Let's say we had a pipe which was closed on one side and open on the other. We can find the wavelength for the first harmonic. To find the second harmonic we just add one more antinode and find a new frequency and wavelength. What confuses me is that both the frequency and wavelength are changing but, this contrasts the experiment done to calculate speed of sound with resonance.

In that experiment, is there a single tuning fork used or are 2 tuning forks? This is because we actually increase the size of the pipe/tube so that the difference is $\lambda/2$. Now if we sub all of that data into $v = f(\lambda)$, which frequency are we using?

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    $\begingroup$ Not clear what experiment you refer to that measures speed of sound with a tuning fork - can you add more details? $\endgroup$
    – Floris
    Commented May 10, 2015 at 19:20

1 Answer 1


A pipe closed on one end with a specific length and containing a specific gas at a specific temperature will have a set of resonant frequencies or partials. The lowest frequency of that set is called the fundamental and all the others (which will be higher) are called overtones. The length of the pipe is 1/4 of the wavelength of the fundamental, neglecting open-end effects. Each succeeding overtone will have a shorter wavelength so that 3/4, 5/4, 7/4 of the new wavelength will fit in the pipe. This results in an odd harmonic series of wavelengths, although we commonly call the frequencies "harmonics." Note the first overtone is actually the 3rd harmonic, not the 2nd.

If one changes any parameter--length, temperature, or gas-- the resonant frequencies will change. Changing temperature or gas will change the speed of the sound wave.

In one constant speed experiment, we could find two differing resonant frequencies, say the fundamental and the 1st overtone. We would need tunable pitch forks for that. In another experiment we could use a single fork and change the length of the tube to match the fork to the fundamental, then change the length again so that fork frequency matches the 1st overtone.

The basic idea for a closed-end tube is that the overtone frequencies follow $$f_n=(2n+1)\frac{v}{4L},$$ where n=0 for the fundamental and n=1, 2, 3, etc for the overtones.

  • $\begingroup$ So the one frequency can match the fundamental mode and the 1st overtone right? $\endgroup$
    – Yulmart
    Commented May 10, 2015 at 19:33
  • $\begingroup$ The fundamental mode of one length and the 1st overtone of a different length. $\endgroup$
    – Bill N
    Commented May 10, 2015 at 19:34

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