The Pendulum Plain and Puzzling

Chris Sangwin

School of Mathematics University of Edinburgh

April 2017

Chris Sangwin (University of Edinburgh) Pendulum April 2017 1 / 38 Outline

1 Introduction and motivation

2 Geometrical physics

3 Classical mechanics

4 Driven pendulum

5 The elastic pendulum

6 Chaos

7 Conclusion

Chris Sangwin (University of Edinburgh) Pendulum April 2017 2 / 38 2 The pendulum illustrates the concerns of each age. 1 Divinely ordered world of classical mechanics 2 Catastrophe theory 3 Chaos 3 Contemporary examples relevant to A-level mathematics/FM. 4 Physics apparatus Illustrated by examples. (Equations on request!)

Connections within and beyond mathematics

The pendulum is a paradigm.

1 Almost every interesting dynamic phenomena can be illustrated by a pendulum.

Chris Sangwin (University of Edinburgh) Pendulum April 2017 3 / 38 3 Contemporary examples relevant to A-level mathematics/FM. 4 Physics apparatus Illustrated by examples. (Equations on request!)

Connections within and beyond mathematics

The pendulum is a paradigm.

Chris Sangwin (University of Edinburgh) Pendulum April 2017 3 / 38 4 Physics apparatus Illustrated by examples. (Equations on request!)

Connections within and beyond mathematics

The pendulum is a paradigm.

1 Almost every interesting dynamic phenomena can be illustrated by a pendulum. 2 The pendulum illustrates the concerns of each age. 1 Divinely ordered world of classical mechanics 2 Catastrophe theory 3 Chaos 3 Contemporary examples relevant to A-level mathematics/FM.

Chris Sangwin (University of Edinburgh) Pendulum April 2017 3 / 38 Illustrated by examples. (Equations on request!)

Connections within and beyond mathematics

The pendulum is a paradigm.

Chris Sangwin (University of Edinburgh) Pendulum April 2017 3 / 38 (Equations on request!)

Connections within and beyond mathematics

The pendulum is a paradigm.

Chris Sangwin (University of Edinburgh) Pendulum April 2017 3 / 38 Connections within and beyond mathematics

The pendulum is a paradigm.

Chris Sangwin (University of Edinburgh) Pendulum April 2017 3 / 38 Mechanics within mathematics

Chris Sangwin (University of Edinburgh) Pendulum April 2017 4 / 38 y¨ = −k 2y? So y(t) = A cos(kt + ρ).

Where is the x-coordinate in the model?

The elastic pendulum

What happens when I displace vertically and release from rest?

Chris Sangwin (University of Edinburgh) Pendulum April 2017 5 / 38 So y(t) = A cos(kt + ρ).

Where is the x-coordinate in the model?

The elastic pendulum

What happens when I displace vertically and release from rest?

y¨ = −k 2y?

Chris Sangwin (University of Edinburgh) Pendulum April 2017 5 / 38 Where is the x-coordinate in the model?

The elastic pendulum

What happens when I displace vertically and release from rest?

y¨ = −k 2y? So y(t) = A cos(kt + ρ).

Chris Sangwin (University of Edinburgh) Pendulum April 2017 5 / 38 The elastic pendulum

What happens when I displace vertically and release from rest?

y¨ = −k 2y? So y(t) = A cos(kt + ρ).

Where is the x-coordinate in the model?

Chris Sangwin (University of Edinburgh) Pendulum April 2017 5 / 38 Modelling

I hope I shall shock a few people in asserting that the most important single task of mathematical instruction in the secondary school is to teach the setting up of equations to solve word problems. [...] And so the future engineer, when he learns in the secondary school to set up equations to solve “word problems" has a ﬁrst taste of, and has an opportunity to acquire the attitude essential to, his principal professional use of mathematics. Polya (1962)

Chris Sangwin (University of Edinburgh) Pendulum April 2017 6 / 38 Pre-Newtonian mechanics

Chris Sangwin (University of Edinburgh) Pendulum April 2017 7 / 38 Chris Sangwin (University of Edinburgh) Pendulum April 2017 8 / 38 Randomness

Chris Sangwin (University of Edinburgh) Pendulum April 2017 9 / 38 A proposition well known to geometers

... it is a proposition well known to geometers, that the velocity of a pendulous body in the lowest point is as the chord of the arc which it has described in its descent. (Newton, Principia, (I), pg 25)

Chris Sangwin (University of Edinburgh) Pendulum April 2017 10 / 38 By conservation of energy, the kinetic energy at O equals the potential energy at P0. 1 gy = 2gl sin2(θ) = v 2. 2 g g v 2 = 4gl sin2(θ) = (2l sin(θ))2 = c2. l l q g Hence v = l c. MEI AS FM 6.02e

y = l(1 − cos(2θ)) = 2l sin2(θ).

Chris Sangwin (University of Edinburgh) Pendulum April 2017 11 / 38 1 gy = 2gl sin2(θ) = v 2. 2 g g v 2 = 4gl sin2(θ) = (2l sin(θ))2 = c2. l l q g Hence v = l c. MEI AS FM 6.02e

y = l(1 − cos(2θ)) = 2l sin2(θ). By conservation of energy, the kinetic energy at O equals the potential energy at P0.

Chris Sangwin (University of Edinburgh) Pendulum April 2017 11 / 38 q g Hence v = l c. MEI AS FM 6.02e

y = l(1 − cos(2θ)) = 2l sin2(θ). By conservation of energy, the kinetic energy at O equals the potential energy at P0. 1 gy = 2gl sin2(θ) = v 2. 2 g g v 2 = 4gl sin2(θ) = (2l sin(θ))2 = c2. l l

Chris Sangwin (University of Edinburgh) Pendulum April 2017 11 / 38 MEI AS FM 6.02e

y = l(1 − cos(2θ)) = 2l sin2(θ). By conservation of energy, the kinetic energy at O equals the potential energy at P0. 1 gy = 2gl sin2(θ) = v 2. 2 g g v 2 = 4gl sin2(θ) = (2l sin(θ))2 = c2. l l q g Hence v = l c.

Chris Sangwin (University of Edinburgh) Pendulum April 2017 11 / 38 y = l(1 − cos(2θ)) = 2l sin2(θ). By conservation of energy, the kinetic energy at O equals the potential energy at P0. 1 gy = 2gl sin2(θ) = v 2. 2 g g v 2 = 4gl sin2(θ) = (2l sin(θ))2 = c2. l l q g Hence v = l c. MEI AS FM 6.02e

Chris Sangwin (University of Edinburgh) Pendulum April 2017 11 / 38 The pendulum as experimental apparatus

MEI FM 6.03e

Chris Sangwin (University of Edinburgh) Pendulum April 2017 12 / 38 The pendulum as experimental apparatus

MEI FM 6.03e

Chris Sangwin (University of Edinburgh) Pendulum April 2017 12 / 38 Issac Newton 1643 – 1727

Chris Sangwin (University of Edinburgh) Pendulum April 2017 13 / 38 Newton’s Laws of Motion

Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it. The relationship between an object’s mass m, its acceleration a, and the applied force F is

F = ma.

For every action there is an equal and opposite reaction.

AS/A level core

Chris Sangwin (University of Edinburgh) Pendulum April 2017 14 / 38 For small θ assume sin(θ) ≈ θ.

mlθ¨ = −mgθ.

Then the solution is θ(t) = A cos(ωt + ρ). where ω2 = g/l

Simple harmonic motion

Point at P. Note s = lθ and a = lθ¨.

mlθ¨ = −mg sin(θ), θ(0) = θ0, θ˙(0) = θ˙0.

Chris Sangwin (University of Edinburgh) Pendulum April 2017 15 / 38 mlθ¨ = −mgθ.

Then the solution is θ(t) = A cos(ωt + ρ). where ω2 = g/l

Simple harmonic motion

Point at P. Note s = lθ and a = lθ¨.

mlθ¨ = −mg sin(θ), θ(0) = θ0, θ˙(0) = θ˙0.

For small θ assume sin(θ) ≈ θ.

Chris Sangwin (University of Edinburgh) Pendulum April 2017 15 / 38 Then the solution is θ(t) = A cos(ωt + ρ). where ω2 = g/l

Simple harmonic motion

Point at P. Note s = lθ and a = lθ¨.

mlθ¨ = −mg sin(θ), θ(0) = θ0, θ˙(0) = θ˙0.

For small θ assume sin(θ) ≈ θ.

mlθ¨ = −mgθ.

Chris Sangwin (University of Edinburgh) Pendulum April 2017 15 / 38 Simple harmonic motion

Point at P. Note s = lθ and a = lθ¨.

mlθ¨ = −mg sin(θ), θ(0) = θ0, θ˙(0) = θ˙0.

For small θ assume sin(θ) ≈ θ.

mlθ¨ = −mgθ.

Then the solution is θ(t) = A cos(ωt + ρ). where ω2 = g/l

Chris Sangwin (University of Edinburgh) Pendulum April 2017 15 / 38 sin(θ) ≈ θ makes it easier. Physics experiment to establish value of g. For SHM the period does not depend on amplitude. For a real pendulum it does. The Hooke’s law spring/mass gives SHM.

MEI AS FM 4.10f

Notes on SHM

Many assumptions: e.g. ignore friction.

Chris Sangwin (University of Edinburgh) Pendulum April 2017 16 / 38 Physics experiment to establish value of g. For SHM the period does not depend on amplitude. For a real pendulum it does. The Hooke’s law spring/mass gives SHM.

MEI AS FM 4.10f

Notes on SHM

Many assumptions: e.g. ignore friction. sin(θ) ≈ θ makes it easier.

Chris Sangwin (University of Edinburgh) Pendulum April 2017 16 / 38 For SHM the period does not depend on amplitude. For a real pendulum it does. The Hooke’s law spring/mass gives SHM.

MEI AS FM 4.10f

Notes on SHM

Many assumptions: e.g. ignore friction. sin(θ) ≈ θ makes it easier. Physics experiment to establish value of g.

Chris Sangwin (University of Edinburgh) Pendulum April 2017 16 / 38 For a real pendulum it does. The Hooke’s law spring/mass gives SHM.

MEI AS FM 4.10f

Notes on SHM

Many assumptions: e.g. ignore friction. sin(θ) ≈ θ makes it easier. Physics experiment to establish value of g. For SHM the period does not depend on amplitude.

Chris Sangwin (University of Edinburgh) Pendulum April 2017 16 / 38 The Hooke’s law spring/mass gives SHM.

MEI AS FM 4.10f

Notes on SHM

Many assumptions: e.g. ignore friction. sin(θ) ≈ θ makes it easier. Physics experiment to establish value of g. For SHM the period does not depend on amplitude. For a real pendulum it does.

Chris Sangwin (University of Edinburgh) Pendulum April 2017 16 / 38 Notes on SHM

Many assumptions: e.g. ignore friction. sin(θ) ≈ θ makes it easier. Physics experiment to establish value of g. For SHM the period does not depend on amplitude. For a real pendulum it does. The Hooke’s law spring/mass gives SHM.

MEI AS FM 4.10f

Chris Sangwin (University of Edinburgh) Pendulum April 2017 16 / 38 Christiaan Huygens 1629 – 1695

Chris Sangwin (University of Edinburgh) Pendulum April 2017 17 / 38 A pendulum which changes its length (cf elastic pendulum)

Christiaan Huygens Horologium oscillatorium, (1673)

The Pendulum Clock or Geometrical Demonstrations Concerning the Motion of Pendula As Applied to Clocks

Chris Sangwin (University of Edinburgh) Pendulum April 2017 18 / 38 Christiaan Huygens Horologium oscillatorium, (1673)

The Pendulum Clock or Geometrical Demonstrations Concerning the Motion of Pendula As Applied to Clocks

A pendulum which changes its length (cf elastic pendulum)

Chris Sangwin (University of Edinburgh) Pendulum April 2017 18 / 38 Multiply by lθ˙ so that,

ml2θ˙θ¨ + mglθ˙ sin(θ) = 0.

Integrating this gives

1 ml2θ˙2 − mgl cos(θ) = E , 2 0 or 1 ml2θ˙2 + mgl(1 − cos(θ)) = E . 2 1 Kinetic and potential energy. MEI FM 6.02

Energy considerations (simple pendulum)

mlθ¨ = −mg sin(θ), θ(0) = θ0, θ˙(0) = θ˙0.

Chris Sangwin (University of Edinburgh) Pendulum April 2017 19 / 38 Integrating this gives

1 ml2θ˙2 − mgl cos(θ) = E , 2 0 or 1 ml2θ˙2 + mgl(1 − cos(θ)) = E . 2 1 Kinetic and potential energy. MEI FM 6.02

Energy considerations (simple pendulum)

mlθ¨ = −mg sin(θ), θ(0) = θ0, θ˙(0) = θ˙0. Multiply by lθ˙ so that,

ml2θ˙θ¨ + mglθ˙ sin(θ) = 0.

Chris Sangwin (University of Edinburgh) Pendulum April 2017 19 / 38 Kinetic and potential energy. MEI FM 6.02

Energy considerations (simple pendulum)

mlθ¨ = −mg sin(θ), θ(0) = θ0, θ˙(0) = θ˙0. Multiply by lθ˙ so that,

ml2θ˙θ¨ + mglθ˙ sin(θ) = 0.

Integrating this gives

1 ml2θ˙2 − mgl cos(θ) = E , 2 0 or 1 ml2θ˙2 + mgl(1 − cos(θ)) = E . 2 1

Chris Sangwin (University of Edinburgh) Pendulum April 2017 19 / 38 Energy considerations (simple pendulum)

mlθ¨ = −mg sin(θ), θ(0) = θ0, θ˙(0) = θ˙0. Multiply by lθ˙ so that,

ml2θ˙θ¨ + mglθ˙ sin(θ) = 0.

Integrating this gives

1 ml2θ˙2 − mgl cos(θ) = E , 2 0 or 1 ml2θ˙2 + mgl(1 − cos(θ)) = E . 2 1 Kinetic and potential energy. MEI FM 6.02

Chris Sangwin (University of Edinburgh) Pendulum April 2017 19 / 38 Writing this as 2g 2E θ˙2 = (cos(θ) − 1) + 1 , l ml2 The form y 2 = α(cos(x) − 1) + β where α, β > 0.

Energy considerations

1 ml2θ˙2 + mgl(1 − cos(θ)) = E . 2 1

Conservation equation: E1, is constant

Chris Sangwin (University of Edinburgh) Pendulum April 2017 20 / 38 Energy considerations

1 ml2θ˙2 + mgl(1 − cos(θ)) = E . 2 1

Conservation equation: E1, is constant Writing this as 2g 2E θ˙2 = (cos(θ) − 1) + 1 , l ml2 The form y 2 = α(cos(x) − 1) + β where α, β > 0.

Chris Sangwin (University of Edinburgh) Pendulum April 2017 20 / 38 Phase plane

y 2 = α(cos(x) − 1) + β

Chris Sangwin (University of Edinburgh) Pendulum April 2017 21 / 38 Blackburn Y pendulum

Chris Sangwin (University of Edinburgh) Pendulum April 2017 22 / 38 If ω1 is rational.... ω2 Lissajous ﬁgure

Blackburn Y pendulum

For small oscillations

x1(t) = A1 cos(ω1t + ρ1).

x2(t) = A2 cos(ω2t + ρ2).

Chris Sangwin (University of Edinburgh) Pendulum April 2017 23 / 38 Lissajous ﬁgure

Blackburn Y pendulum

For small oscillations

x1(t) = A1 cos(ω1t + ρ1).

x2(t) = A2 cos(ω2t + ρ2). If ω1 is rational.... ω2

Chris Sangwin (University of Edinburgh) Pendulum April 2017 23 / 38 Blackburn Y pendulum

For small oscillations

x1(t) = A1 cos(ω1t + ρ1).

x2(t) = A2 cos(ω2t + ρ2). If ω1 is rational.... ω2 Lissajous ﬁgure

Chris Sangwin (University of Edinburgh) Pendulum April 2017 23 / 38 Add in damping

Chris Sangwin (University of Edinburgh) Pendulum April 2017 24 / 38 Wilberforce’s spring (1894)

1 my¨ = −ky − φ, 2 1 Iφ¨ = −τφ − y. 2 These are weakly coupled linear oscillators.

k τ ω2 = , ω2 = . s m t I

Coupled oscillators

Chris Sangwin (University of Edinburgh) Pendulum April 2017 25 / 38 1 my¨ = −ky − φ, 2 1 Iφ¨ = −τφ − y. 2 These are weakly coupled linear oscillators.

k τ ω2 = , ω2 = . s m t I

Coupled oscillators

Wilberforce’s spring (1894)

Chris Sangwin (University of Edinburgh) Pendulum April 2017 25 / 38 k τ ω2 = , ω2 = . s m t I

Coupled oscillators

Wilberforce’s spring (1894)

1 my¨ = −ky − φ, 2 1 Iφ¨ = −τφ − y. 2 These are weakly coupled linear oscillators.

Chris Sangwin (University of Edinburgh) Pendulum April 2017 25 / 38 Coupled oscillators

Wilberforce’s spring (1894)

1 my¨ = −ky − φ, 2 1 Iφ¨ = −τφ − y. 2 These are weakly coupled linear oscillators.

k τ ω2 = , ω2 = . s m t I

Chris Sangwin (University of Edinburgh) Pendulum April 2017 25 / 38 Techniques

Many methods: Direct substitution of candidate y(t) = eωt . Set up whole system as x˙ = Ax. Laplace transforms

Chris Sangwin (University of Edinburgh) Pendulum April 2017 26 / 38 α + β β − α cos(α) − cos(β) = 2 sin sin . 2 2

Solution

After some work ... q 2 2 2 2 22 2 2ω = ωs + ωt ± ωs − ωt + /(mI). (1)

and y φ( ) = 0 ( (ω ) − (ω )) , t 2 2 cos F t cos St 2I(ωF − ωS) y ( ) = 0 ( ω2 − τ) (ω ) − ( ω2 − τ) (ω ) . y t 2 2 I F cos F t I S cos St I(ωF − ωS)

Chris Sangwin (University of Edinburgh) Pendulum April 2017 27 / 38 Solution

After some work ... q 2 2 2 2 22 2 2ω = ωs + ωt ± ωs − ωt + /(mI). (1)

and y φ( ) = 0 ( (ω ) − (ω )) , t 2 2 cos F t cos St 2I(ωF − ωS) y ( ) = 0 ( ω2 − τ) (ω ) − ( ω2 − τ) (ω ) . y t 2 2 I F cos F t I S cos St I(ωF − ωS)

α + β β − α cos(α) − cos(β) = 2 sin sin . 2 2

Chris Sangwin (University of Edinburgh) Pendulum April 2017 27 / 38 y ω + ω ω − ω φ( ) = 0 F S F S t 2 2 sin t t I(ωF − ωS) 2 2 which is the product of two sinusoids. Special case (ωF = ωS) when

ωs = ωt

complete energy transfer between torsion and vertical modes. Physics experiment to measure the value of τ.

α + β β − α cos(α) − cos(β) = 2 sin sin . 2 2

Chris Sangwin (University of Edinburgh) Pendulum April 2017 28 / 38 Special case (ωF = ωS) when

ωs = ωt

complete energy transfer between torsion and vertical modes. Physics experiment to measure the value of τ.

α + β β − α cos(α) − cos(β) = 2 sin sin . 2 2

y ω + ω ω − ω φ( ) = 0 F S F S t 2 2 sin t t I(ωF − ωS) 2 2 which is the product of two sinusoids.

Chris Sangwin (University of Edinburgh) Pendulum April 2017 28 / 38 Physics experiment to measure the value of τ.

α + β β − α cos(α) − cos(β) = 2 sin sin . 2 2

y ω + ω ω − ω φ( ) = 0 F S F S t 2 2 sin t t I(ωF − ωS) 2 2 which is the product of two sinusoids. Special case (ωF = ωS) when

ωs = ωt

complete energy transfer between torsion and vertical modes.

Chris Sangwin (University of Edinburgh) Pendulum April 2017 28 / 38 α + β β − α cos(α) − cos(β) = 2 sin sin . 2 2

y ω + ω ω − ω φ( ) = 0 F S F S t 2 2 sin t t I(ωF − ωS) 2 2 which is the product of two sinusoids. Special case (ωF = ωS) when

ωs = ωt

complete energy transfer between torsion and vertical modes. Physics experiment to measure the value of τ.

Chris Sangwin (University of Edinburgh) Pendulum April 2017 28 / 38 Elastic pendulum

I ﬁrst noticed the mode coupling in the swinging pendulum when I was a Teaching Assistant as a graduate student when (by chance) a student could not make the lab experiment work. This classical experiment, as you know, ﬁrst measures the spring constant and then predicts the period. After stewing about this for a while I was able to work it out. (Olsson, M. 2008, private communication)

See Olsson 1976.

Chris Sangwin (University of Edinburgh) Pendulum April 2017 29 / 38 Approximate the Lagrangian L = T − V as

Deriving equations

Kinetic energy m T = (x˙ 2 + y˙ 2), (2) 2 Potential energy k V = mgy + (l − l )2, (3) 2 0 2 2 2 l = (l1 − y) + x .

Chris Sangwin (University of Edinburgh) Pendulum April 2017 30 / 38 Deriving equations

Kinetic energy m T = (x˙ 2 + y˙ 2), (2) 2 Potential energy k V = mgy + (l − l )2, (3) 2 0 2 2 2 l = (l1 − y) + x . Approximate the Lagrangian L = T − V as

Chris Sangwin (University of Edinburgh) Pendulum April 2017 30 / 38 where 2 g 2 k ωp = , ωs = l1 m l k l λ = ω2 0 = 0 . s 2 2 l1 m l1

Approximate equations of motion

2 x¨ = −ωpx + λxy, λ y¨ = −ω2y + x2. s 2

Chris Sangwin (University of Edinburgh) Pendulum April 2017 31 / 38 Approximate equations of motion

2 x¨ = −ωpx + λxy, λ y¨ = −ω2y + x2. s 2 where 2 g 2 k ωp = , ωs = l1 m l k l λ = ω2 0 = 0 . s 2 2 l1 m l1

Chris Sangwin (University of Edinburgh) Pendulum April 2017 31 / 38 Numerical solutions

Chris Sangwin (University of Edinburgh) Pendulum April 2017 32 / 38 So 2 x¨(t) + (ωp + λa cos(ωst))x(t) = 0. Mathieu’s equation ωs = 2ωp.

Start with a vertical motion

|x| << 1 y(t) ≈ a cos(ωst).

Chris Sangwin (University of Edinburgh) Pendulum April 2017 33 / 38 Mathieu’s equation ωs = 2ωp.

Start with a vertical motion

|x| << 1 y(t) ≈ a cos(ωst). So 2 x¨(t) + (ωp + λa cos(ωst))x(t) = 0.

Chris Sangwin (University of Edinburgh) Pendulum April 2017 33 / 38 Start with a vertical motion

|x| << 1 y(t) ≈ a cos(ωst). So 2 x¨(t) + (ωp + λa cos(ωst))x(t) = 0. Mathieu’s equation ωs = 2ωp.

Chris Sangwin (University of Edinburgh) Pendulum April 2017 33 / 38 Instability of the elastic pendulum Stability of the driven inverted pendulum

Mathieu’s equation

x¨(t) + (α + β cos(t))x(t) = 0

Chris Sangwin (University of Edinburgh) Pendulum April 2017 34 / 38 Mathieu’s equation

x¨(t) + (α + β cos(t))x(t) = 0

Instability of the elastic pendulum Stability of the driven inverted pendulum

Chris Sangwin (University of Edinburgh) Pendulum April 2017 34 / 38 Driven Double Pendulum

Chris Sangwin (University of Edinburgh) Pendulum April 2017 35 / 38 complex behavior! Double pendulum. Complexity very different from irrational Lissajous.

Chaos

Simple systems: simple behavior.

Chris Sangwin (University of Edinburgh) Pendulum April 2017 36 / 38 Double pendulum. Complexity very different from irrational Lissajous.

Chaos

Simple systems: simple behavior. complex behavior!

Chris Sangwin (University of Edinburgh) Pendulum April 2017 36 / 38 Complexity very different from irrational Lissajous.

Chaos

Simple systems: simple behavior. complex behavior! Double pendulum.

Chris Sangwin (University of Edinburgh) Pendulum April 2017 36 / 38 Chaos

Simple systems: simple behavior. complex behavior! Double pendulum. Complexity very different from irrational Lissajous.

Chris Sangwin (University of Edinburgh) Pendulum April 2017 36 / 38 Magnetic pendulum

Release the pendulum from rest. Which magnet does it end up above?

Chris Sangwin (University of Edinburgh) Pendulum April 2017 37 / 38 Magnetic pendulum

Release the pendulum from rest. Which magnet does it end up above?

Chris Sangwin (University of Edinburgh) Pendulum April 2017 37 / 38 2 The pendulum illustrates the concerns of each age. 3 Contemporary examples relevant to the new A-level. 4 Physics apparatus

Connections within and beyond mathematics

The pendulum is a paradigm.

1 Almost every interesting dynamic phenomena can be illustrated by a pendulum.

Chris Sangwin (University of Edinburgh) Pendulum April 2017 38 / 38 3 Contemporary examples relevant to the new A-level. 4 Physics apparatus

Connections within and beyond mathematics

The pendulum is a paradigm.

1 Almost every interesting dynamic phenomena can be illustrated by a pendulum. 2 The pendulum illustrates the concerns of each age.

Chris Sangwin (University of Edinburgh) Pendulum April 2017 38 / 38 4 Physics apparatus

Connections within and beyond mathematics

The pendulum is a paradigm.

1 Almost every interesting dynamic phenomena can be illustrated by a pendulum. 2 The pendulum illustrates the concerns of each age. 3 Contemporary examples relevant to the new A-level.

Chris Sangwin (University of Edinburgh) Pendulum April 2017 38 / 38 Connections within and beyond mathematics

The pendulum is a paradigm.

1 Almost every interesting dynamic phenomena can be illustrated by a pendulum. 2 The pendulum illustrates the concerns of each age. 3 Contemporary examples relevant to the new A-level. 4 Physics apparatus

Chris Sangwin (University of Edinburgh) Pendulum April 2017 38 / 38