Fall 2007 19 Misc

Small transverse oscillations u(z, t) of an inextensible but otherwise perfectly flexible cable suspended at one end and hanging under gravity are described by the equation

(1)
\begin{align} \frac{\partial^2 u}{\partial t^2} = \frac{\partial}{\partial z}\left(gz\frac{\partial}{\partial z} \right),\;\;u(L,t)=0 \end{align}

where L = const is the length of the cable and 0 < z < L is the vertical coordinate.

(a) Following Bernoulli (1732), one can seek eigenmodes of the system in the form $u(z,t)=\psi(z)\cos{\omega t}$, where $\psi$ is given by Taylor series,

(2)
\begin{align} \psi(z)=\sum_{j=0}^{\infty}c_j z^j. \end{align}

Find all the coefficients cj assuming $\omega$ is given. Sketch the expected behavior of function $\psi(z)$ for a few first eigenmodes.

(b) A rigorous bound on the smallest eigenfrequency $\omega$ can be found from comparison with that of a perfectly rigid cable (which is a pendulum). Do so and explain whether this is a lower or an upper bound.

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(a) Substituting the power series into the equation of motion and equating the coefficients for equal powers, we get

(3)
\begin{align} c_{j+1}=-\frac{1}{(j+1)^2}\frac{\omega^2}{g}c_j, \end{align}

so that

(4)
\begin{align} c_{j}=\frac{(-1)^j}{(j!)^2} \left( \frac{\omega^2}{g} \right)^j c_0, \;\;j \geq 1. \end{align}

The arbitrary coefficient c0 controls the overall amplitude of the harmonic oscillations. We can set it to unity. The sketches of a few eigenmodes are shown below. According to the general theory of the Sturm-Liouville problem, the lowest frequency mode $\psi_0$ is nodeless (except z = L, of course); the next one, $\psi_1$, has one node, the third one, $\psi_2$, has two nodes, etc.

(b) The eigenfrequency of a rigid cable is

(5)
\begin{align} \omega_{R}=\sqrt{\frac{MgL}{2I}}=\frac{3g}{2L}\approx 1.22 \sqrt{\frac{g}{L}}. \end{align} This is a strict upper bound: $\omega < \omega_R$. Indeed, our Sturm-Liouville eigenvalue problem obeys a variational principle. The oscillations of the rigid cable, which are described by

(6)
\begin{align} \psi_R(z)=L-z, \end{align}

can be considered a trial function. Hence, $\omega_R^2$ is in fact a variational estimate of $\omega^2$.

Note: Those who are familiar with special functions would recognize that $\psi(z) = c_0J_0(2\omega\sqrt{z/g})$. Accordingly, the exact result for the lowest eigenfrequency is $\omega=(r_1/2)\sqrt{g/L}\approx1.20\sqrt{g/L}$, where $r_1\approx2.40$ is the first root of the Bessel function J0. As we can see, $\omega_R$ is within 2% of this value.

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