The pulse that climbs a rope at half of g

The medium decides the speed gave a formula for the wave speed on a stretched string: the square root of tension over mass per unit length, $c = \sqrt{T/\mu}$. It is the first formula almost anyone learns in this subject and every example afterwards has the tension fixed. But a rope hanging by one end has a tension that goes from its whole weight at the top to nothing at all at the bottom. A rope has no choice in this. The chain that swings in Bessel’s notes looked at its standing waves, the first problem ever solved with what became known as Bessel functions. Now take one single travelling pulse, which needs nothing from Daniel Bernoulli’s calculation at all. A few things happen on its way down and back that turn out to be simpler than expected, as well as one that is stranger.

The system could not be simpler. Let $x$ be the height above the free end, with $L$ the rope’s length. The tension at height $x$ holds up the rope below it, so $T = \mu g x$. Thus the wave speed is

$$c(x) = \sqrt{\frac{\mu g x}{\mu}} = \sqrt{gx}.$$

The mass per unit length has gone. There is no property of the rope in that. A nylon cord, a steel cable and a rope of old sheets would all carry a pulse at exactly the same speed at a given height. Anyone who has climbed in a gym knows how a shake at the top takes a noticeable part of a second to reach the bottom, and it does not depend on the rope.

A pulse that falls at a steady rate #

Follow the pulse down from the top. Its speed depends on how far it has left to go, and $dx/dt = -\sqrt{gx}$. Integrate:

$$\sqrt{x} = \sqrt{L} - \tfrac{1}{2}\sqrt{g}\,t,$$

and so $x$ is a perfect square of something linear in time. The motion is a parabola, with a constant acceleration, and differentiating twice gives a magnitude of

$$\ddot{x} = \frac{g}{2}.$$

So a pulse moves exactly like a falling object, but with the acceleration halved and the direction of its acceleration opposite to its motion. It starts at the top with some speed and steadily slows at $g/2$, arriving at the free end with zero speed. Equally, a pulse running from the bottom to the top behaves as though it were falling upward at half of g.

The picture has the two on top of each other. One curve integrates the partial differential equation for a rope with varying tension, with a pulse sent in from the top and its peak tracked at every step. The other is a one-line calculation. There is no reason to doubt the second, yet a check is reasonable, since a local speed in a varying medium is not always what a pulse follows. Waves in a varying medium have phase and group speeds, and they scatter. In this case, they agree to about the width of the pulse, so the pulse’s path can be taken as settled.

The total time is $2\sqrt{L/g}$: about 2.02 s for 10 m, a time depending on only $L$ and $g$. It is an interesting fact that this is $\sqrt{2}$ times the fall time of a stone dropped from the same height, as a stone takes $\sqrt{2L/g}$. A stone starts slowly and speeds up, whereas a pulse starts fast and slows down, and they do not take the same path through time. The relation $c = \sqrt{gx}$ also looks identical to the speed of waves in shallow water, $\sqrt{gh}$, a coincidence that has a real basis in the two equations having the same structure.

Shortening and growing #

The other simple thing about the rope’s pulse is its shape. A wave in a medium whose speed changes has a fixed duration in time, as seen passing any point, because nothing in a time-independent medium can change the frequency.

The pulse’s duration is fixed. So its length in space, which is its speed multiplied by that duration, is proportional to the local speed. As the speed falls near the bottom, it shortens. So it arrives at the free end as a short, sharp thing. That is why a rope’s end whips. The growth comes from the flow of energy. A wave’s power is $\frac{1}{2}\sqrt{T\mu}\,\omega^2 A^2$, with an impedance $\sqrt{T\mu}$ in it. Holding that power constant through a changing tension requires

$$A \propto (T\mu)^{-1/4} \propto x^{-1/4}.$$

The numerics and the formula are almost indistinguishable over the rope’s range. This is called Green’s law in other contexts. It was given for tidal waves in channels in 1837, for water with changing depth, and it has the same form there: amplitude going as depth to the minus a quarter. The waves that always arrive square to the beach found it for a swell at a beach. The crack at the end of a whip is a much more extreme case of a similar concentration, in which a taper of mass as well as tension brings about far more growth in speed than a rope gets. So a hanging rope is the simplest example that can be held in the hand.

The fourth root is a mild function. Even near the bottom the amplitude is only a few times its starting value. It takes extreme tapers to focus much energy, and the rope does not get close to the speed of sound. The notion of a rope’s free end snapping like a whip is not really correct for a hanging rope alone. It needs tapering mass.

No reflection on the way #

The rope’s tension varies by a thousand-fold in its range, without a sharp boundary. As the taper that matches every note discussed in the case of horns, a slow change in impedance reflects hardly anything, and it is only a sudden change that does. That holds if the change over one wavelength is small. With a short pulse on a long rope, this is well satisfied for most of the path, and it is the condition under which the two approximations already used, the speed following $\sqrt{gx}$ and the energy flow staying constant, are valid. All of it fails near the bottom, where the wavelength becomes comparable to the distance from the free end. There, the tension’s fractional change per wavelength is large, and WKB breaks down.

That point has a simple location. A pulse of duration $\tau$ has a length $\tau\sqrt{gx}$, which equals $x$ when $x = g\tau^2$. For a tenth-second pulse, that is 10 cm. Below that distance, the rope cannot be treated as gradually varying, and what happens is decided by an exact solution.

What a heavy load does #

Add a weight to the end, equal to a length $\ell$ of the same rope, and the tension at height $x$ becomes $\mu g(x + \ell)$. Then the same integral gives a transit time

$$t = \frac{2}{\sqrt{g}}\left(\sqrt{L + \ell} - \sqrt{\ell}\right).$$

A load equal to the rope’s weight brings the time down to 0.84 s, and a large load approaches the ordinary string’s formula $L/\sqrt{T/\mu}$. This is the one place in the problem where the rope’s own mass reappears: the ratio of the two weights can be read from the transit time, without any weighing. It is a practical measurement, for checking the tension in cables with a flick and a stopwatch. Engineers do that on guy-wires and stays, using a form of this expression for a cable with its own weight in it. For a very heavy load, where the rope’s own weight can be neglected, it reduces to a uniform string with the familiar formula.

It is a simple example of how a physical parameter can disappear from a formula and return when something else is present to compare with. With nothing else, only $g$ and $L$ are left. Add a load, and a ratio has appeared.

What comes back #

The more surprising part happens at the free end. A uniform string with a free end reflects a pulse the same way up. A uniform string with a fixed end reflects it upside down. It is one of the very earliest wave demonstrations with a rope, and it is correct for uniform strings. The bottom of a hanging rope is a free end, in the sense that it cannot exert any force. So it should return the pulse the same way up.

It does not.

What is coming back at this height has a positive lobe and a negative lobe. It carries the same energy and it has the same duration, and it has a different shape. In fact, it is close to what mathematicians call the Hilbert transform of the original. The reason lies in the exact solution near the end, in which the rope’s equation is Bessel’s equation, as the chain that swings in Bessel’s notes described. The Bessel function $J_0$ of $2\omega\sqrt{x/g}$ has, at large argument, the form of a cosine with a phase shift of $\pi/4$. So the combination of an incoming wave and an outgoing wave in that form has a phase difference between them of $\pi/2$, a quarter cycle, and it is the same at every frequency.

That is the same quarter cycle that appears in the quarter cycle a turning point costs, for a quantum particle at a smooth turning point or a light ray at a caustic, as in the half cycle a focus adds. The rope’s free end is a turning point. As a wave nears it, its speed goes to zero, and it is turned back smoothly rather than bounced. A quarter-cycle shift at every frequency is not a time delay. A delay would shift each frequency by a phase proportional to that frequency. This has the same phase shift at all frequencies, so it cannot be undone by waiting. Applied to a pulse, it gives a shape that is no longer the same pulse, and that is the case on the figure, with a tail that extends some distance away from the pulse’s main body.

No uniform string can do this, regardless of its boundary conditions. It takes a medium with a smoothly vanishing speed, and those are the conditions for a whole family of transformations. A pulse going down and coming back has been transformed once, and if it reflects from the fixed top and goes down and back again, it undergoes a further reversal and another quarter cycle, and it returns again with a different shape. After four such trips it comes back almost to what it was, apart from its dispersion. Demonstrating that with a rope is not easy, as real ropes damp strongly, but with care the odd shape of the first return can be seen with a high-speed camera.

The same equation on a sloping beach #

A wave’s speed going as the square root of distance from an end is not special to ropes. It is exactly what long waves do on a uniformly sloping beach. Shallow water has speed $\sqrt{gh}$, as the speed that depends on the length found, and on a plane slope of gradient $s$ the depth is $h = sx$, with $x$ the distance from the shoreline. So the speed is $\sqrt{gsx}$, the rope’s formula with $g$ reduced by the slope. The governing equation, for small waves, is the same as well, term for term: the water’s depth plays the role of tension. All the rope’s results transfer. A long wave running up a beach of slope 1 in 50 slows at a constant rate of $gs/2$, about 0.1 m/s², and takes a time of $2\sqrt{L/gs}$ to cross a shelf of width $L$. For a shelf 100 km wide with that slope, a tsunami would need about an hour.

The wave’s height grows as $x^{-1/4}$, the same as the rope and the same as Green’s law. And the shoreline has the rope’s free end in it. Water at the shoreline has no restoring force of its own, and so the reflection there has the quarter-cycle phase. George Carrier and Harvey Greenspan in 1958 solved this problem exactly, including the nonlinear motion of the shoreline, by a transformation that turns the full nonlinear equations into a linear equation with the same Bessel functions. It remains one of the few exact solutions for water waves reaching a shore. A tsunami approaching a straight coast is therefore mathematically a rope being flicked, up to the point where the wave breaks or floods. A reflected tsunami has a shape with lobes of both signs, as the rope’s does. Its form as a leading depression followed by a crest, which drew people out onto the exposed sand in 2004, has other causes as well, including the shape of the sea floor’s displacement, though the reflection makes such forms more common in the waves that then go back out.

Wave equations with speeds that vanish like this at a point, in other words turning points, are everywhere. The equation that lets a shape travel is the uniform case. A tide in a tapering estuary has the same structure as a pulse in a hanging rope, with a different power, as has a sound wave going up through a stratified atmosphere towards a place where its speed falls. So the same family of results, in which there is a law of amplitude growth, a quarter-cycle phase and a Bessel function of some order, turns up at every scale up to a planet. It would be hard to choose a cheaper or more convenient place to see them. A rope of 10 m and a phone camera at a high frame rate will do.

What the rope is not #

The figures use an ideal rope, without stiffness, with small amplitude, no damping and no sag. Real ropes have bending stiffness, which matters near the bottom where tension vanishes and where the wave equation would imply infinite curvature, and they also have air drag and internal friction. Any of those may greatly change the behaviour at the very end. A heavy rope’s bottom end is seldom a tidy free end: it twists, and it has a knot in it.

The transverse pulse is taken to be small. A large-amplitude wave would stretch the rope and change the tension locally, leading to effects of its own, and if the end swings wide enough, the tension at the bottom would no longer be determined by the weight alone. Those are the conditions under which whip-cracking occurs. The grid has its own limitation, too: near the free end the wavelength becomes shorter, and a fixed grid cannot follow it, so the amplitude at the very tip has not been established here. The check against the fourth-root law uses a range that stops about a metre short of the end, where the pulse still has many grid cells to its width.

The return figure has a tail that goes on after the main part, and the quarter-cycle argument is one for high frequencies. For a pulse short compared with the rope’s full time scale, this is an excellent description, and for a long smooth pulse, the rope’s response is more like its slowest normal modes, in which case the quarter-cycle picture does not help. The reader should not take the exact values in the plotted curves as universal: they depend on the pulse’s shape.

Still open: what a chain does when its end goes slack #

Rope and swell share the same simple equation in the linear limit, and they part company where amplitude matters. A swell approaching a beach follows the $h^{-1/4}$ law until, at some point, it gets too steep and breaks, in a way that is not fully understood even with today’s numerical simulation. Whether a transverse wave on a real heavy chain has anything analogous, in the way that a chain’s motion goes nonlinear near its free end, has been studied much less. High-speed experiments on hanging chains near their ends, where the tension goes to zero and any amplitude at all counts as large, show irregular behaviour that goes beyond the wave equation. The chain’s end can begin to whirl or to go slack, sending back something different in each trial. A rope’s free end is the one location in this smooth problem where the equation has no way of telling what will happen next. It may be the simplest place to study what becomes of a wave whose medium disappears at its feet.

On a hanging rope the wave speed is $\sqrt{gx}$, with the rope’s mass cancelled, so a pulse moves on a parabola with a steady acceleration of g/2, grows as $x^{-1/4}$ without reflecting and takes 2.02 s for 10 m; but at the free end it turns back through a quarter cycle at every frequency and returns with lobes of both signs, neither itself nor inverted. Two simple rules hold for most of the way. The third, which applies to the last few centimetres, is the most interesting. The habit worth carrying is to look at where a smooth medium’s coefficient vanishes, because that is where the familiar boundary rules stop being the right ones.