# Fundamental frequency , wavelength and the length?

What is the concept behind when it is said that for the first fundamental f=c/λ , λ should be equal to 2L . I have read the page http://en.wikipedia.org/wiki/Fundamental_frequency but I am still confused about why 2L ?

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The image above (shamelessly cribbed from Google images!) shows various features of a sine wave. The wavelength is the distance between two crests. Note that there are two nodes every wavelength, so the distance between nodes is $\lambda/2$.

Suppose you're plucking a guitar string. A guitar string is fixed at either end, and because the fixed ends can't move there must be a node at the ends. The fundamental frequency is the one with the fewest number of nodes, so it's the one with only two nodes, one at each end of the string. This means that if the string length is $L$, the distance $L$ must be equal to $\lambda/2$ so $\lambda = 2L$.

However we've concluded that the fundamental has a wavelength of $2L$ only because the guitar string has a node at each end, and this is not true for all instruments. For example an organ pipe is closed at one end and open at the other. This means it has a node at the closed end but a crest at the open end. The fundamental of an organ pipe therefore has a wavelength of $\lambda/4$ (the minimum distance between a node and a crest), so if $L$ is the length of the organ pipe the wavelength of the fundamental is $4L$ not $2L$.

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Thanks a lot . It presents a clear picture now. So can it be generalized that for a wave in closed structure the λ corresponding to fundamental will always be the length of wavelength between two nodes and hence λ=2L and for a structure open at one end it will be between a crest and a node λ=4L. And if we have both ends open then should it be the one corresponding to two consecutive nodes ? – user12448 Sep 24 '12 at 21:35
In case of a rectangular room , is it so that the node corresponding to first fundamental will be at the center of the room. If so then how ? – user12448 Sep 24 '12 at 22:41
A pipe open at both ends has a node in the middle, and a crest/trough at the ends. So its fundamental also has a wavelength of 2L. The site phys.unsw.edu.au/jw/flutes.v.clarinets.html has a good explanation of this. – John Rennie Sep 25 '12 at 5:46
When you move from 1D to 2D and 3D things get more complicated because you have to satisfy the rules for nodes etc in all directions. sengpielaudio.com/calculator-roommodes.htm has a calculator that gives the first few overtones for a room. – John Rennie Sep 25 '12 at 5:51
Thanks a ton . it helps now – user12448 Sep 26 '12 at 22:16

$L$ is the length of the tube. The fundamental frequency looks like $\sin (\pi x / L)$, one upper wave of a sine (or the same with cosine if it's the other kind of the wave). However, the function $\sin (\pi x / L)$ has periodicity $\Delta x = 2L$, and the periodicity of the wave is what we call the wavelength, so $\lambda = 2L$.

The number 2 just means that there are 2 half-waves in a period – and one half-wave is exactly the minimum that is needed to be squeezed in between the ends of the tube. That's how the fundamental frequency is defined.

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Thanks for your answer but I have some doubts.I am not able to understand when you say " sin(πx/L) has periodicity Δx=2L, and the periodicity of the wave is what we call the wavelength, so λ=2L" .I understand that it is just another way of saying that the time period is 2 pi . Why is that we need to have one half wave as the one corresponding to fundamental frequency. – user12448 Sep 24 '12 at 21:18