# Harmonics in an open organ pipe

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? $$L=\frac{(2n+1)v}{4f}$$

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).
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 $v=f\lambda$.
• Sorry, I do not understand. For two open end, why not $L=\frac{nv}{2f}+\frac{v}{2f}=\left(\frac{n}{2}+\frac{1}{2}\right)\left(\frac{v}{f}\right)=\left(\frac{n+1}{2}\right)\left(\frac{v}{f}\right)$? Thank you for explaination. Oct 19, 2013 at 9:48