classical mechanics - 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.)

Answer

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