A = B = 1

Figure 2: A plot of the solution θ(t) in t for the given initial conditions

Figure 3: The phase portrait of θ(t) versus θ(t)

From The plot it can clearly be seen that at first the angle changes quite a bit, but very quickly it

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settles into the particular solution. The phase portrait shows the same, it starts of quite chaotically,but quickly spirals inwards and settles in a stable circular orbit.

Figure 4: The Poincaré map of the linearized system

The Poincaré map also shows a stable orbit around the origin. The high density of points in thecenter is due to the points that start out close to the origin that move outwards.

3 Forced damped pendulumThe damped, periodically forced pendulum system can be written as:

θ + αθ + sin θ = A sinωt. (20)

This equation has three independent parameters: the strength of the damping α, and the amplitudeA and the frequency ω of the external field. For chaotic solutions, two conditions has to be satisfied.The first one is that the equation must be nonlinear, which is the case in (20) because of sin(θ). Thesecond condition is that we must have at least three dynamical variables, i.e. a three-dimensionalstate space, which is also the case, since we can rewrite (20) to

χ+ αχ+ sin θ = f sinφ,

θ = χ,

φ = ω.

So we have non-linearity and the minimum dimension of three of a differential equation for chaoticbehaviour, but there are no chaotic solutions for the small driven amplitude A and large dampingcoefficient α. This is because in the first case, the system will become lineal (small angle) where inthe last case we have that the differential equation system will reduce to dimension two. Chaoticbehaviour exists for the driven frequency to the resonant eigenfrequency of a pendulum. So chaoticeffects will occur near ω ≈ 1.

From the comparison of the driven amplitude A and the magnitude of the torque acting on thependulum, which is equal to sin θ, we can conclude the general properties of the solutions of (20).For a weak external force, A < 1, the pendulum will stay at θ < π

2 , where it will perform oscilla-tions near the downward equilibrium point. But if A > 1, the strong driven force leads towardsfull rotations.For different initial values, we get big differences in the solutions, from which three different pathsto chaos have been found:

1. Period-doubling cascade preceded by the appearance of symmetric-breaking solutions. Thiscascade holds for both locked and running solutions.

2. The loss of phase locking and random transitions between two locked states.

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3. The intermittency form of the transition to chaos occurring for an external field of largeamplitude. The paths are then a combination of clockwise and counter-clockwise rotationswith damped oscillations in-between.

(20) has been solved by using a hybrid digital-analog computer (a computer that is uses changeableaspects of physical phenomena to model a problem being solved) for α = 0.5 by varying ω andA. The differences in the behaviour for different values of the driving amplitude and frequency isgiven in the bifurcation diagram (Figure 5) [8].

Figure 5: Bifurcation diagram for a periodically driven damped pendulum for α = 0.5.(Note that Ω = ω)

This graph gives us the different types of dynamic behaviour of the system (20). We get periodicsolutions if ω << 1 or if ω >> 1. In region A, for large amplitudes A, we get that the pendulummakes rotations corresponding to the periodic motion from one potential well to another, whereis each well the pendulum performs simple oscillations around a certain point. In region B weobserve oscillations near the downward position. In this region, the set of period-doubling bifur-cations appears what will lead to a chaotic state.

The chaotic behaviour is here characterized by strange attractors, which are attractors with fractalstructure [10]. The structure of these strange attractors can be understood in terms of two distincttime scales of the motion: the shorter time scale corresponds to fast oscillations between someclose wells; the longer time scale corresponds with a slower diffusion throughout the lattice. Thischaotic motion is best characterized by the power spectral density in Figure 6 [9] for the parametersA = 3.8 and ω = 0.64. Even though the sharp peak observed at the driving frequency is narrow,it broadens at large driving amplitude.

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Figure 6: Power spectrum of the attractors for a periodically driven damped pendulum withA = 3.8, ω = 0.64 and α = 0.5

For a specially constructed experimental setup[11], the solutions of (20) have been obtainedfor fixed values of the damping constant α = 0.24 and the driving frequency ω = 0.67. If thedriven amplitude A has small values, then the motion of the pendulum will be periodic, where thependulum oscillates back and forth where the spectrum of θ = θ(t) has a principal peak at thedriven frequency ω and smaller peaks at the harmonics of ω. We obtain a period-two orbit forA = 0.69, where the power spectrum contains peaks at integer multiples of ω/2. Then anotherperiod-doubling appears, and chaos will fully develop at A ≈ 0.70.

By changing the amplitude A for fixed α = 0.5 and ω = 0.47, a intermittent transition to chaoswas found from the numerical and analog solutions of (20). A critical value of A exists, denotingAC , such that if A < AC , we get stable oscillations having the driven frequency. In the casewhere Ais slightly larger than AC , the system becomes chaotic via the intermittent transition asA is decreased. At this transition, θ becomes unstable and periodic oscillations appear which arerandomly interrupted by bursts of θ. These bursts are shown in Figure 7[8].

Figure 7: Phase trajectory for a periodically driven damped pendulum with A = 0.9045, ω = 0.47and α = 0.5.

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The elaborated analysis of this intermittency case of the transition to chaos in (20) was donein [12], where α = 0.5, ω = 2/3 and for different values of the amplitude A of an external field. IfA ≤ 1.5 there are two different periodic running solutions, clockwise and counter-clockwise. Thesetwo modes remain separate and will become chaotic as A increases. When A = 1.4945, there willoccur intermittent switching between the two modes, which produces a large amount of noise ifthe frequency is smaller than the driving frequency.

4 Numerical solutionsTo determine how good the solutions of the small angle approximation are the problem is solvednumerically. To solve this problem the Euler approximation is used. Before the code is shown, thenumerical method is explained.

4.1 Euler’s methodEuler’s method is one of the most simple numerical approximations to make. It is based on thedefinition of the derivative and is very useful for programming. The essential thing of the methodis using the definition of the derivative and instead of a limit use very small steps. To make moresense of this, first a derivation for the first derivative is given.

4.1.1 The first derivative

The definition of the first derivative in the limit is given by:

f ′(x) = limh→0

f(x+ h)− f(x)

h(21)

For the numerical approximation according to Euler’s method the limit has to disapear and insteaduse a very small value for h, most in the order of 0.0001. So for the numerical approximation thefollowing formula is used for the first derivative:

f ′(x) ≈ fn+1 − fnh

(22)

But there is a second order differential equation, so there must be found a numerical approximationfor the second order derivative. Again, using the definition of the derivative and put this in thedefinition of the first derivative, the definition of the second derivative is given by:

f ′′(x) = limh→0

f(x+ 2h)− 2f(x+ h) + f(x)

h2(23)

For the numerical approximation, using the same technic as for the first derivative the following isfound:

f ′′(x) ≈ fn+2 − 2fn+1 + fnh2

(24)

Numerical solutions are never perfect, there is always introduced a small error since it is an approx-imation. This error can be estimated using the way of derivation of the method. The method isbased on a Taylor expansion, which is said to be n-th order accurate. This means that the highestorder of h is included in hn. For the case studied in this essay it means that the largest error madein each step is given by hn+1 where n is equal to two since there are up to second order differentialequations.

4.1.2 Code for the numerical approximation

The Euler method is used in the following code.

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%matp lo t l i b i n l i n eimport numpy as npimport mathfrom matp lo t l i b . pyplot import f i gu r e , showimport numpy as np

f i g = f i g u r e ( f i g s i z e =(16 ,16))frame = f i g . add_subplot ( 2 , 1 , 1 )

the_0 = 0 .3the_1 = 0 .3to t = 0w = (9 . 8 1 )∗∗0 . 5h = 0.001∗2∗math . p i /wwd = 1.5b = 0.02Fd = 2

s o l = [ the_0 , the_1 ]tim = [ 0 , 0 .001∗2∗math . p i /w]an = [ ]tdot = [ 0 , ( ( the_1 − the_0 )/h ) ]andot = [ ]ntdot = [ ]

whi l e tot <100:the_2 = 2∗ the_1 − the_0 − h∗b∗( the_1 − the_0 ) − h∗h∗w∗w∗np . s i n ( the_0 )+ h∗h∗Fd∗np . cos (wd∗ to t )the_dot = ( the_2 − the_1 )/hthe_0 = the_1the_1 = the_2tot = tot + htim . append ( to t )s o l . append ( the_2 )tdot . append ( the_dot )

s o l u t i o n = np . array ( s o l )time = np . array ( tim )thetadot = np . array ( tdot )newthetadot = np . array ( ntdot )

frame . g r id ( )

frame . p l o t ( time , s o lu t i on , l a b e l = ’ numerica l s o lu t i on ’ )

frame . s e t_x labe l ( ’ time ’ )frame . s e t_y labe l ( ’ theta ’ )

l = 1g = 9.81omega = ( g/ l )∗∗0 . 5omegad = 1 .5A = 1B = 1

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beta = 0 .1F = 2R = (( omega∗∗2 − omegad∗∗2)∗∗2 + (2∗ beta ∗omegad )∗∗2)∗∗0 .5bigomega = ( beta ∗∗2 − omega ∗∗2)∗∗0 .5phi = np . arctan (2∗ beta ∗omegad /(omega∗∗2 − omegad ∗∗2))theta = [ ]time = np . l i n s p a c e (0 ,100 ,800)f o r t in time :

the = np . r e a l (F/R∗np . cos ( omegad∗ t − phi ) )theta . append ( the )

frame . p l o t ( time , theta , l a b e l = ’ smal l ang le approximation ’ )frame . legend ( )show ( )

4.1.3 Results

To check how good the numerical solutions are there is done an approximation of one of the mostsimple linear second order differential equations, which can also be solved by hand. Below thederivation of the equation:

y + 2y + y = 0 (25)

The most simple solution for this equation is e−x. The plot with this together with the numericalsolution is shown in figure 8.

Figure 8: Numerical solution and analytic solution of the simple linear second order differentialequation.

From this code, the following figure comparing the numerical solution with the small angle ap-proximation is found. This plot is showing the results on the long term, the start of the pendulumis not shown.

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Figure 9: A numerical solution of the full equation and the small angle approximation of the fullequation

As shown in figure 9 the small angle approximation solution approximates the numerical solu-tion very well. This is very logical because the angular distance from the origin is around 0.3 radwhich is equal to approximately 17 degrees which is just outside the maximum angle of the smallangle approximation. If the same code is used, but a bigger angle is used, the following is obtained:

Figure 10: Another numerical solution and the small angle approximation

As seen in the figure the small angle approximation is not a good approximation of the numericalsolution anymore. This is the result of an angle of 0.8 rad which is equal to approximately 45degrees.

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5 Bibliography

References[1] Figure 1: Chetvorno, 2008, Simple gravity pendulum, wikimedia, accessed 24-12-2018

https://commons.wikimedia.org/wiki/File:Simple_gravity_pendulum.svg

[2] Thornton and Marion, 2008 Classical dynamics of particles and systems pp. 117-123

[3] David Morin, 2007 Classical mechanics with problems and solutions pp. 109-111

[4] Needham, Science and Civilisation in China: Volume 3, Mathematics and the Sciences of theHeavens and the Earth, pp627-629

[5] Murdin 2008 Full Meridian of Glory: Perilous Adventures in the Competition to Measure theEarth, pp 41.

[6] Wikipedia, pendulum accessed 24-12-2018https://en.wikipedia.org/wiki/Pendulum#History

[7] Thomson and Poynting, 1907, A textbook of physics, pp 10

[8] M. Gitterman, 2010, The chaotic pendulum, pp27-30.

[9] B. A. Huberman, J. P. Crutchfield, and N. H. Packard, 1980, Noise phenomena in Josephsonjunctions pp 750-752.

[10] Wikipedia, attractors accessed 5-2-2019https://en.wikipedia.org/wiki/Attractor#Strange_attractor

[11] J. A. Blackburn, Z. J. Yang, S. Vik, H. J. T. Smith, and M. A. H. Nerenberg,Physica D26,385 (1987)

[12] E. G. Gwin and R. M. Westervelt, 1984 Intermittent Chaos and Low-Frequency Noise in theDriven Damped Pendulum Phys. Rev. Lett.54, 1613 (1985); Phys.Rev. A33,4143 (1986).

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