What is the relation between the fundamental frequency and a harmonic? I am currently busy with a physics report about determining the speed of sound in air. In order to do this, I was told to use a tube that can extend or shorten in order to find the different harmonics of the pipe for different frequencies. I am, however, struggling to understand how to execute the project correctly, what the purpose of the pipe is and how to use the data to get the speed of sound. I should also give the relation between the fundamental frequency and a harmonic and then use that relation to formulate an equation for determining the expected tube lengths that will resonate using a specific wavelength. The work is new to me and I am finding it very difficult to grasp, so if anyone can help me I would really appreciate it!
 A: Well, you kinda ask a lot of questions together here... I will do my best to help here, but I believe you should divide that into smaller chunks for your own convenience.
Starting with the question related to the title... As very well mmesser314 has pointed to in their comment you can find in Wikipedia (same link as in the comment) that a harmonic (we refer to frequencies here) is one that has a positive integer-multiple relation to the fundamental (which is the first frequency of the harmonic series). In other words (as stated also in the link) the frequency of ay harmonic will be
$$f_{n} = n \cdot f_{0} ~, n = 2, 3, 4, ...$$
where we have "reserved" the $n = 1$ for the fundamental frequency. You could very well include it in the equation and consider the fundamental to be $f_{n}$ for $n = 0$.
Now, regarding the "tube/pipe problem". You can find in the literature (see books like "Fundamentals of Acoustics" by Kinsler et al., Foundations of Acoustics" by Skudrzyk, "Foundations of Engineering Acoustics" by Fahy or really any other book related to acoustics) that tubes support only specific harmonic series bases on their boundary conditions (that is, in the simplest case, if their ends are open or closed). So, copying relations from the "Fundamentals of Acoustics" book by Kinsler et al. the frequencies supported by the pipes are as follows:


*

*Open-Closed: $f_{n} = \frac{2n - 1}{4} \frac{c}{L}$

*Open-Open: $f_{n} = \frac{n}{2} \frac{c}{L}$

*Closed-Closed: $f_{n} = \frac{n}{2} \frac{c}{L}$
where $L$ is the length of the tube, $c$ the speed of sound in the material contained in the tube (probably air in your case) and $n = 1, 2, 3, ...$ in this case. As you can see, the open-open tube supports the same frequencies as the closed-closed one, with the exception that the mode patterns are translated by $\frac{\pi}{2}$, or $90^{o}$, or $\frac{\lambda}{2}$, with $\lambda$ being the wavelength (for some nice visuals with some extra text visit Wikipedia).
Now, by measuring the harmonic frequencies you can use them with the appropriate equation to solve for $c$. One minor thing you would like to keep in mind is that your results will not be very accurate, because the formulas provided above do NOT use the correct terminating condition for the open end. The correct one is to use an acoustic radiation impedance which will give the result
$$f_{n} = \frac{n}{2} \frac{c}{L + \left( \frac{8}{3} \pi\right) \alpha}$$
where, in this case, $\alpha$ is the radius of the pipe (I believe that you will get similar results for a rectangular pipe but I don't remember exactly the formula and don't have a reference at the moment).
