I have a series of doubts regarding the principle of organ pipes. For a given length of closed organ pipe there are various modes of vibration for the standing waves.I dont know if this is silly but what decides the frequency of standing wave in the column,is it the frequency with which a person blows into pipe or are the frequencies or various harmonics dependent only on the length of column (the ones we calculate by drawing the number of waves in the column). What happens if the frequency of source cannot be matched by any of the possible variations of waves (harmonics), do we get low intensity of sound in such setups due to negative superpositions, how do we draw the standing wave inside the column then, will the closed end still be a node?
I was studying the resonance tube experiment to determine the speed of sound in air that lead me here. I understood till the part where loudness was maximum because the air in that particular length vibrated at the same frequency, the problem comes when we find the frequency at the second resonance length, why is that wave not in its first harmonic as depicted in any available picture of the experiment, does this imply that frequency of wave in column is determined by source?

  • $\begingroup$ Blowing does NOT have a single frequency. $\endgroup$
    – user196418
    Commented Dec 28, 2019 at 15:22

2 Answers 2


There are several questions here.

First, the factors that determine the resonance frequency of a piece of pipe are 1) the speed of sound waves in the pipe, 2) the length of the pipe, and 3) the nature of the termination of the undriven end of the pipe. 1) and 2) tell you how long it takes a sound wave to travel from one end of the pipe to the other and 3) tells you the phase of the wave reflected off the undriven end in comparison with the phase of the incoming wave. Organ pipe acoustical physics is well-documented; any well-written source will tell you how to combine 1), 2), and 3) to get the fundamental and the overtone series.

Next, the issue of what overtones the pipe will contain. This depends on the frequency content of the signal that is driving the pipe. If the driving frequency is random noise that contains a broad range of frequencies, then the pipe will select out the frequencies that are multiples of the fundamental and resonate at them. If the driving signal contains no multiples of the fundamental, then there will be no resonances excited in the pipe.

I cannot comment on your last question because I do not know what was depicted in your experimental materials.


Your question is a very good one and is related to, of all things, vowel sounds.

Fant's approximation method, a general approach:

In the 1960s Fant Gunner used the principles established in transmission line theory of electronics as an analogy to the physics of acoustics occurring in the human vocal tract. Essentially, a open-close tube of 17cm is able to differentiate its overtone ratios by differentiating the series of its cross sectional areas, as demonstrated by the human vocal tracts ability to generate different vowel sounds.

Here are some examples that will help give an intuition of the physical principles that are involved here. Two well known demonstrations of resonance are the use of a rope and suspended balls. In the rope analogy, variation in the cross sectional area of an acoustics pipe are analogous to difference of density along the length of a rope. So if you tie a thread to a piece of yarn and wiggle it back and forth you can observe differences happening to the fundamental and overtone resonances. Similarly, if you line a series of suspended balls and place balls of different size on the tract it can represent the changes differing areas will have on resonance.

You can see the full RCLG circuit analogy derived from differential equations using complex numbers in "Speech Analysis Synthesis and Perception" (2008) Third Edition James L. Flanagan, Jont B. Allen, Mark A. Hasegawa-Johnson Chapter 3.

For practical purposed though, the differentials reduce to two recursive equations. One modeling pressure and the other modeling air flow along the segments of the vocal tract (acoustic pipe). The literature uses P and U for this pressure and air flow. The recursive equations are as follows: $$P_i=cos(\omega)P_{i-1}-sin(\omega)U_{i-1}/A_i$$ $$U_i=cos(\omega)U_{i-1}+sin(\omega)P_{i-1}A_i$$ $$\omega=\frac{2\pi Hz l_i}{c}$$ $$P_0=0$$ $$U_0=1$$

  • Hz is frequency.
  • $l_i$ is segment i length
  • $A_i$ is the cross sectional area of segment i
  • c is the speed of sound

This is formula is the lossless case. This lossless case is the case of the $$f(n)=\frac{c}{2nL}$$ of open-open tubes $$f(n)=\frac{c}{4L(2n-1)}$$ of open-close tube. Thus, with Fant's source filter method, you take an ordered list of cross sectional areas and find resonant frequencies. You do so by using numerical methods, such as goal seek on excel, to alter Hz until the final U = 0. If you use total length is 17cm, the speed of sound as 34000 cm/sec, and each cross sectional area the same; then 500 Hz, 1500 Hz, and 2500 Hz each will make U = 0.

Phibert's equation for conic pipes:

Now in the 1931 book by G. OSCAR RUSSELL, titled "SPEECH AND VOICE" on pg 102 an equation for the standing wave lengths of open ended conic horns is given: $\lambda = L(D+d)/D.$

  • L is length of the tube.
  • D is diameter of the mouth.
  • d is diameter of the entrance.

A Philbert Ch. is credited for this equation(1877). Though the equation in the book is inaccurately described as being a frequency rather then a wave length. You just place it under c, f=c/$\lambda$, to get frequency. This is just the first frequency though. I use Fant's model to figure out the relation of the overtone frequencies and it works out as follows:

$$f(n)=\frac{c}{2L}(n-\frac{1}{1+\frac{d}{D}})$$ for open-close tube, as in a vocal tract

Technical correction:

There are formulas are used to find the actual value of $U_o$. It has no practical difference from using 1 though except when finding bandwidths not resonance.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.