Wave speed of a hanging rope

Question

Let us consider a homogeneous rope hanging from the ceiling. I will call the vertical direction $x$ and the horizontal displacement $y$. When we apply the second Newton's Law to a portion of mass $\Delta m$ and proceed in the same way we do for a horizontal string we get $$\mu\frac{\partial^2 y(x,t)}{\partial t^2}=\frac{\partial}{\partial x}\left[T(x)\frac{\partial y(x,t)}{\partial x}\right].$$ The difference now is that the tension is not constant. Defining $x=0$ at the free end of the rope and orienting it upwards we have the tension $T=\mu gx$. Therefore, $$\frac{\partial^2 y}{\partial t^2}=gx\frac{\partial^2y}{dx^2}-g\frac{\partial y}{\partial x}.$$

So, my issue is that I have seen in a couple of physics books (for instance) that the wave speed of this rope is simply $$v=\sqrt{\frac{T}{\mu}}=\sqrt{gx}.$$ In my opinion these books are either:

i) Cheating the students: they know it is wrong but assume it is right just to show some nice features (the speed would increase as the wave goes upwards).

ii) Had misconception assuming that the usual wave equation are still valid.

iii) Is doing some obscure approximation, which is another cheat if you do not reveal it.

My questions are:

1) Is there an expression for the wave speed for the "wave equation" above?

2) Is there an approximation leading to the wave speed given above?

Answer 1

@diracology writes in a comment above that the solutions are $$y(x,t)=A\cos(\omega_n t+\phi)J_0(2\omega_n\sqrt{x/g})$$ where $\omega_n$ are "certain discrete frequencies".

For small $x$ (up to the first zero), $$J_0(x) \sim \cos\frac{x}{\sqrt 2}$$ and for higher $x$, $$J_0(x) \sim \sqrt{\frac{2}{\pi x}} \cos(x-\frac{\pi}{4}).$$

In the latter case we thus get $$\begin{align} y(x,t) & \sim A\cos(\omega_n t+\phi) \sqrt{\frac{2}{\pi 2\omega_n\sqrt{x/g}}} \cos(2\omega_n\sqrt{x/g}-\frac{\pi}{4}) \\ & = A \sqrt{\frac{\sqrt{g}}{\pi\omega_n\sqrt{x}}} \cos(\omega_n t+\phi) \cos(\frac{2\omega_n}{\sqrt g}\sqrt{x}-\frac{\pi}{4}) \end{align}$$

Using the formula $\cos u \cos v = \frac12 (\cos(u+v) + \cos(u-v))$ this can be rewritten as $$\begin{align} y(x,t) & \sim A \frac12\sqrt{\frac{\sqrt{g}}{\pi\omega_n\sqrt{x}}} \left(\cos(\omega_n t+\phi + \frac{2\omega_n}{\sqrt g}\sqrt{x}-\frac{\pi}{4})+\cos(\omega_n t+\phi - \frac{2\omega_n}{\sqrt g}\sqrt{x}+\frac{\pi}{4})\right) \\ & = A \frac12\sqrt{\frac{\sqrt{g}}{\pi\omega_n\sqrt{x}}} \left(\cos(\omega_n(t+ \frac{2}{\sqrt g}\sqrt{x}) + \phi_1)+\cos(\omega_n(t-\frac{2}{\sqrt g}\sqrt{x})+\phi_2)\right) \end{align}$$

The wave speed is given by constant phases, i.e. by calculating $dx/dt$ where $$t \pm \frac{2}{\sqrt g}\sqrt{x} = t_0,$$ where $t_0$ is some constant.

This gives $$x = \frac{g}{4} (t-t_0)^2$$ so $$\frac{dx}{dt} = \frac{g}{2} (t-t_0) = \mp\sqrt{gx}.$$

Thus, the wave speed is $$u = \sqrt{gx}$$ when $x$ is not too close to $0$.

Answer 2

2) Prof Carl E Mungan of the US Naval Academy derives a wave speed of $\sqrt{gx}$ by assuming the length of a pulse travelling along the rope is much smaller than the length of the rope. He states that this relation has been verified in experiments (reference 2). http://www.usna.edu/Users/physics/mungan/_files/documents/Scholarship/HangingPulse.pdf

Answer 3

If you have a pulse with short wavelength $\Delta x$ passing by the position $x_0$ and $\Delta x \ll x_0$, $$\frac{\partial^2 y}{\partial t^2}=gx_0\frac{\partial^2y}{dx^2}+g(x-x_0)\frac{\partial^2y}{dx^2}-g\frac{\partial y}{\partial x}\approx gx_0\frac{\partial^2y}{dx^2}$$ where we neglect the term involving $(x-x_0)$ since the pulse is within $\Delta x$ of $x_0$, and we neglect the first order derivative in comparison with the first term since $$x_0\frac{\partial^2y}{dx^2}\sim \frac{x_0}{\Delta x}\frac{\partial y}{dx}.$$

So for a short wavelength pulse near $x_0$ this equation is approximately the wave equation with speed $\sqrt{g x_0}.$

Of course this was an approximation and if the wavelength gets too short other physical effects (e.g. links on a chain) might come into play.