# Will a wave packet undergo dispersion when traveling down a hanging rope?

Suppose I tie one end of a rope to my ceiling and the other end to a spot on my floor directly underneath it. Because the rope has some mass, the tension varies along the rope, from highest at the ceiling to lowest at the floor.

If a wave packet begins propagating down the rope, will its shape change? If so, it is possible to calculate the shape of a Gaussian wave packet as it travels down the rope?

Intuitively, it seems to me that the wave packet will change shape for two reasons. First, the portions higher up the rope have higher tension, and therefore higher speed. They will "catch up" to portions of the wave packet further down. Second, because the tension is changing, the wave equation now has a term related to the first derivative of the displacement of the rope.

Specifically, I tried assuming that the displacement of the rope is only horizontal and that the slope of the rope is never far from vertical. I made $y$ a coordinate measuring up from the floor and $x$ a coordinate to the right. Letting the tension be $T(y) = T_0+\lambda g y$, with $\lambda$ the mass per unit length, I got the wave equation (edit: important typo corrected)

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

but I don't know what to do with it.

• Rope displacement would generate two waves traveling in both directions. We need boundary conditions for x(0) and $\frac{\partial x}{\partial t}\mid _{t=0}$. – gigacyan Dec 20 '10 at 13:35

I tried to calculate the solution to your problem in your related question.

If I did not miscalculate, the dispersion relation should be implicitly given by the boundary conditions at $y=0$ and $y=h$. The solution was

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

but it is very unhandy. Performing a coordinate transformation

$$\rho = \frac{T_0 + \lambda y}{\lambda g}$$

we can drop the $Y_0$ part of the solution since it is not limited at the origin and $J_0$ is already zero there.
We thus find that the dispersion relation can be stated as

$$2\omega_n \cdot \rho{(y = h)} = \mathrm{"root\ of\ }J_0"$$

Sincerely

Robert

• @Mark Eichenlaub: Added the dispersion relation to the new solution. Is this along your lines of thought? – Robert Filter Dec 21 '10 at 14:13
• Thanks, Robert. The expression is not necessarily easy to work with, but that's how the equations go, I guess. – Mark Eichenlaub Dec 23 '10 at 21:40
• @Mark: Thank you for accepting :) I think Bessel functions are as fundamental for physics as e.g. $sin$ for which we know that it has roots every $2\pi$. It might be a good starting point to get to know some of its properties by using the question you posed :) – Robert Filter Dec 26 '10 at 16:02

I don't think it's possible to get an analytical solution to that, but I could be wrong. Conceptually, what you have is a situation where the speed of the wave varies as you move along the rope, as the square root of the local tension ($v\propto \sqrt{T(y)}$). This is analogous to things like water waves moving up a beach (where the speed of the wave depends on the depth), so that might be a place to look for tips on the math.

This scenario is also sort of similar to the case of the Schrodinger equation in a linear potential (a ball bouncing vertically under the influence of gravity, say), though the Schrodinger equation is only first-order in time, so that's another place you might look for solutions.