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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

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)Find all the coefficients *c _{j}* 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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Answer

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

(3)so that

(4)The arbitrary coefficient *c _{0}* 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)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

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 *J _{0}*. As we can see, $\omega_R$ is within 2% of this value.

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