Why is it that two transverse waves travelling in same medium have same wave speed? What does the medium has to do with it? I have searched through the web, but I couldn't understand. I have found that wave speed is a property of the medium $$\sqrt{\frac{\text{tension}}{\text{mass per unit length}}}$$ but also that the velocity of wave determined by $$\text{velocity}=\frac{\text{angular frequency}}{\text{wavenumber}}$$ Are both of these equations the same ? Or are they different?

  • $\begingroup$ Re, "What [the] medium has to do with it...?" A wave is not a physical thing. When you see a wave propagating through a medium, the only physical thing that is present is the medium itself. The wave is just the medium reacting to a disturbance in itself, and propagating that disturbance to other parts of itself. $\endgroup$ Commented May 6, 2020 at 12:21
  • $\begingroup$ Can you explain why you think they should be different? I presume we are talking about an isotropic medium. Both of your equations are true; are you saying that they appear to you to be inconsistent with each other? $\endgroup$
    – garyp
    Commented May 6, 2020 at 12:33
  • $\begingroup$ Like wave velocity depends on tension and mass per unit length so by just seeing the formula how can we say that different transverse waves in same medium will have same ratio of tension and mass per unit length.ok I wanted to confirm both forms of velocity are correct or not $\endgroup$
    – SM Sheikh
    Commented May 6, 2020 at 12:36
  • $\begingroup$ Waves don't have "mass per unit length." That's a property of the medium. $\endgroup$ Commented May 6, 2020 at 13:04
  • $\begingroup$ But the string has mass per unit length. $\endgroup$
    – SM Sheikh
    Commented May 6, 2020 at 13:10

3 Answers 3


Your first equation determines the wave speed based on the properties of the medium the wave is propagating through. Your equation is specifically for a wave on a string. The equation makes sense qualitatively. If the string tension is greater, then the wave will propagate faster because there is a larger restoring force in the system. If the density of the medium is larger, then the wave will propagate slower because there is a larger inertia in the system that "resists" the restoring force. Any wave in this medium is subject to these same considerations, so any wave will have the same speed.

Your second equation is different. It is a true relationship between wave speed, angular frequency, and wave number.$^*$ This equation does not not say that frequency or wavelength determines the wave speed, for the wave speed is physically determined by your first equation. The intuition here is that since the wave covers a certain distance in some time as determined by its speed, then there has to be a set ratio between how long it takes in both space and time for the wave to "repeat itself". In other words, this equation is just a statement about what must be true for waves that propagate at a certain speed. It is not a statement of how the frequency or wavelength of a wave sets its speed.

An appropriate application of these two equations would be something like (I am not sure how physically realistic some of these numbers are, but that isn't the point here)

A string is kept at a tension of $10\,\mathrm N$, and the string has a linear mass density of $10\,\mathrm{kg/m}$. Therefore, a wave on the string will have a speed of $$v=\sqrt{\frac{10\,\mathrm N}{10\,\mathrm{kg/m}}}=1\,\mathrm{m/s}$$ If we then drive one end of the string with an angular frequency of $\omega=5\,\mathrm{rad/s}$, we know that the resulting wave will have a wave number of $k=5\,\mathrm{rad/m}$.

An incorrect application from the above example would then be

If we then drive one end of the string with double the angular frequency $\omega=10\,\mathrm{rad/s}$, then because $v=\omega/k$ we will then double the wave speed to $2\,\mathrm{m/s}$

The above is incorrect because our second equation does not determine the wave speed. Since we have not changed any property about the medium itself, the wave speed will remain constant. Correcting our mistake then:

If we then drive one end of the string with double the angular frequency $\omega=10\,\mathrm{rad/s}$, then because $v=\omega/k$ remains constant, we will then have a wave with double the wave number: $k=10\,\mathrm{rad/m}$. The wave speed is still $v=1\,\mathrm{m/s}$.

$^*$Or others might be more familiar with $v=f\lambda$ where $f$ is the frequency of the wave and $\lambda$ is the wavelength of the wave.

  • $\begingroup$ Great answer at explaining what the equations mean. I wonder if you also found this hard to approach due to the "why?". It's one thing to state an equation and say "yeah, that seems to agree with what I would expect", but to really know that the wave speed doesn't depend on something is quite difficult. For instance, this speed of a wave on a string wouldn't apply if the string were in treacle due to the different restoring force (frictional term in the DE), I would expect. Is the treacle part of the medium? What do you think? $\endgroup$
    – Joe Iddon
    Commented May 6, 2020 at 14:55
  • $\begingroup$ @JoeIddon "...but to really know that the wave speed doesn't depend on something is quite difficult" Not really. All you have to do is change the property you are proposing. If the wave speed doesn't change, then you have shown that the wave speed does not depend on that thing. Of course this question an answer assumes an ideal case with no dissipation. If you have a question about how this analysis changes with dissipation, then I would recommend asking a new question. The comments of an answer is not the place to have such discussions. $\endgroup$ Commented May 6, 2020 at 14:59
  • $\begingroup$ Sure, feels slightly misleading to state definitively that this is the speed of a wave on a string, when there are caveats. I suppose this is the case with most equations. $\endgroup$
    – Joe Iddon
    Commented May 6, 2020 at 15:02
  • 1
    $\begingroup$ @JoeIddon Idealizing a situation isn't misleading. There are certainly situations where we can ignore dissipative effects. The question is obviously considering the idealized case, so I stuck with that. Would you say that people ignoring air resistance in projectile motion problems are being misleading? There are always caveats. You have to choose what to keep and what to ignore. You will never be able to take everything into account. $\endgroup$ Commented May 6, 2020 at 15:05

I would question: what else would a transverse wave's speed depend on, apart from the properties of its medium?

Would you expect the wave to travel faster if it was night rather than day? Would you expect the wave to travel slower if it was a Tuesday rather than a Wednesday? Obviously not.

If you had any reasonable suspicion of some other condition that could be determining the waves speed, then either imagine or actually set up an experiment where that condition changes, and see if the wave speed would change.

So far as experiments have shown, for the simple case of, say, a ripple travelling along a rope, the only thing that determines the speed of the wave is - how thick and dense the rope is (mass per unit length), and how taught it is pulled (the tension). Nothing else changes the speed of the wave. (Obviously we are accepting a slight idealistic approach here where we don't consider factors like "hairiness of rope" which may play a tiny factor, but usually negligible).

Maybe you would be convinced reading this passage:

For example, there have been many experiments running rats through all kinds of mazes, and so on—with little clear result. But in 1937 a man named Young did a very interesting one. He had a long corridor with doors all along one side where the rats came in, and doors along the other side where the food was. He wanted to see if he could train the rats to go in at the third door down from wherever he started them off. No. The rats went immediately to the door where the food had been the time before.

The question was, how did the rats know, because the corridor was so beautifully built and so uniform, that this was the same door as before? Obviously there was something about the door that was different from the other doors. So he painted the doors very carefully, arranging the textures on the faces of the doors exactly the same. Still the rats could tell. Then he thought maybe the rats were smelling the food, so he used chemicals to change the smell after each run. Still the rats could tell. Then he realized the rats might be able to tell by seeing the lights and the arrangement in the laboratory like any commonsense person. So he covered the corridor, and, still the rats could tell.

He finally found that they could tell by the way the floor sounded when they ran over it. And he could only fix that by putting his corridor in sand. So he covered one after another of all possible clues and finally was able to fool the rats so that they had to learn to go in the third door. If he relaxed any of his conditions, the rats could tell.

Now, from a scientific standpoint, that is an A‑Number‑l experiment. That is the experiment that makes rat‑running experiments sensible, because it uncovers the clues that the rat is really using—not what you think it’s using. And that is the experiment that tells exactly what conditions you have to use in order to be careful and control everything in an experiment with rat‑running.


The defintion of medium , or more precisely the true sense of the word Medium over here in physics is not what you encounter normally. In waves on String, medium means a string of same mass per unit length and of same average tension, in sound waves in a gas, medium means a gas with same adiabatic constant and temperature , you must take the feel of the statement rather than the literal meaning. So what you grasp from here is that what is medium in Physics is not precisely the same in language. But the converse is also true. Consider a steel wire and a jute wire with same tension and mass per unit length. If the definition of medium is taken in a literal sense then a wave should not have same speed in them since they are different material, but in reality they do. I hope you understand now. Critisisms welcomed


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