Wave speed of a hanging rope

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?

• Every book that pretends this is 100% false. A short look for resources on the internet shows that the solutions should be the Bessel functions of order zero, which, despite the rustiness of my math, sounds about right. It seems that this problem was already solved by Bernoulli and Euler in the 18th century... which really leaves no excuses on the table. – CuriousOne May 11 '16 at 0:19
• @CuriousOne Yes, the solutions are of the type $y(x,t)=A\cos(\omega_n t+\phi)J_0(2\omega_n\sqrt{x/g})$, where $\omega_n$ are certain discrete frequencies. However I have not found anything about the speed of the wave/pulse. – Diracology May 11 '16 at 0:21
• @Diracology, it appears that you have seen the free response questions on the AP Physics 1 exam. Those questions have already been thoroughly vetted by the professors associated with APCentral. – David White May 11 '16 at 1:22
• @DavidWhite I don't know about AP Physics. The subject of these question came about when I was preparing a lecture. I colleague told me about the interesting example of calculating the time a pulse would take to go upwards a hanging rope. Then I realized that a couple of undergrad physics books are taking this wrong (I suppose). – Diracology May 11 '16 at 1:29
• @Diracology, thanks for the prompt reply. The problem you are asking about was tested on 5/03/16 during the AP Physics 1 exam, for high school students across the country, and internationally. The conversation that you had with your colleague may have been prompted by this question. If not, I apologize for posting my original response. – David White May 11 '16 at 1:56

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.

@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 $0$.

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

• Actually, if you read the paper in detail, he derives that the velocity of the pulse varies along the rope, just as one would assume. While one can derive an average velocity from this, it is not correct to say that the average velocity is the wave velocity. – CuriousOne May 11 '16 at 4:36
• @CuriousOne : That is what I've written : that wave speed varies along the rope as sqrt(gx) approx. This is not an average wave speed; I have not claimed that it is, neither has Prof Mungan. Diracology's question does not ask for an average. – sammy gerbil May 11 '16 at 12:27