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 ...
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
2668 JOSEPH B. KELLER
vertical displacement a(t) along the y-axis. One end of her ponytail is attached to her
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/
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/
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
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
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)
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 sx
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)
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
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
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)