This paper was written by D.J. Acheson while he was trying to give ...
In 1908 Stephenson found that the upper vertical position of the pe...
This paper is from 1993! I had thought it was much older on my firs...
As we can see in this video by Steve Mould the chain of pendulums c...
To prove this we are going to use the Lagrangian formulation and wi...
To prove that the natural frequencies are proportional to $g$ we st...
Near $\alpha = 0$, $ \sqrt{-2\alpha} <\beta <0.450$. Since $\alpha_...
Also the larger the value of $N$, the smaller the value of $l$ beca...
A
pendulum
theoremt
BY D.
J.
ACHESON
Jesus
College,
Oxford
OX1
3DW,
U.K.
We
consider N
linked
pendulums
which are inverted and
balanced
on
top
of
one
another,
and
establish
a
general
theorem which shows
how
they may
be stabilized
by
small vertical
oscillations
of the
support.
1.
Introduction
Stephenson
(1908a, b)
showed
that it
is
possible
to
stabilize
a
single rigid
pendulum
in
its
inverted,
or
upside-down, equilibrium position by
subjecting
the
pivot
to small
vertical oscillations
of
suitably
high frequency.
He confirmed his
theoretical
predictions by
a
practical
demonstration of
the
phenomenon.
While
this
is a well-known
curiosity
of classical mechanics it does not seem to
be
generally
known
that an inverted
double,
or
even
triple
pendulum
can be stabilized
in
the same
way.
This
was,
again,
first
predicted theoretically by Stephenson,
in a
comparatively
overlooked
paper
of
1909,
though
the idea has
reappeared
in a
number of
subsequent
studies
(Lowenstern
1932;
Hsu
1961;
Kalmus
1970;
Otterbein
1982;
Leiber
&
Risken
1988).
Here we
present
a
simple
but
general
theorem on the linear
stability
of an
inverted
N-pendulum
of
any
kind.
The
generality
of the theorem is achieved
by
relating
the
stability
question
to
just
two
elementary
properties
of the
system
as
a
whole when
it
is
in
its
non-inverted,
or
downward-hanging,
state.
2.
The
stability
of
upside-down
pendulums
Theorem.
Let
N
pendulums
hang
down,
one
from
another,
under
gravity
g,
each
having
one
degree
of
freedom,
the
uppermost
being
suspended
from
a
pivot point
O.
Let
O)max
and
wmin
denote
the
largest
and the smallest
of
the natural
frequencies
of
small
oscillation about
this
equilibrium
state.
Now
turn the whole
system
upside-down.
The
resulting configuration
of
the
pendulums
can be stabilized
(according
to linear
theory,
at
least)
if
we
subject
the
pivot point
0
to
vertical
oscillations
of
suitable
amplitude
e
and
frequency
(o.
When
(O2
>
o)ax
the
stability
criterion
is
/V2g/oo
tmin
< e
<
0.450g/2ma.
(2.1)
Note.
When
several
pendulums
are
involved,
T2ax
is
typically
much
greater
than
2i.
The condition
o2
>
Omax
is
then
necessary
for the
stability
of the inverted
state,
so
(2.1)
then
gives
the
whole
stability region
in the
e-o)0
plane,
as
in
figure
1.
t
This
paper
was
accepted
as
a
rapid
communication.
Proc. R. Soc.
Lond. A
(1993)
443,
239-245
?
1993
The
Royal
Society
Printed in Great Britain
239
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8*
?
Figure
1.
Typical region
of linear
stability
for an
upside-down
N-pendulum,
as
given
by
(2.1).
Here
*
=
0.450
g/Oax
and
co
=
3.1432
ax/(min.
The
sketches
indicate theoretical
predictions
of the
behaviour
just
outside this
stability region.
Thus,
if
at
a
given frequency
w
the
amplitude
e
is
just
too
large,
the
system
is unstable to
rapidly growing buckling
oscillations
at
frequency
-o0.
If,
on
the
other
hand,
e is
a
little too
small,
the
pendulums
fall over
comparatively slowly,
keeping
to the
same side
of
the vertical.
Proof.
Consider
first small
disturbances
to the
system
about
the
original,
i.e. non-
inverted
equilibrium
state,
the
pivot point
0
being
fixed. Let the
natural
frequencies
be
oi,
and let
Xi
be the
corresponding
normal
coordinates,
each of
which will
be some
linear
combination
of
the
(small)
angles
which the
pendulums
make
with
the
downward vertical.
We
then
have
Xi+?
Xi,=0,
i=l,...,N.
(2.2)
Now,
each of
the
quantities
()2
will be
simply
proportional
to
g;
this
follows from
the standard
theory
of small oscillations based on
Lagrange's
equations
of motion
(see,
for
example,
Landau
&
Lifschitz
1976),
given
that the uniform
gravitational
field
g
is
the
sole
source of
potential
energy
in this
problem.
If
we tackle the
stability
of the inverted state
by changing
the
sign
of
g
we then find that small
disturbances
are
governed
instead
by
Xi-o
Xi
i=0,
i==l,...,N.
(2.3)
Suppose
now that the
pivot point
0 oscillates
up
and
down so that its
coordinate
in
the direction
of
the
upward
vertical
is h
=
-ecoso0t.
We
may
allow for
this
simply by
replacing
g
in the
above
argument by apparent
gravity
g
+h
=
g
+
e()
cos
o0
t.
In this
way
we
find that
X?--0~(?
1
+
(eo)/g)
coS
ot)Xi
=
0,
i
=
1,...,N.
On
introducing
the
scaling
T
=
(ot
we
therefore obtain N
uncoupled
Mathieu
equations:
d2Xi/dT2
+
(ci
+
ficosT)Xi
=
0,
i
=
1,...,N,
(2.4)
where
i
=-2/o2,
/=-i
=-
/g. (2.5)
Proc.
R.
Soc. Lond.
A
(1993)
D. J. Acheson
240
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11
1_
01
(
4
4
Figure
2.
Stable
(shaded)
and unstable
regions
for the X= 0
solution of Mathieu's
equation
d2X/dT2
+
(a
+
fP
cos
T)
X
=
0.
The
points
(2.6)
lie on the
straight
line
OP,
and for the inverted state
to be stable
they
must
all lie within the
shaded
region
with a <
0,
in
the manner indicated
(for
N=2).
Now,
the
stability
regions
of Mathieu's
equation
are
well
known
and indicated in
figure
2.
We
achieve
stability
of
our inverted
system
if,
by
choosing
e
and
ow
appropriately,
we
can
place
all the
points
2(-/2,
6g),
i
=
..
,N
(2.6)
in the shaded
part
of
the
region
a <
0.
To see
that this
can indeed
be done
it
is
simplest
to
pass directly
to the
high-
frequency
limit,
i.e. 02 > 02ax
so
that
w0)/02
is small for all i
=
1
...V,N.
Now,
when
lal
is
small the
stability boundary
which
passes
through
the
origin
is
given
asymptotically by
a
=
_fl2
(see,
for
example,
Jordan &
Smith
1987,
p.
257).
Moreover,
the
upper
boundary
to the
shaded
region
in
a
<
0 is known to meet
the
a
=
0
axis
at
I0fI
=
0.450.
Thus all
the
points
(2.6)
will be in the
stable
region
if
(2
1222/ 2 1
<
e/g
<
0.450,
i=
1,
...,N,
i.e.
if
/2g/oi
< e <
0.450g/0
(2.7)
for all
i
=
1,...,N.
This is
ensured
by
(2.1).
While
this
completes
the
proof,
it is worth
considering
a
more
general
and
geometric
approach
to the matter.
All
the
points (2.6)
lie on a line
through
the
origin
with
slope
-eo0/g.
Imagine,
then,
that we attach an elastic band
OP
to the
origin
in
figure
2,
stretch
it
out,
and mark
points
along
it
so that their distances from
the
origin
are in
the
proportion
o1'
:
:
...:
(o.
By choosing
e
and
w0
we
may
vary
the
length
of the
band,
and its
slope
-
eO2/g,
at
will.
Now,
if
)max
and
o)min
are
almost
equal,
so that the
marked
points
are
tightly
clustered
along
the
band,
we
may evidently
steer those
points
into
the
stable
region
in
figure
2
even
if
c1a
is not
particularly
small,
so that 02
is
not
large compared
with
2ax.
If,
on
the
other
hand,
COmax
is
substantially
greater
than
omin
-
as
will
typically
be the
case
-
we
may
still steer all the
points
into
the
stable
region,
but
only
by
moving
the
band
very
close to the a
=
0
axis in
figure
2
and
by shrinking
it
accordingly.
The
stability
criterion
(2.1)
will then
apply,
and the
region
of
stability
in
the
e-oo
plane
will
be as indicated in
figure
1.
Proc.
R. Soc.
Lond. A
(1993)
A
pendulum
theorem
241
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D.
J.
Acheson
3.
Examples
(a) Stephenson's system
inverted
Stephenson's
original
double-pendulum
analysis
was for two identical rods
of
uniform
density
and
equal length
1.
An
elementary
exercise
shows
that the
natural
frequencies
are then
Wmax,min
=
(3
+6//7)1
(g/l),
(3.1)
and
(2.1)
becomes
(1.653/o)
(g/l)
<
e/l
<
0.085.
(3.2)
The
first of these
inequalities
may
be written as
(1ew6)2
>
0.683gl, (3.3)
which is
Stephenson's
original
criterion,
derived
by
a different
method.
His formula
for the inverted
triple pendulum appears,
unfortunately,
to be
in
error;
the
theorem of
?2
gives
(1.738/0o)
(g/l)
<
6l1
<
0.03
(3.4)
as the
high-frequency stability
criterion for an inverted
triple pendulum
consisting
of three
equal heavy
rods.
(b)
The
point-mass
double
pendulum
Consider
now a double
pendulum
consisting
of two
point
masses
ml,
m2
and
two
light
rods
of
equal
length
1
(as
sketched
in
figure
1).
The normal
modes of small
oscillation about the
lower,
inherently
stable
equilibrium
position
(with
O
fixed)
have
frequencies
(g/1),
(g/l)
max
=
)
1
and
min
=
(3.5)
(1-
mi)
(1
+
mi)'
where
m
=
m2/(m
+m2),
(3.6)
ml
denoting
the
point
mass which is attached
directly
to
O.
The
high-frequency
criterion
(2.1)
for the
stability
of
the inverted
equilibrium
position
then
reduces to
V2(
)i(0
<
<
0.450(1-m).
(3.7)
For
comparison,
we show
in
figure
3
the
stability
boundaries for
this inverted
double
pendulum
obtained
by
direct numerical
integration
of the
equations
of
motion,
linearized
about
the
upward
vertical. At
high
frequencies
o0
the
stable
range
of
e/l
is
given
well
by
(3.7),
but
if
the
upper
mass
m2
is
much
smaller than the
lower
mass
m1
it is
possible
to
stabilize
the
system
with
a
comparatively
small
driving
frequency
o0
and
(3.7)
does
not then
apply.
This case
corresponds,
of
course,
to
Omax
and
tmin
being
almost
equal
(see (3.5)
and the
end
of
?2);
indeed,
when
m
=
0
the
two
coincide and
(3.7)
reduces to
the
stability
criterion for an inverted
single
pendulum.
More
typically,
rather
high
frequencies
w)
are needed to
give stability,
and
(2.1)
then
provides
a
good
approximation
to the whole
stability
region
in
the
e-w0
plane.
The
case of
equal
masses
(m
=
0.5)
is a
good
example:
(1.85/(0o)
(g/l)
<
e/1
<
0.132,
(3.8)
Proc. R. Soc.
Lond.
A
(1993)
242
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50
-
0.8
oo
40-
0.5
(gll)112
30
-
0.2
20
-
11
10-
I
I
I
,
I
,
0.1
0.2 0.3
0.4
0.5
0.6
elI
Figure
3.
Computed
linear
stability
boundaries for
the
inverted
point-mass
double
pendulum
for
four
values of m
=
m2/(ml+m2).
for
the
right-hand
stability
boundary
in
figure
3 is
indeed
almost a
straight
line
parallel
to
the
o0-axis,
save at
the
very
lowest
frequencies
giving
stability.
The
larger
the
value
of
m,
clearly,
the
smaller
we
have to
take
e
and the
higher
we
have
to take
o0,
but there
is in
theory
no
upper
limit to
the
mass
ratio
m2/m1
which
can
be
stabilized
by
vertical
oscillations of
the
support.
(c)
An
inverted
N-pendulum
We
now
apply
the
theorem
of
?2
to
a
pendulum
with N
equal
point
masses
connected
by
N
equal
light
rods of total
length
L
=
Nl.
The
natural
frequencies
of
oscillation
about the
lower
equilibrium
position
are
i
=
(/l,
i=
1,...,N,
(3.9)
where
i
denotes the ith
zero of the
Laguerre
polynomial
LN(x);
this was
shown
by
Daniel Bernoulli
and
by
Euler in
the late
1730s
(see
Cannon
&
Dostrovsky
1981).
In
the
case N
=
3 this
gives
1
=
0.64(g/l)i,
2
=1.51(g/l)i,
()3=2.51(g/l),
(3.10)
and on
taking
the
greatest
and
least of
these we find
that
the
criterion
(2.1)
becomes
(2.21/0l)(g/l)i
<
e/l
<
0.07.
(3.11)
For
comparison,
figure
4
shows the
stability
boundary
obtained
by
numerical
integration
of the
linearized
equations
of motion.
When
N
is
large
we can
use the
asymptotic
expression
$
X2/4N,
where
Xi
denotes
the ith
zero of the
Bessel
function
J0(x).
With
X1
2.4 and
XN
(N-)
)n
we
then
have
min-
1.2(glL)
()
*(N- 1)
n(gL)H
Omin
--
Omax
and
therefore
1.18
g
e
0.182
1.18(q-
ie<
.<
1)-82
(3.12)
NO
)
L
4(N-
as the criterion
for the
stability
of
this
particular
inverted
N-pendulum,
when N
is
large.
Proc.
R.
Soc. Lond. A
(1993)
A
pendulum
theorem
243
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40-
30-
\
S
(g/1)1/2
20
-
10-
I
I
I I
0.05
0.1
0.15 0.2
eli
Figure
4.
Computed
regions
of linear
stability
for the
upside-down N-pendulum,
for N
=
1,2,3,
in
the case
when the
various
(point)
masses are
equal.
Suppose
finally
that L is
fixed.
Then,
the
larger
the value
of
N,
the
smaller
we
must
take
e,
and,
consequently,
the
larger
we
must take
to
if
stability
is to
be
achieved.
The limit
N-* oo
for
fixed L
corresponds
to a
perfectly
flexible
string,
and
was
of
major
interest
in
the
original
investigations
of Bernoulli and
Euler. The
region
of
stability
of the
inverted state
vanishes
altogether
in
this
limit,
so we
cannot
perform
the Indian
rope
trick.
4.
Concluding
remarks
While
the
theorem of
this
paper
concerns linear
stability,
numerical
integrations
of the
full
equations
of motion
have
also been carried
out,
together
with
laboratory
experiments
(Acheson
&
Mullin
1993).
These further
studies
show the
stability
of
these inverted
states to
be
remarkably
robust,
at
least in the
cases N
=
1,2,
3.
In
particular,
we
have seen an
inverted
triple pendulum
recover from
very
severe
initial
disturbances and
gradually
wobble
back to the
upward
vertical.
I am
grateful
to Peter
Clifford,
Don
Fowler,
Janet
Mills,
Tom Mullin
and
John Roe for
helpful
discussions and
encouragement.
References
Acheson,
D.
J. &
Mullin,
T.
1993 On
upside-down
pendulums.
Nature,
Lond.
(Submitted.)
Cannon,
J.
T.
&
Dostrovsky,
S. 1981
The
evolution
of dynamics:
vibration
theory
from
1687 to
1742.
Springer-Verlag.
Hsu,
C.
S.
1961
On
a
restricted class of
coupled
Hill's
equations
and some
applications.
J.
appl.
Mech.
28,
551-556.
Jordan,
D.
W. &
Smith,
P.
1987 Nonlinear
ordinary
differential
equations,
2nd
edn.
Oxford:
Clarendon
Press.
Kalmus,
H. P.
1970
The
inverted
pendulum.
Am. J.
Phys.
38,
874-878.
Landau,
L.
D. &
Lifshitz,
E. M.
1976
Mechanics,
3rd
edn.
Pergamon.
Leiber,
T.
&
Risken,
H.
1988
Stability
of
parametrically
excited
dissipative
systems.
Phys.
Lett.
A
129,
214-218.
Lowenstern,
E.
R.
1932 The
stabilizing
effect
of
imposed
oscillations of
high
frequency
on a
dynamical
system.
Phil.
Mag.
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Otterbein,
S.
1982
Stabilisierung
des
n-Pendels und der
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Analysis
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Proc. R. Soc.
Lond. A
(1993)
D. J.
Acheson
244
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All use subject to JSTOR Terms and Conditions
A
pendulum
theorem
245
Stephenson,
A.
1908a On a new
type
of
dynamical
stability.
Mem.
Proc. Manch.
Lit. Phil.
Soc.
52
(8),
1-10.
Stephenson,
A.
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Phil.
Mag.
15,
233-236.
Stephenson,
A.
1909 On induced
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Mag.
17,
765-766.
Received 28
May
1993;
accepted
16
July
1993
Proc. R. Soc. Lond. A
(1993)
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Discussion

This paper is from 1993! I had thought it was much older on my first read through. Near $\alpha = 0$, $ \sqrt{-2\alpha} <\beta <0.450$. Since $\alpha_i=\frac{-w_i^2}{w_0^2}$, $ \sqrt{2\frac{w_i^2}{w_0^2}} <\beta <0.450$ Also the larger the value of $N$, the smaller the value of $l$ because $L$ is a constant, which means that it will push $w_0$ and $\epsilon$ out of the stability zone. To prove this we are going to use the Lagrangian formulation and will start with the $N=2$ case (two coupled pendulums). ![](http://i.imgur.com/FPrkfpN.png?1) The Lagrangian $L$ of the system is given by $L=T-V$, where $T$ and $V$ are the kinetic energy and the potential energy of the system. $$ T=\frac{1}{2}m_1(x_1^2+y_1^2)+\frac{1}{2}m_2(x_2^2+y_2^2) \\ V= m_1gy_1+m_2gy_2 $$ For small angles ($\theta_i\approx 0 \rightarrow \cos(\theta_i)\approx 1$ and $\sin(\theta_i)\approx \theta_i$ ) we can write $T$ and $V$ as a function of the angles: $$ T\approx \frac{1}{2}m_1l_1^2\dot \theta_1^2+\frac{1}{2}m_2l_1^2\dot \theta_1^2+\frac{1}{2}m_2l_2^2\dot \theta_2^2+m_2l_1l_2\dot \theta_1\dot \theta_2 \\ V\approx \frac{1}{2}(m_1gl_1\theta_1^2+m_2gl_1\theta_1^2+m_2gl_2\theta_2^2) $$ If we calculate the Lagrange equations $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot \theta_i}\right)=\frac{\partial L}{\partial \theta_i}$ we end up with a system of coupled differential equations of the form: $$ \left( \begin{array}{ccc} \frac{l_1}{g} & \frac{m_2l_2}{(m_1+m_2)g} \\ \frac{l_1}{g} & \frac{l_2}{g} \\ \end{array} \right)\ \left( \begin{array}{ccc} \ddot \theta_1 \\ \ddot \theta_2 \\ \end{array} \right)\ = - \left( \begin{array}{ccc} \theta_1 \\ \theta_2 \\ \end{array} \right)\ $$ It is helpful to find a new basis in which the matrix $M$ is diagonal. The basis that diagonalizes $M$ is what we call the normal coordinates. In this basis we end up with a set of homogeneous second order linear differential equations, each of which corresponds to the motion of a simple harmonic oscillator (SHO). $$ \left( \begin{array}{ccc} \frac{1}{w_1^2} & 0 \\ 0 & \frac{1}{w_2^2} \\ \end{array} \right)\ \left( \begin{array}{ccc} \ddot X_1 \\ \ddot X_2 \\ \end{array} \right)\ = - \left( \begin{array}{ccc} X_1 \\ X_2 \\ \end{array} \right)\ $$ where $X_1$,$X_2$ are the normal coordinates and $w_1$,$w_2$ are the natural frequencies. Finally we get $$ \ddot X_i+w_i^2 X_i = 0 \\ i=1,2 $$ The generalization for $N$ pendulums is a similar process that just involves calculating a more complex lagrangian and dealing with a bigger matrix, but in the end we always end up with a diagonalized matrix if we use the normal coordinates. As we can see in this video by Steve Mould the chain of pendulums can be balanced if the pivot is vibrated up and down by a small enough amount and at a high enough frequency. [![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") In 1908 Stephenson found that the upper vertical position of the pendulum might be stable when the driving frequency is fast. Yet no one could scientifically answer this unusual and counterintuitive phenomenon until almost fifty years later, a Russian physicist Pyotr Kapitza successfully analyzed the stability of this system by separating the motions into fast and slow, and introduced a new concept called Effective Potential (to read more about Kapitza's demonstration click [here](https://en.wikipedia.org/wiki/Kapitza%27s_pendulum)) Once Kapitza found the effective potential, he then use it to analyze the motion and stability of the system. Systems always tend to move towards positions of minimum potential energy, so by finding the minimum values of the potential energy expression, we can find the stability points. In particular Kapitza found that the vertical position was stable. To prove that the natural frequencies are proportional to $g$ we start again with the double pendulum and with the matrix: $$ \left( \begin{array}{ccc} \frac{l_1}{g} & \frac{m_2l_2}{(m_1+m_2)g} \\ \frac{l_1}{g} & \frac{l_2}{g} \\ \end{array} \right)\ \left( \begin{array}{ccc} \ddot \theta_1 \\ \ddot \theta_2 \\ \end{array} \right)\ = - \left( \begin{array}{ccc} \theta_1 \\ \theta_2 \\ \end{array} \right)\ $$ Since we expect oscillatory solutions let's start with the following ansatz: $$ \theta_1 = A_1e^{iwt} \\ \theta_2 = A_2e^{iwt} $$ If we now substitute $\theta_1,\theta_2$ with the expressions $$ \left( \begin{array}{ccc} \frac{l_1w^2}{g}+1 & \frac{m_2l_2w^2}{(m_1+m_2)g} \\ \frac{l_1w^2}{g} & \frac{l_2w^2}{g}+1 \\ \end{array} \right)\ \left( \begin{array}{ccc} A_1 \\ A_2 \\ \end{array} \right)\ = 0 $$ For this to be true for any amplitude, the matrix on the left must be singular i.e. must not be invertible, such that one cannot multiply both sides of the equation by the inverse, leaving the right matrix equal to zero. It follows that the determinant of the matrix must be equal to 0, so: $$ \left(\frac{l_1w^2}{g}+1 \right)\left(\frac{l_2w^2}{g}+1 \right)-\frac{l_1w^2}{g}\frac{m_2l_2w^2}{(m_1+m_2)g}=0 $$ And finally: $$ w_1=\frac{\sqrt{g/l}}{\sqrt{1-\sqrt m}} \\ w_2=\frac{\sqrt{g/l}}{\sqrt{1+\sqrt m}} $$ which means that $w_i^2 \propto g$ This paper was written by D.J. Acheson while he was trying to give a new twist to an old problem in dynamics, first studied by Daniel Bernoulli in 1738. He had considered a hanging chain of several linked pendulums, all suspended from one another, and discovered various different modes of oscillation. 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. 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. Although is theorem doesn't hold for the case $N\rightarrow \infty$ (a perfectly flexible string can be approximated by a series of infinite pendulums all linked together), thus not explaining the Indian rope trick it is still a surprising mathematical result that has a certain magic on its own! ![](http://www.stageillusion.com/news/wp-content/uploads/Indian-Rope-Trick.jpg)