Joseph Keller is a professor at Stanford where he used to go joggin...

In order to derive this equation of motion, one can use the express...

You might think that if you run faster, the jogging frequency will ...

D.J. Acheson studied the behaviour of a hanging chain of several li...

In order to derive the equation of motion for a perfectly flexible ...

For more on how to find the instability intervals, you can look up ...

Copyright © by SIAM. Unauthorized reproduction of this article is prohibited.

SIAM J. APPL. MATH .

c

2010 Society for Industrial and Applied Mathematics

Vol. 70, No. 7, pp. 2667–2672

PONYTAIL MOTION

∗

JOSEPH B. KELLER

†

Abstract. A jogger’s ponytail swa ys from side to side as the jogger runs, although her head

does not move from side to side. The jogger’s head just moves up and down, forcing the ponytail to

do so also. We show in two ways that this vertical motion is unstable to lateral perturbations. First

we treat the ponytail as a rigid pendulum, and then we treat it as a ﬂexible string; in each case, it is

hanging from a support which is moving up and down periodically, and we solve the linear equation

for small lateral oscillation. The angular displacement of the pendulum and the amplitude of each

mode of the string satisfy Hill’s equation. This equation has solutions which grow exponentially

in time when the natural frequency of the pendulum, or that of a mode of the string, is close to

an integer multiple of half the frequency of oscillation of the support. Then the vertical motion is

unstable, and the ponytail sways.

Key words. instability, parametric resonance, Hill’s equation

AMS subject classiﬁcations. 34F15, 70J40

DOI. 10.1137/090760477

1. Introduction. The ponytail of a running jogger sways from side to side, but

the jogger’s head generally does not move from side to side. The head just moves

up and down, so the ponytail also moves up and down with it. But, as we shall

show, this vertical motion of the hanging ponytail is unstable to lateral perturbations.

The resulting lateral motion, the swaying, is an example of parametric excitation, a

phenomenon which is common in oscillating mechanical and electrical systems.

We shall demonstrate this instability, and analyze the resulting motion, in two

diﬀerent ways. First, in section 2, we shall represent the ponytail as a rigid pendulum

hanging from a support which is moving up and down periodically. The pendulum also

moves up and down periodically. Any small angular deviation θ(t) from the vertical

position satisﬁes Hill’s equation, a linear second order ordinary diﬀerential equation

with a periodic coeﬃcient (Stoker [1]). This equation has one solution, which grows

exponentially in time if the natural frequency of the pendulum is close to an integer

multiple of half the frequency of oscillation of the support (Magnus and Winkler [2]).

Then the purely vertical motion of the pendulum is unstable, and it sways.

Next, more realistically, in section 3 we represent the ponytail as a ﬂexible string

hanging from a vertically oscillating support. Again a purely vertical motion of the

string is possible. As was shown by Belmonte et al. [3], the linear equation for small

lateral perturbations of this motion has an inﬁnite number of modes of periodic vibra-

tion. Each mode amplitude satisﬁes Hill’s equation. Therefore, just like the pendulum,

a mode is unstable when its natural frequency is close to an integer multiple of half

the frequency of oscillation of the support.

A still more realistic model of a ponytail is an inextensible rod with small bending

stiﬀness, described in section 5.

2. Ponytail as a rigid ro d. Suppose that a runner moves with constant speed

U along the positive z-axis, and that her head moves up and down with the periodic

∗

Received by the editors May 29, 2009; accepted for publication (in revised form) May 6, 2010;

published electronically July 29, 2010.

http://www.siam.org/journals/siap/70-7/76047.html

†

Departments of Mathematics and Mechanical Engineering, Stanford University, Stanford, CA

94305-2125 (k eller@math.stanford.edu).

2667

Downloaded 11/25/20 to 77.54.204.61. Redistribution subject to SIAM license or copyright; see https://epubs.siam.org/page/terms

Copyright © by SIAM. Unauthorized reproduction of this article is prohibited.

2668 JOSEPH B. KELLER

vertical displacement a(t) along the y-axis. One end of her ponytail is attached to her

head, at the position

x =0,y= a(t),z= Ut.(2.1)

We assume that the ponytail can move in the transverse plane z = Ut but not normal

to that plane, because it would hit the runner’s head or neck.

Let the ponytail be represented as a uniform rigid rod of length L,sothatits

center of mass is at its midpoint. Then its position in the plane z = Ut is determined

by the angle θ(t) which it makes with the downward pointing vertical line through

the point (2.1). If it is free to rotate about the point (2.1), it is a pendulum, and the

angle θ(t) satisﬁes the following equation of motion:

θ

tt

+

2

L

(g + a

tt

)sinθ =0.(2.2)

This is the usual equation for a pendulum with a ﬁxed endpoint, but with the vertical

acceleration a

tt

of the endpoint added to the acceleration of gravity g.

Equation (2.2) has two constant solutions, θ

0

=0andθ

0

= π,intheinterval

0 ≤ θ<2π. The ﬁrst, θ

0

= 0, represents the pendulum hanging straight down,

and the second, θ

0

= π, represents it balanced pointing upward. The stability or

instability of either solution is determined by the equation for the perturbation

˙

θ(t)

obtained by linearizing (2.2) about θ

0

. The linearized equation is

˙

θ

tt

±

2

L

(g + a

tt

)

˙

θ =0, +forθ

0

=0, − for θ

0

= π.(2.3)

When a

tt

= 0, the solution for

˙

θ is sinusoidal for θ

0

= 0 and exponentially growing

or decaying for θ

0

= π. Then the hanging pendulum is stable (a small perturbation

stays small), and the balanced pendulum is unstable (a small perturbation can grow

to be large).

When a

tt

is not zero but is a periodic function of t, (2.2) is called Hill’s equation

after the American astronomer G.W. Hill. He derived and studied it to determine

whether the periodic motion of the Moon about Earth is stable. The results for that

equation can be expressed in terms of the dimensionless parameter 2g/Lω

2

,whereω

is the angular frequency of a(t). For a(t) given, there are inﬁnitely many intervals of

2g/Lω

2

throughout which Hill’s equation has solutions that grow exponentially with

t [2]. When 2g/Lω

2

lies in one of these intervals, the hanging pendulum is unstable.

The resulting large amplitude motion is the observed swaying of the ponytail.

For |a

tt

| small, each instability interval contains a point 2g/Lω

2

= k

2

/4, with

k an integer. At this point (2g/L)

1/2

= kω/2, so there the natural frequency of

the pendulum (2g/L)

1/2

is an integer k times ω/2, which is half the frequency of the

vertical motion of the head. For a ponytail of length L = 25 cm, the natural frequency

is (2 × 980/25)

1/2

≈ 8.85 radians/sec. ≈ 1.41 cycles/sec. If the frequency of the head

motion is twice this (ω =17.71 radians/sec. ≈ 2.82 cycles/sec.), then the condition

given above for instability will hold with k =1.

A cycle corresponds to a step with one leg, so 2.82 cycles/sec. corresponds to

(60)(2.82) = 169 cycles/minute = 169 steps/minute. According to the Web site

RunGearRun.com, elite runners’ cadence is between 85 and 95 right-foot strikes per

minute at all distances from 800 meters to 26 miles, corresponding to 170 to 190

steps/minute. They vary their step length to change their speed. Joggers report 140

Downloaded 11/25/20 to 77.54.204.61. Redistribution subject to SIAM license or copyright; see https://epubs.siam.org/page/terms

Copyright © by SIAM. Unauthorized reproduction of this article is prohibited.

PONYTAIL M OTION 2669

to 160 steps/minute. These values indicate that a ponytail of length 25 cm can be

expected to sway at a typical running cadence.

Stability of the vertical position of the pendulum θ

0

= π was studied by Stephen-

son [4] when a(t)=A cos ωt. (See also Stoker [1].) He found conditions on A, L,

g,andω for which the position is stable. This result was extended to an N-link

pendulum by Acheson [5], and veriﬁed experimentally by Acheson and Mullin [6] for

N =1, 2, 3.

3. Ponytail as a ﬂexible string. We now model the ponytail as an inextensible

ﬂexible string of length L and constant density ρ hanging in the plane z = Ut.As

in section 2, the y-axis points vertically upward, and the x-axisishorizontalandis

normal to the direction of running. The top end of the string is attached to the head

at the point given in (2.1).

Let x

(s, t)=(x(s, t),y(s, t)) be the position at time t of the point at arclength

distance s from the top of the string in the plane z = Ut. It satisﬁes the equation of

motion

ρx

tt

=(Tx

s

)

s

+ ρg, 0 <s<L.(3.1)

Here T (s, t) is the tension in the string, and g

=(0, −g) is the acceleration of gravity.

The condition that s is arclength requires

x

2

s

=1, 0 <s<L.(3.2)

At the end s = 0, the position in the plane z = Ut is given by (2.1):

x

(0,t)=(0,a(t)) .(3.3)

At the end s = L, the tension vanishes:

T (L, t)=0.(3.4)

One solution of (3.1)–(3.4), representing a vertically hanging string moving up

and down, is

x

0

(s, t)=[0,a(t) − s] .(3.5)

The corresponding tension T

0

(s, t) is obtained by using (3.5) in the y-component

of (3.1), integrating with respect to s, and using (3.4) to eliminate the constant of

integration. The result is

T

0

(s, t)=ρ(g + a

tt

)(L − s).(3.6)

The stability of this solution is governed by the linearized problem for the per-

turbation ˙x,

˙

T . That problem, obtained by linearizing (3.1)–(3.4) around the solution

x

0

, T

0

,is

ρ ˙x

tt

=

˙

Tx

0

s

+ T

0

˙x

s

s

= −

˙

T

s

ˆy +(T

0

˙x

s

)

s

,(3.7)

x

0

s

· ˙x

s

= − ˙y

s

=0,(3.8)

˙x

(0,t)=0,(3.9)

˙

T (L, t)=0.(3.10)

Downloaded 11/25/20 to 77.54.204.61. Redistribution subject to SIAM license or copyright; see https://epubs.siam.org/page/terms

Copyright © by SIAM. Unauthorized reproduction of this article is prohibited.

2670 JOSEPH B. KELLER

In (3.7), ˆy is a unit vector in the positive y direction.

Upon integrating (3.8) with respect to s, and using the y-component of (3.9), we

obtain

˙y(s, t)=0.(3.11)

When (3.11) is used in the y-component of (3.7), it yields

˙

T

s

= 0. Integration of this

equation using (3.10) yields

˙

T (s, t)=0.(3.12)

We must still determine the lateral displacement ˙x(s, t), which satisﬁes the x-components

of (3.7) and (3.9).

4. Solution for the lateral displacement. The x-component of (3.7) be-

comes, when (3.6) is used for T

0

,

ρ ˙x

tt

= ρ(g + a

tt

)[(L − s)˙x

s

]

s

.(4.1)

We seek a solution of the product form

˙x(s, t)=u(t)v(s).(4.2)

Substitution of (4.2) into (4.1), and separation of variables, yields

u

tt

(g + a

tt

)

−1

u

−1

=[(L − s) v

s

]

s

v

−1

= −λ.(4.3)

Here λ is a constant.

From (4.3) we get two equations:

[(L − s) v

s

]

s

+ λv =0, 0 <s<L,(4.4)

u

tt

+ λ(g + a

tt

)u =0.(4.5)

The boundary condition (3.9) requires that

v(0) = 0.(4.6)

The only solution of (4.4) which is regular at s = L is a constant multiple of the

Bessel function J

0

:

v(s)=J

0

2λ

1/2

(L − s)

1/2

.(4.7)

Requiring this solution to satisfy (4.6) yields

J

0

2λ

1/2

L

1/2

=0.(4.8)

The function J

0

has an inﬁnite increasing sequence of positive roots j

n

, n =1, 2,....

Thus (4.8) shows that λ has one of the values λ

n

deﬁned by

λ

n

= j

2

n

/4L, n =1, 2,....(4.9)

We call the solution (4.7) with λ = λ

n

the nth mode v

n

(s):

v

n

(s)=J

0

2λ

1/2

n

(L − s)

1/2

= J

0

1 −

s

L

1/2

j

n

,n=1, 2,....(4.10)

Copyright © by SIAM. Unauthorized reproduction of this article is prohibited.

PONYTAIL M OTION 2671

Then (4.2) becomes

˙x(s, t)=u (t, λ

n

) J

0

(1 − s/L)

1/2

j

n

.(4.11)

This result was obtained by Belmonte et al. [3] for a(t) a cosine function, in which

case (4.5) is a Mathieu equation.

The amplitude u(t, λ

n

) in (4.11) satisﬁes (4.5), which is Hill’s equation, with

λ = λ

n

= j

2

n

/4L. The product λ

n

g = j

2

n

g/4L is the square of the frequency of

the nth mode. As was stated in section 2, for any periodic function a

tt

(t) with

frequency ω there are ranges of λ

n

g/ω

2

for which Hill’s equation has solutions which

grow exponentially with t.Whenλ

n

g/ω

2

lies in one of these instability intervals, the

vertical motion (3.5) is unstable to the lateral perturbation (4.11).

Just as in the last paragraph of section 2, if |a

tt

| is small, each instability interval

contains a point λ

n

g/ω

2

= k

2

/4 with k an integer. There (λ

n

g)

1/2

= k (ω/2), so at

this point the frequency of the nth mode is an integer multiple of ω/2. For the lowest

mode n =1andj

1

≈ 2.4,themodefrequencyisj

1

/2(g/L)

1/2

≈ 1.2(g/L)

1/2

.This

is slightly smaller than the pendulum frequency of a rigid rod of length L discussed in

section 2, which is (2g/L)

1/2

≈ 1.4(g/L)

1/2

. The lowest mode will become unstable

for a ponytail with L =25cmwhenω is around twice the lowest mode frequency,

i.e., 2(1.2)(980/25)

1/2

≈ 15.0 radians/sec. = 2.39 cycles/sec. = 143.5 steps/minute.

This is slightly less than the cadence required for swaying of the rigid pendulum of

the same length, but still within the range of joggers’ cadences.

5. Ponytail as ﬂexible rod. A still more realistic model of a ponytail is an

inextensible ﬂexible rod with small bending stiﬀness B. This model is intermediate

between the rigid rod of section 2, which has inﬁnite stiﬀness, and the string of section

3, which has zero stiﬀness.

One advantage of this model is that its slope at the point of attachment can

be speciﬁed to have any value. If it is horizontal there, then when the runner is

not moving, the ponytail will extend away from the head and hang downward in its

characteristic shape. This is just the well-known shape of a cantilever beam, which is

obtained by solving the equilibrium equations governing a thin rod.

When the runner is moving, and her head is bobbing up and down, the cantilever

beam shape is no longer a solution, except when it hangs straight down. In any other

case, the ponytail will oscillate in the vertical y, z plane. This motion has not been

calculated. The instability of this motion to lateral perturbation would determine

when swaying occurs, and would determine the swaying mode shape.

In the special case in which the rod hangs straight down, that shape remains a

solution when the runner is moving. The linear partial diﬀerential equation governing

the small lateral displacement ˙x(s, t) is (4.1) with the bending term −B ˙x

ssss

added

to the right-hand side. Since it is of fourth order in s, it requires four boundary

conditions, which are

˙x (0,t)=0, ˙x

s

(0,t)=0, ˙x

ss

(L, t)=0, ˙x

sss

(L, t)=0.(5.1)

The ﬁrst two of these conditions denote that the ponytail is clamped at the top, and

thesecondtwodenotethatitisfreeatthebottom.

Thepresenceofa

tt

(t) as a coeﬃcient prevents the use of separation of variables

to solve the modiﬁed (4.1). Instead (4.1) and (5.1) could be solved by treating the

stiﬀness B as small. This would lead to a singular perturbation problem, such as was

Copyright © by SIAM. Unauthorized reproduction of this article is prohibited.

2672 JOSEPH B. KELLER

treated by Handelman and Keller [7] using matched asymptotic expansions, and by

Champneys and Fraser [8] using both two-timing and a numerical implementation of

Floquet theory. It would be interesting to apply these methods to this problem.

REFERENCES

[1] J. J. Stoker, Nonlinear Vibrations, Interscience, New York, 1950.

[2] W. Magnus and S. Winkler, Hill’s Equation, Interscience, New York, 1966.

[3] A. Belmonte, M . J. Shelley, S. T. Eldakar, and C. H. Wiggins, Dynamic patterns and

self-knotting of a driven hanging chain, Phys. Rev. Lett., 87 (2001), pp. 114301–114304.

[4] A. Stephenson, On a new type of dynamical stability, Mem. Proc. Manch. Lit. Phil. Soc., 52

(1908), pp. 1–10.

[5] D. J. Acheson, A pendulum theorem, Proc. Roy. Soc. London Ser. A, 443 (1993), pp. 239–245.

[6] D. J. Acheson and T. Mullin, Upside-down pendulums, Nature, 366 (1993), pp. 215–216.

[7] G. H. Handelman and J. B. Keller, Small vibrations of a slightly stiﬀ pendulum, in Proceed-

ings of the 4th U.S. National Congress on Applied Mechanics, Amer. Soc. Mech. Eng., New

York, 1963, pp. 195–202.

[8] A. R. Champneys and W. B. Fraser, The “Indian rope trick” for a parametrically excited

ﬂexible rod: Linearized analysis,R.Soc.Lond.Proc.Ser.AMath.Phys.Eng.Sci.,456

(2000), pp. 553–570.

Joseph Keller is a professor at Stanford where he used to go jogging around the campus. He was struck by all the women running around campus and the bouncing motion of their ponytails from side to side.
In this paper Keller tries to understand how a vertical displacement of the head creates an horizontal swing of the ponytail. This paper was also awarded the [2012 Ignobel prize in Physics](https://blogs.scientificamerican.com/scicurious-brain/ignobel-prize-winner-in-physics-the-amazing-ponytail/)
![](https://s2.gifyu.com/images/ezgif.com-gif-maker-254b373559599e61b.gif)
You might think that if you run faster, the jogging frequency will increase, making it harder for the ponytail to swing, but that's not the case. In most of the situations, running faster will result in longer steps, while keeping the frequency constant and thus maintaining the ponytail swing.
For more on how to find the instability intervals, you can look up the annotations on ["A Pendulum Theorem" by D.J. Acheson](https://fermatslibrary.com/p/a3733a74) that has been annotated on Fermat's Library as well.
In order to derive this equation of motion, one can use the expression for the Torque
![](https://i.imgur.com/mYp8gJ5.png)
$$
\tau = I\alpha \equiv -m(g+a(t))\frac{L}{2} \sin \theta = m\frac{L^2}{2^2} \frac{d^2 \theta}{dt^2} \\
\frac{d^2 \theta}{dt^2} + (g+a(t))\frac{2}{L} \sin \theta = 0
$$
where $\tau$ is the torque, $\alpha$ is the angular acceleration and $I$ is the moment of inertia.
In order to derive the equation of motion for a perfectly flexible string (a string that offers no resistance to bending) we start by considering an infinitesinally thin segment of string with mass density $\rho$ and mass $\rho \Delta x $. Using Newton's law for a point mass for the vertical component:
![](https://i.imgur.com/YPy35zV_d.webp?maxwidth=760&fidelity=grand)
$$
F=ma \\
\rho \Delta x \frac{\partial^2 u}{\partial t^2} = T(x+\Delta x, t) \sin \theta(x+\Delta x, t)-T(x, t) \sin \theta(x, t)+\rho \Delta x g
$$
where $T(x,t)$ is the magnitude of the tensile force at point $x$ and time $t$ and $g$ is the acceleration of gravity. Dividing the previous expression by $\Delta x$ and taking the limit as $\Delta x \rightarrow 0$ we get
$$
\rho \frac{\partial^2 u}{\partial t^2} = \frac{\partial}{\partial x}(T(x, t) \sin \theta(x, t))+\rho g
$$
D.J. Acheson studied the behaviour of a hanging chain of several linked pendulums, all suspended from one another, and discovered various different modes of oscillation (his paper [has been previously annotated on Fermat's Library](https://fermatslibrary.com/p/a3733a74))
Acheson proved that it was possible to take all these linked pendulums, turn them upside-down, so that they are all precariously balanced on top of one another, and then stabilise them in that position by vibrating the pivot up and down, as we can see in this awesome demonstration by Steve Mould.
[![IMAGE ALT TEXT](https://slack-imgs.com/?url=https%3A%2F%2Fi.ytimg.com%2Fvi%2Fgnn21smGVrQ%2Fhqdefault.jpg&width=400&height=300)](https://youtu.be/gnn21smGVrQ?t=56 "Inverted Pendulum")
He started calling this gravity-defying experiment "Not quite the Indian rope trick" because of a famous magic trick performed in India during the 19th century. During this trick, a magician would hurl a rope into the air. The rope would stand erect, with no external support. His boy assistant would climb the rope and then descend.
In the context of running, this means for "punk rock" hairstyle the hair will be mostly in a stable equilibrium!
![](https://menhairstylesworld.com/wp-content/uploads/2017/10/Fanned-Mohawk-Punk-Hairstyles-for-Guys.jpg)