harmonicshomework-and-exercisesresonancewaves
How to solve the following question?
An open organ pipe has two adjacent natural frequencies of 500 and 600 Hz. Assume the speed of sound in the air 340m/s. The length of the organ pipe is?
What is open organ pipe? It means two open ends? Like this? Or one closed and one open?
Should I use this formula? $L=\frac{(2n+1)v}{4f}$
Sound is a compression wave, that is the air vibrates in a direction along the direction of the wave. I found this image that shows you what a standing sound wave looks like in a half open pipe (nicked from this site). The black dots represent some small volume of the air:
The key point to note is that there are some points where the air is not moving, called nodes, and some points where the air motion is at a maximum, called antinodes.
At the closed end of the pipe the air cannot move, obviously so because the end of the pipe is in the way. This means there must be a node at the closed end. In contrast at the open end of the pipe the air can move, and in fact for a standing wave to form at an open end the wave must have an antinode.
So for a closed pipe, as in the picture above, we have a node at one end and an antinode at the other, and the distance between a node and antinode is $(2n + 1)\lambda/4$, where $n = 0$ gives you the fundamental, $n = 1$ is the first overtone and so on (the diagram actually shows the fifth overtone). For the fundamental frequency, $n = 0$, the length of the pipe is equal to $\lambda/4$ so the wavelength is four times the length of the pipe as you say.
If you take an open pipe the air will be moving, i.e. there is an antinode, at both ends. The distance between two antinodes is $n\lambda/2$, where this time $n = 1$ gives us the fundamental, so for the fundamental the length of the pipe is equal to $\lambda/2$ i.e. the wavelength is twice the length of the pipe.
There's an error in that the type of pipe for each of the two fundamental frequencies as described in your comment don't match the problem description. The pipe with a fundamental frequency of 440Hz is open-closed, and the pipe with a fundamental frequency of 660Hz is open-open. You actually said "closed-closed", which isn't even an option, but even if that's taken to mean "open-open", you've still got the fundamental frequencies backwards from what they are in the problem description.
Assuming you meant to say "open-open" instead of "closed-closed", you've got it right that open-open produces all integer multiples of the fundamental frequency, and open-closed produces only odd multiples; see the Cylinders section of Wikipedia's Acoustic resonance article.
The term "overtone" does indeed mean any pitch higher than the fundamental frequency. But you can assume for this problem that the only overtones involved are harmonic overtones (overtones whose frequency is an integer multiple of the fundamental frequency), and ignore the fact that musical instruments can also produce inharmonic overtones.
Best Answer
Open organ pipe is the one with two open ends, and instead of the formula you mention you need to use $$L=n\frac{v}{2f_n}$$ where $f_n$ is the frequency of the ${n^{th}}$ mode, and $n=1,2,3,...$ your formula is for a closed organ pipe (with one open and one closed end).
EDIT
Because the number of half-wavelengths ($\lambda /2$) need to be an integral multiple in case of a open pipe. This is because both the ends of an open organ pipe are pressure nodes (or displacement antinodes), and the difference between two successive nodes (or antinodes) is $\frac{\lambda}{2}$. Therefore, to meet the resonance-condition, the number of half wavelengths between the ends need to be an integral value, therefore $L=n\frac{v}{2f_n}$ as $v=f\lambda$.