3.1 The Hanging Cable

3.1.1 The Equation of Motion

The continuous limit of the hanging chain considered in Section 1.4.2 is a hanging cable or rope of linear mass density $\mu$ .

In the limit of small displacements from equilibrium, the tension in the cable acts mainly in the vertical direction and is given by

$\begin{displaymath} T(y) = \mu g y \end{displaymath}$

(3.1)

where

is the vertical position relative to the free end of the cable.

Any net transverse force on an infinitesimal mass element of the cable

is due to a difference

in the

components of the tensions in the cable just above and just below the cable element, due to a bending of the cable element through an angle $\theta$ . Newton's second law yields

$\begin{displaymath} dm \ddot{x} = dT_x = d(\mu g y \sin \theta) \end{displaymath}$

(3.2)

which can be approximated via $\tan \theta = \frac{dx}{dy} \approx \sin \theta$ by

$\begin{displaymath} dm \ddot{x} = d(\mu g y \frac{dx}{dy}) \end{displaymath}$

(3.3)

Differentiating Eq. 3.3 with respect to

yields

$\begin{displaymath} \frac{dm}{dy} \ddot{x} = \mu g \frac{d}{dy} y \frac{dx}{dy} \end{displaymath}$

(3.4)

Recognizing $\frac{dm}{dy}$ as the linear mass density $\mu$ of the cable, we obtain the equation of motion of the cable

$\begin{displaymath} \ddot{x} = g \frac{d}{dy} y \frac{dx}{dy} \end{displaymath}$

(3.5)

3.1.2 The Solutions : Bessel Functions of Order Zero

Like Eq. 1.55, the equation of motion for waves on a string, Eq. 3.5 is a second order linear partial differential equation, and we will try solving it using the method of separation of variables.

You will show in homework that the product solution

$\begin{displaymath} x(y,t) = f(y) g(t) \end{displaymath}$

(3.6)

leads rapidly to the separation of Eq. 3.5 into

$\begin{displaymath} \ddot{g}(t) = -\omega^2 g(t) \end{displaymath}$

(3.7)

which gives the usual time dependence, and

$\begin{displaymath} \frac{d}{dy} y \frac{d}{dy} f(y) = -\frac{\omega^2}{g} f(y) \end{displaymath}$

(3.8)

You will also show that with the ``enlightened'' substitution

$\begin{displaymath} u = \sqrt{\alpha y} \end{displaymath}$

(3.9)

and with

$\begin{displaymath} \alpha = \frac{4 \omega^2}{g} \end{displaymath}$

(3.10)

that Eq. 3.8 becomes

$\begin{displaymath} \frac{d^2}{du^2} f(u) +\frac{1}{u}\frac{d}{du} f(u) + f(u) = 0 \end{displaymath}$

(3.11)

Eq. 3.11 is Bessel's differential equation of order zero. It has two independent solutions

, the zeroth order Bessel function, and

, the zeroth order Neumann functions.

goes to $-\infty$ at

, so it does not contribute to the solutions for the hanging cable.

Figure 3.1: The zeroth order Bessel function

and its first six roots.

$\scalebox{0.7}{ \includegraphics{Jo.eps} }$

Root	u
1	2.404825558
2	5.520078110
3	8.653727913
4	11.79153444
5	14.93091771
6	18.07106397

The Bessel and Neumann functions cannot be simply written in terms of other elementary functions, but they have been extensively studied and their properties are readily available in the literature.^3.1 Many scientific and mathematics software packages include them as mathematical functions like sine and cosine.

To obtain standing wave solutions for the hanging cable, we apply the boundary condition

, that the end of the cable at

is fixed. This implies that the argument $\sqrt{\alpha L}$ of the Bessel function must be one of the infinite set of roots of

Just as for the stretched string, we have a countably infinite set of standing wave modes. For the

th standing wave mode,

$\begin{displaymath} \alpha_M = \frac{r_M^2}{L} \end{displaymath}$

(3.12)

where

is the

th root of

Eq.'s 3.10 and 3.12 combine to give

$\begin{displaymath} \omega_M = \frac{r_M}{2} \sqrt{\frac{g}{L}} \end{displaymath}$

(3.13)

**Figure 3.2:** The first 4 standing wave modes of a hanging cable.
$\scalebox{0.7}{ \includegraphics{cable.eps} }$

The general solution, combining the time dependent and position dependent parts is

$\begin{displaymath} x(y,t) = Re\left\{ C J_o\left(\sqrt{\frac{y}{L}} r_M\right) e^{-i (\omega_M t - \phi)} \right\} \end{displaymath}$

(3.14)

Shapes of the first four standing wave modes are shown in Fig. 3.2.

The resemblance between Fig.'s 3.2 and Fig. 1.7 showing the first three normal modes of a 20-link hanging chain is expected. However, the two systems are not identical. Fig. 1.8 shows the progression of the discrete system toward the cable solution with increasing link density for the second mode. The solid curve in that figure is the cable solution of Eq. 3.14. Note that the 100 link chain is already quite close to the continuous limit.

Demo : a driven, hanging, beaded chain 1.0 m long. The predicted frequencies (Eq. 3.13) and periods for the first three modes are

$\begin{eqnarray*} \omega_1 = 3.76 \mathrm{\frac{rad}{s}}, T_1 = 1.67 \mathrm{s}... ... \omega_3 = 13.54 \mathrm{\frac{rad}{s}}, T_3 = 0.46 \mathrm{s} \end{eqnarray*}$

Updated Wed Jan 18 09:51:28 2006