Imagine, we have an air column and the the temperature of the air was the same outside and inside the column. If sound waves from a tuning fork then enters the air column, the speed of the standing wave would have to be the same as the waves outside the pipe as well. We know, that for a STANDING WAVE, its wavelength is determined by the geometry of the pipe. Since frequency must be constant, in this case, how would then the wavelength of the standing wave change to ensure the wave speed is constant, assuming resonance takes place? What if the geometry of the air column was such that the wavelength could not be setup to keep the wave speed constant? Or is this scenario not even possible? If so, why not?


2 Answers 2


Since frequency must be constant, in this case, how would then the wavelength of the standing wave change to ensure the wave speed is constant, assuming resonance takes place?

The wavelength of the standing wave won't change. What will change will be how efficiently the wave from outside couples to the standing wave resonance.

The acoustic impedance at the point of excitation of the column will depend on length of the column relative to the standing wave wavelength. The coupling from free air to the standing wave will depend on the ratio between the free air impedance and the impedance of the column. If the coupling is inefficient, part of the wave will scatter away from the column rather than contribute to the standing wave.


I believe your reasoning is fundamentally wrong.

What is wrong?

The fundamental wave equation relating wave propagation speed, $c$, frequency, $f$ and wavelength $\lambda$ is

$$ c = \lambda f \tag{1} \label{wave-speed} $$

Despite the relation shown in equation \eqref{wave-speed} the quantities included cannot change arbitrarily. It is known that wave propagation speed $c$ depends solely on the medium of propagation for linear waves in non-dispersive media. For completeness stating the generic form of the equation to be that of equation \eqref{generic-wave-speed} below

$$ c = \sqrt{\frac{B}{\rho}} \tag{2} \label{generic-wave-speed} $$

where $B$ represents the internal forces of the medium in some sense (for example in fluids it is the bulk modulus and strings the tension) and $\rho$ the density of the medium (in the generic 3D case it is the volume velocity but for 1D strings it is the linear density).

This means that for the speed of sound to change one should change the medium of propagation. This means that the left hand side of equation \eqref{wave-speed} won't change in your problem.

What is left to change is the wavelength $\lambda$ and frequency $f$. For the equation to hold though they must change in a reciprocal way so that their product will remain constant (for the equation to hold). As you already stated in the question, the frequency $f$ is a characteristic of the source and cannot change arbitrarily (at least for linear waves). This leads us with the conclusion that neither the wavelength $\lambda$ can change since $f$ does not change.

What is happening

In order to reach a description of the phenomenon that will take place we have to resort to forced oscillations since this is what is actually happening (you excite a system with some external excitation function).

From a macroscopic point of view, the description of the phenomenon is this: The source will excite the pipe system at whatever frequency the source radiates. If the frequency happens to coincide with one of the natural frequencies of the pipe then the amplitude will be maximised. Otherwise the amplitude will be smaller.

One could very well ask "how smaller is smaller"? For an "ideal" system with no attenuation the amplitude will be infinite at a very specific frequency but this is as far from reality as you can get. There is no system without losses, which when present in the system will both reduce the maximum attainable amplitude and broaden the resonance curve. This behaviour is depicted in the following image (source: Acoustics - An Introduction by Heinrich Kuttruff).

Forced oscillation amplitude

Note that higheq $Q$ values represent lower losses in the system. As you may be able to see when there are high losses in the system the amplitude does not change significantly when you move away from the resonant frequency. Thus, for a realistic system (at least in a first order approximation sense) you will see amplitude being increased when closing in to the resonant frequency and dropping when you move away, with the exact amplitude being dependent on the characteristics of the system.


To parallelise this explanation to the (very well expressed) one in another answer, when you reach the frequency of excitation comes closer to the natural frequency of the system, the impedance presented by the pipe will become smaller and will go closer to that of the surrounding air. In the ideal case with no losses, the impedance will become equal to that of the air and all energy will go "into" the pipe system and this is why the amplitude will become infinite.

In the more realistic system with losses, at resonance (excitation frequency becomes identical to the natural frequency of the system) the impedance presented by the pipe will go as close as possible to that of the surrounding (or that of the source if the latter is attached to the pipe) and the energy transfer towards the system will become maximum.


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