Solution: Using the indicated substitution~. find a general

Chapter 5, Problem 5.1.98

(choose chapter or problem)

Modeling a Vibrating Cable (Fig. 108). A flexible cable, chain, or rope of length L and density (mass per unit length) \(\rho\) is fixed at the upper end (x = 0) and allowed to make small vibrations (small angles \(\alpha\) in the horizontal displacement u(x, t), t =  time) in a vertical plane.

(a) Show the following. The weight of the cable below a point x is \(W(x)=\rho g(L-x)\). The restoring force is \(F(x)=W \sin \alpha \approx W u_{x}, u_{x}=\partial u / \partial x\). The difference in force between x and \(x+\Delta x\) is \(\Delta x\left(W u_{x}\right)_{x}\). Newton's second law now gives

\(\rho \Delta x u_{t t}=\Delta x \rho g\left[(L-x) u_{x}\right]_{x}\).

For the expected periodic motion \(u(x, t)=y(x) \cos (\omega t+\delta)\) the model of the problem is the ODE

\((L-x) y^{\prime \prime}-y^{\prime}+\lambda^{2} y=0 . \quad \lambda^{2}=\omega^{2} / g\).

(b) Transform this ODE to \(\ddot{y}+s^{-1} \dot{y}+y=0\), \(\dot{y}=d y / d s, s=2 \lambda-=^{-1 / 2}, z=L-x\), so that the solution is

\(y(x)=J_{0}(2 \omega \sqrt{(L-\bar{x}) / g})\) .

(c) Conclude that possible frequencies \(\omega / 2 \pi\) are those for which \(s=2 \omega \sqrt{L / g\) is a zero of \(J_{0}\). The corresponding solutions are called normal modes. Figure 108 shows the first of them. What does the second normal mode look like? The third? What is the frequency (cycles/min ) of a cable of length 2m? Of length 10 m?

Text Transcription:

rho

alpha

W(x) = rho g(L - x)

F(x) = W sin alpha approx Wu_x, u_x = partial u/partial x

x + Delta x

Delta x (Wu_x)_x

rho Delta xu_tt = Delta x rho g[(L - x) u_x]_x

u(x, t) = y(x) cos (omega t + delta)

(L - x) y “ - y ‘ + lambda^2 y = 0,     lambda^2 = omega^2/g

ddot{y} + s^{-1} dot{y} + y =0

dot{y} = dy/ds, s = 2 lambda z^{-1/2}, z = L - x

y(x) = J_0(2 omega sqrt {(L - bar{x})/g)

omega/2pi

s = 2 omega sqrt{L /g

J_{0}