If we have let us say fixed air column of length 'L', in a open-closed column problem, lamba is equal to 4*L/(2n-1).

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n = number of nodes / anti nodes in air column

How does 'n' changes with respect to the temperature? Is the "number of nodes / anti nodes" at room temperature equal to "number of nodes/anti nodes" in same air column at let us say T =1200C? or do they increase or decrease with temperature?

As speed = frequency * wavelenght

and speed and temperature have reciprocal relationship, that means

increase Temperature will increase speed, which will result in increase of frequency but is there any relationship that after certain frequency(or temperature indirectly) the quarter standing waves turn into 3 quarters ?

  • $\begingroup$ What do you mean by "number of standing waves"? $n$ may be any natural number, I don't see how temperature is related to this at all. $\endgroup$
    – ACuriousMind
    Commented Mar 23, 2015 at 19:00
  • $\begingroup$ thanks for your comment, I edited the post, it will be better now. please let me know if you still have confusion with it $\endgroup$
    – Omer
    Commented Mar 23, 2015 at 19:05
  • $\begingroup$ It still doesn't really make sense to me - the relationship $\frac{4L}{2n-1}$ describes which wavelengths are the possible standing waves. $n$ doesn't change with temperature, it's an input you set to find e.g. the "first overtone" at $n=1$. $\endgroup$
    – ACuriousMind
    Commented Mar 23, 2015 at 19:14
  • $\begingroup$ Thanks for your reply. What I am observing is that after certain temperature around T>500, using v = f*wavelength relationship, the wave length decreases suddenly and that happens again at around T>1000, I think that in my system, the mode of oscillation change with temperature, as the wavelength is dependent on only L and n. Please let me know if the question is still not clear? $\endgroup$
    – Omer
    Commented Mar 23, 2015 at 19:24
  • $\begingroup$ Oh, you are talking about an actual system you are observing? Then you should describe how you are exciting it to generate the standing wave - I think the reason for the behaviour you see is not in the formula for the standing wave, but in your experimental setup! $\endgroup$
    – ACuriousMind
    Commented Mar 23, 2015 at 19:26

2 Answers 2


The standing waves you introduced are models for how air moves as it resonates. In fact, each is called a mode (of oscillation). In the continuum approximation of air being an infinitely divisible, continuous fluid, you need infinitely many of them to simultaneously model arbitrary resonating movement.

In practice, you can hope to neglect all but low numbered modes. How low depends on what criteria you use. The way you phrased your question, I must guess because you did not elaborate on what purpose you intend to serve.

You can certainly stop when you reach such short wavelengths that the air does not support: If they get too close to the typical intermolecular distance, the motion will essentially be heat, not sound. As you approach that limit, you turn your sound (oscillatory motion) into unordered motion or a temperature increase. That happens sooner as air gets less dense and the distance between molecules larger: Your relevant number of standing waves goes down as temperature increases. In practice, other effects may cause higher modes to become uninteresting, and perhaps too dissipative, much earlier, such as interaction with a perhaps frequency-dependently dissipative tube wall.

If it is sound that you are after, there will likely be another more relevant effect: We cannot hear sound above a certain frequency. Since the speed of sound varies only slightly with temperature, that may mean the number of interesting standing waves does not significantly chance with temperature.

  • $\begingroup$ Thank you for your great answer, you completely understand mz problem. Don't you think that the increase in temperature should increase number of modes, as increase in temperature results in increase in energy, which results in increase in its velocity, with an increasing velocity, it makes it easier for the wave to reflect much faster and hence creating more standing waves ? $\endgroup$
    – Omer
    Commented Mar 23, 2015 at 19:29
  • $\begingroup$ No, frankly, I think you are stuck in some misconception. I'm afraid I do not know how better to help you than to suggest the possibility that you look upon this backwards. The increase in air molecule's kinetic energy makes them go faster, but this is an effect that does not help them to mmove collectively as modes, so if anything, it means you need to consider less modes. And heating won't change the mode number of any excited mode; if you actually heat fast (but adiabatically and uniformly) enough to maintain an oscillating mode without further stimulation,it will only change its frequency. $\endgroup$
    – user73762
    Commented Mar 23, 2015 at 19:42

The wavelength is fixed by the dimensional length of the tube, but since wavelength is equal to the speed of sound divided by the wave frequency, temperature will affect the frequency because temperature determines the speed of sound.

  • $\begingroup$ Thank you for the answer, I just edited my question further, you are right that the wavelength is fixed because the length is fixed, but if there is high change in temperature, will it change the structure from 1/4 standing wave to 3/4 standing wave $\endgroup$
    – Omer
    Commented Mar 23, 2015 at 19:07
  • $\begingroup$ No, the modes are fixed by the geometry and occur upon proper excitation by the energy input and it's frequency content. But pitch (frequency ) will change with temperature. $\endgroup$
    – docscience
    Commented Mar 24, 2015 at 19:27

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