# Why is the phase velocity of waves on a rope independent of the wave length?

A rope is described by the wave equation $$(\partial_t^2 - \partial_x^2) \, \varphi = 0$$ It is well known that there is no dispersion for resulting wave solutions. This implies that all plane waves we can generate in a rope have exactly the same phase velocity. For example, if we focus on one specific point on the rope, say one maximum, it will travel during the time interval $$\Delta t$$ the distance $$\Delta x$$. This fact is independent of the wave length of the wave in question. Thus the phase velocity is always $$v = \frac{\Delta x}{\Delta t}$$.

To analyze the system, let's imagine that a rope consists of lots of mass points that are connected to their neighbors by small strings. For a wave with large wave length the strings between neighboring points are almost not distorted. In contrast, if the wave length is small, the strings between neighboring points are strongly distorted.

Thus, intuitively I would have guessed that the mass points get pulled down much more quickly if the wave length is small. However, as mentioned above, this is not the case as the phase velocity does not depend on the wave length. How can this be understood in intuitive terms?

PS: I recently asked a closely related question which sparked this question.

• Re. "rope consists of lots of mass points that are connected to their neighbors by small strings" Should this be 'springs' instead of 'strings'? – user45664 Oct 14 at 17:25
• @jak Your intuition is right - masses oscillate faster for a low wavelength wave because of greater distortion; it's just that in this case the wave also repeats faster in space...so phase velocity remains the same. If you like, mathematically, low $\lambda$ means low time period $\tau$ of oscillations of a particular point on the string, so that the phase velocity equal to $\lambda/\tau$ stays constant. – Vivek Oct 14 at 22:06
• @Vivek thanks a lot! If you copy your comment into an answer, I can accept it. – jak Oct 15 at 4:17
• "This implies that all plane waves" A plane wave is a wave in 3D space. You may be going for something along the lines of "eigenstate". "For example, if we focus on one specific point on the rope, say one maximum, it will travel during the time interval Δt the distance Δx." The rope only travels in the vertical direction, not the horizontal direction. The phase moves horizontally, but the rope doesn't. – Acccumulation Oct 15 at 19:43
• @Acccumulation Just a matter of semantics in 1D. "Plane" maybe superfluous but isn't wrong obviously. – Vivek Oct 16 at 12:13

Your intuition is right - masses oscillate faster for a low wavelength wave because of greater distortion; it's just that in this case the wave also repeats faster in space...so phase velocity remains the same. If you like, mathematically, low $$\lambda$$ means low time period $$\tau$$ of oscillations of a particular point on the string, so that the phase velocity equal to $$\lambda/\tau$$ stays constant.

### Extra Commentary

That the ratio $$\omega/k=\lambda/\tau$$ is completely independent of $$k$$ is not easy to answer a priori. What we can say is that high $$k$$ means more distortion, and hence a more violent normal mode operating at a higher frequency. How high $$-$$ i.e. proportional to $$k^\alpha$$ for which value of $$\alpha$$? That's impossible to claim without the exact formula for the potential energy of distortion. In this case small local length extension is given by $$\frac{1}{2}(\frac{\partial y}{\partial x})^2$$, which gives $$\alpha = 2/2 = 1$$.

Also, for a rope, the aforementioned approximation for the potential energy only works for small amplitudes, and for every amplitude there is a UV cut-off (for $$k$$) in-built in this system, beyond which the approximation breaks down and/or the rope breaks.

### Dimensional Analysis

Another heuristic insight is dimensional analysis, which is the next best thing to a field Lagrangian. The only dimensionless combination one can construct in this case is $$\omega/ck$$, where $$c = \sqrt{T/\text{mass density}}$$. However, this is assuming the system supports waves and that $$T$$, mass density are the only relevant physical parameters (case in point: nobody knew vacuum supports EM waves until Maxwell & Hertz found them. Incidentally, this also happens to be an example of dispersionless waves across a very wide range of the EM spectrum as far as we can tell).

Edit

### After-thought: A complete, slick and rigorous proof

I realized this cute argument while answering this question on stackexchange. The usual wave equation is invariant under Lorentz transformations in which $$c$$ is equal to the speed of the wave. While this procedure is purely mathematical and doesn't carry a physical interpretation in terms of frames with it in this case, one can still do it and argue that that $$(\omega, k)$$ transforms like the relativistic four-vector (two vector in 1+1 D) under these transformations, implying $$\omega^2-k^2$$ is a constant. Using the fact that Goldstone's theorem predicts massless modes, we've $$\omega \to 0$$ as $$k \to 0$$, implying that $$\omega^2-k^2$$ is actually equal to zero (this means we're relying on the existence of "Lorentz transformations" that can produce arbitrary red-shifts, making $$k$$ as small as we like)!

• Well, this answer is about the wave equation that jak wrote on top of his question. But this is supposing the answer to start with. I am asking a totally different question. Why should this wave equation describe the physical problem of waves travelling along a stressed string ? There is no deep physical reason fo that, one has to do the analysis of the physical situation detail by detail. – Alfred Oct 19 at 18:43
• The dimentional analysis is your best attempt. To tell the truth, I buy it. But htat is because I have a long experience of physics. I don't think a layman will be convinced that the only relevants quantities are really T and the linear mass density and so the only quantity with the dimension of a speed is the square root of the ration ! – Alfred Oct 19 at 18:47
• In fact there is a dimentionless parameter: the ration of the wavelength to the (necessarily finite) radius of the string. Who knows ? The velocity might depend on sqrt(T/linear mass) AND this number. In fact at very short wavelength it will be the case. But who can say it does not enter evne at long wavelength, at some power (possibly fractional) ? Dimentional analysis cannot disprove this. Only exact aclculation. – Alfred Oct 19 at 20:12
• @Alfred You've to start somewhere to describe your system. So one would either start with the Lagrangian/Hamiltonian or the equation of motion of the string. Assuming that this then goes on to show that the fact that the phase velocity of waves on string is independent of $k$ can be thought to arise intuitively from relativistic insights and Goldstone's theorem. (I remember you saying in a previous post that for a string with springs, the fact that the waves on a string are dispersive is a result of a calculation and so is the fact that without the springs are non-dispersive.) – Vivek Oct 20 at 1:11
• Of course, there's a mathematical proof using partial derivatives that such waves are non-dispersive, but that would have been a purely mathematical argument, which neither @jak nor any of us wanted. – Vivek Oct 20 at 1:13