Examples: Hanging chain
The hanging chain problem was solved
around the year 1732 by Daniel Bernoulli
(not Bernoulli again!) You would think it is
a done deal, solved 270+ years ago…. but in
the process of solving the problem, ol’ DB
discovered a class of functions known as
Bessel functions. So historically, this
problem is of great importance.
It is also a great example of vibrational
modes.
Who needs a stamp when you have an effect named for you?
The vertical problem
A flexible uniform chain or cable of length L and constant linear density (mass/unit
length) of ρ gm/cm is fixed at the
upper end (which we will call
x = L). We let the x-axis be
vertical, measured up from the
equilibrium position of the free
end of the chain. The function
u(x,t) represents the displacement
function for a point x on the
chain. Positive values of u are to
the right of the vertical
equilibrium position. The
displacements are small
compared with the length of the
chain; we do not need to consider
the difference between distances
measured along the chain and
distances measured along x, ie
| x 2 + u 2 − x |≈ 0 .
The chain’s equilibrium position (hanging vertically) is due to the gravitational force, and
the gravitational acceleration g may be measured in m/sec2. Making a few
simplifications (no internal friction, the chain is perfectly flexible, no air resistance), we
can determine the subsequent motion of the chain.
The tension in the chain is due to the weight below point x, w(x) = ρgx, and the
difference in the horizontal components of the tension at the ends of a small interval Δx
of chain is the accelerating force.
For any displacement of angle α, the restoring force (which tends to accelerate the chain
back towards equilibrium) is F(x) = W sin α ~ W ux. The difference in force between
points on the change at x and
x + Δx is thus ΔF = Δx(W ux)x.
From Newton’s 2nd Law, f = ma = m utt :
Δx ρ g[x ux]x = ρ Δx utt
Or utt = g(ux + xuxx)
The vertical solution
Separate the variables with the solution function of the form u(x,t) = F(x)G(t).
(t ), (super-dots are time derivatives)
utt = F ( x)G
u x = G (t ) F '( x)
So
u xx = G (t ) F ''( x)
(t ) = gxG (t ) F ''( x) + gG (T ) F '( x)
F ( x)G
(t ) gxF ''( x) + gF '( x)
G
=
G (T )
F ( x)
As always, these are each also equal to a separation constant, which we will set to –ω2.
The time function is thus just a cosine of the form G(t) = cos(ωt + φ). The angular
frequency ω has units of time-1. The chain will probably begin its oscillation at
maximum displacement (initial velocity = 0), so that the phase angle φ will be 0.
ω2
The ODE for x is “just” xF ''( x) + F '( x) +
F ( x) = 0 , which can also be written in a
g
“collapsed” form,
.
d
dF ω 2
(x
)+
F =0
dx dx
g
This ODE is an ugly form of Bessel’s equation, which has a typical form:
x2
d2y
dy
+ x + ( x2 − p2 ) y = 0
2
dx
dx
where p is the order of the Bessel function (in our case p = 0).
The coefficient of F’’(x) is supposed to be x2, but we can’t just multiply our ODE by x
because that wouldn’t put an x2 in front of F(x). So we make a clever substitution by
letting z2 = 4x/g and seek a transform of our equation into one of Bessel form.
This will take a few steps:
a. Use a Chain Rule approach (
dy dy dz
dF
) to find an expression for
in
=
dx dz dx
dx
dF
; substitute accordingly.
dz
d
d
b. Find an expression for the operator
in terms of
-- again, using the
dx
dz
Chain Rule.
terms of g, z and
c. Take the indicated derivative and simplify to obtain
d 2 F 1 dF
+
+ ω2F = 0
dz 2 z dz
d 2F
dF
d. Multiply the result by z , obtaining the desired z
+z
+ z 2ω 2 F = 0
2
dz
dz
2
2
x
and A is just
g
the maximum amplitude. The symbol J0 stands for the “zeroth order Bessel function.”
The solution to this ODE is of the form F ( z ) = AJ 0 (ω z ), where z = 2
Plot of the first four orders of Bessel
functions. In this case, we only have J0
to worry about.
Boundary conditions require that the full solution
x
u ( x, t ) = F ( x)G (t ) = AJ 0 (2ω
) cos(ωt + φ )
g
has u(x,0) = A and thus φ = 0 as we guessed above. In addition, we require F(L) = 0 (the
upper end of the chain cannot move). Our hanging chain will therefore only vibrate at
specific “modes,” which correspond with frequencies that match the zeros of this
particular Bessel function for the given value of L. We can obtain these zeros quite
readily, thanks to a handy built-in Mathematica function:
Use these values to solve for the first three possible values of ω, taking L = 1.
Our complete solution is u ( x, t ) = AJ 0 (2ωn
x
) cos ωnt , with values of ω dependent on
g
L
and the mode number n, corresponding to the specific zero of J0 we use. You
g
L
, so this
g
result is not completely a surprise. However, there aren’t any higher modes of a simple
pendulum (which is why it is simple).
should recall from Physics that the period of a simple pendulum, T = 2π
It is also possible that the chain has some initial shape and some initial velocity; these
would provide initial conditions u(x,0) = f(x) and ut(x,0) = g(x).
Who was this Bessel guy?
See http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Bessel.html
Well don’tcha know, he got his pic on a stamp!
As depicted in the stamp, there are all kinds of
shapes and flavors of Bessel functions, which
show up in astrophysics and waves in cylindrical
coordinates, among other cool places.
A Bessel function can also be written as an infinite series:
x
( )2n+ p
J p ( x) = ∑ (−1) n 2
.
n !(n + p )!
n=0
∞
( 2) − ( 2) + ( 2)
4
x
With p = 0, the first few terms are just J 0 ( x) = 1 − ( x ) 2 +
2
4
x
62
6
x
242
8
− ...
More details: http://www.math.hmc.edu/~dyong/papers/strings.pdf
animations: http://www.kettering.edu/~drussell/Demos/HangChain/HangChain.html
Problem
Find the first three modes of a hanging chain of length L = 2 meters. Produce an
animation of the motion of this chain when these three modes are added according
to M0 + .5M1 + .25 M2