# What are the normal modes of a vertical rope?

Closely related to this question on traveling waves on a hanging rope, I would also like to know what the normal modes are on a rope that hangs vertically, fixed at both ends.

Tension in the rope increases with height, so I expect that the wavelength of standing waves decreases as we move up, and nodes are clustered somewhat more towards the top of the rope than the bottom. Is this correct?

The wave equation I found was (note: important typo corrected)

$$\frac{\partial^2{x}}{\partial t^2} = g \frac{\partial x}{\partial y} + \frac{T_0+\lambda g y}{\lambda} \frac{\partial^2 x}{\partial y^2}$$

so mathematically, my question is about how to find the eigenfunctions of this equation with the boundary conditions $x(0) = x(h) = 0$ with $h$ the height of the ceiling.

(I was able to locate a reference, but don't have access to it.)

• Does this link work? If not, I'll upload the file somewhere. – Raskolnikov Dec 20 '10 at 13:32
• @Raskolnikov No, that link is behind a pay wall. However, if I make a trip to the university library today I can access the article from there. – Mark Eichenlaub Dec 20 '10 at 14:12
• @Keenan You are correct that I left out a factor of y, but that doesn't make it nonlinear. I'll fix the difEq. – Mark Eichenlaub Dec 21 '10 at 0:55
• Your differential equation is incorrect. There should be a factor of y in the d^2x/dy^2 term, so it's not just a harmonic oscillator equation with constant coefficients. (Still linear though, you're right.) – Keenan Pepper Dec 21 '10 at 1:57

I want to try to give a solution to your equation using a separation ansatz of the form
$$x(y,t) = T(t)\cdot Y(y)$$

We find $$\frac{T''}{T} = \frac{1}{Y}\left( gY' + \left(\frac{T_0 + \lambda g y}{\lambda}\right) Y'' \right)$$

Both sides do depend on different variables, hence the given equation must equal a constant, say $-\omega_n^2$.

We see directly that the fundamental system for $T$ can be stated as oscillations (if this notation is convenient for you): $$eig(T) = \{ \sum_n \left[ e^{\mathrm{i}\omega_n t}, e^{-\mathrm{i}\omega_n t} \right]\}$$

## New solution for Y

The differential equation for $Y$ now takes the form $$\frac{\omega_n^2}{g} Y + Y' + \left(\frac{T_0}{\lambda} + y\right) Y'' = 0$$

and it is obvious that one should first of all make a substitution of the form $$z = \frac{T_0}{\lambda} + y$$

Such that we find with $Y(y) = Z(z)$

$$\frac{\omega_n^2}{g} Z + Z' + z Z'' = 0$$

The solution to this equation are indeed Bessel functions

$$eig(Z) = \{J_0\left(2\omega_n \sqrt{z/g} \right) , Y_0\left( 2\omega_n \sqrt{z/g}\right) \}$$

So, finally

$$x{(t,y)}=\sum_n e^{\{+,-\}\mathrm{i}\omega_n t}\left[ a_n J_0 \left( 2\omega_n \sqrt{\frac{T_0 + \lambda y}{\lambda g}} \right) + b_n Y_0 \left( 2\omega_n \sqrt{\frac{T_0 + \lambda y}{\lambda g}} \right) \right]$$

The overall solution is then given due to the starting and boundary conditions you specified. Since this is a discussion on its own, I will leave this for the other thread :)
Sincerely

Robert

• @Robert Thanks! I remember this method from my quantum mechanics course now that you've written it out. All your calculations look correct to me. – Mark Eichenlaub Dec 20 '10 at 14:11
• @Mark: You're welcome :) As you can see, this was straight forward. Now, actually, comes the tricky part: the interpretation of the solution at hand. – Robert Filter Dec 20 '10 at 14:16
• This solution is incorrect because the original differential equation is incorrect. The eigenfunctions are not sinusoids, but Bessel functions. – Keenan Pepper Dec 20 '10 at 23:11
• @Keenan You are right, there was a typo. @Robert Sorry - that made you do all the work on the wrong equation. – Mark Eichenlaub Dec 21 '10 at 1:00
• @Mark: I updated my answer due to the new equation. I hope I might find the time to do as well for the other thread :) Greets – Robert Filter Dec 21 '10 at 10:41

The normal modes are related to Bessel functions.