jinsan Posted March 28, 2008 Share Posted March 28, 2008 Hi, I'm stuck on a question that asks why is the wavelength in a closed pipe longer than the wavelegth in a open-end pipe. I know by diagram the wavelength of a closed pipe is 4 times as long as the pipe (natural harmonic) and the wavelegth of an open ended pipe is twice as long. does someone have a good explanation (in words) why this is? Thanks Oh and I'm new, is there a list of symbols I can put into posts? like lambda and the f for frequency.. I looked thru the latex ref but cant seem to find any. Thanks for all the help Link to comment Share on other sites More sharing options...
thedarkshade Posted March 28, 2008 Share Posted March 28, 2008 In a closed pipe the wave is reflected creating so the standing wave as seen here: and in an opened pipe it would look like this: Now you know that wavelength is the distance from one peak to another. And when we have the wave in an opened pipe, there is no reflection so the wave keeps moving all the time. But in a closed pipe the wave is reflected and now you have extra peaks between the existing ones and so the wavelength gets shorter. Focus on the first picture and try to take out the wave that comes from right to left. Now the wave length gets bigger right? well that's basically it, in a closed pipe the peaks of the wave are closer due to refection of the wave, and that makes it have a shorter wavelength. Hope this helps, Shade Link to comment Share on other sites More sharing options...
Klaynos Posted March 28, 2008 Share Posted March 28, 2008 You must consider the boundary conditions for both situations.... Link to comment Share on other sites More sharing options...
jinsan Posted March 28, 2008 Author Share Posted March 28, 2008 Yep Sorry I must have not explained everything properly The picture below tells us what it should look like with open ended on the left and closed on the right. I'm still @ high school so just speaking in general thanks Link to comment Share on other sites More sharing options...
swansont Posted March 28, 2008 Share Posted March 28, 2008 So, do you understand the difference in the boundary conditions? The wave must be a node where the pipe is closed — no air can move there. Link to comment Share on other sites More sharing options...
timo Posted March 28, 2008 Share Posted March 28, 2008 What exactly is your question? How to understand the factors 2 and 4 from the pictures, what the pictures actually show, why the waves have to have the form sketched in the pictures or something else? Oh and I'm new, is there a list of symbols I can put into posts? like lambda and the f for frequency.. I looked through the latex reference but cant seem to find any. Thanks for all the help You didn't find any latex reference or it wasn't mentioned? - A guide is here: http://www.scienceforums.net/forum/showthread.php?t=4236 - The TeX code for any capital greek letter is \Letter, the code for a small greek letter is \letter, e.g. \Lambda for [math]\Lambda[/math] and \lambda for [math]\lambda[/math]. So, do you understand the difference in the boundary conditions? The wave must be a node where the pipe is closed - no air can move there. Got a good explanation why there's a maximum amplitude for the open end, too? Link to comment Share on other sites More sharing options...
jinsan Posted March 28, 2008 Author Share Posted March 28, 2008 So, do you understand the difference in the boundary conditions? The wave must be a node where the pipe is closed — no air can move there. yeah I know the node and the antinode. Boundary conditions? explanation could help thanks Well, the question directly from the book: "What aspect of the standing wave in a closed pipe makes it impossible for it to have the same wavelength as the standing wave in an open pipe of the same length?" Link to comment Share on other sites More sharing options...
swansont Posted March 29, 2008 Share Posted March 29, 2008 Got a good explanation why there's a maximum amplitude for the open end, too? Because you're at a resonance you have to have either a node or antinode, in an ideal case. They don't end up being at that point because of the change in coupling to the free space — there's basically an impedance matching issue, and the node or antinode point shifts a little bit. yeah I know the node and the antinode. Boundary conditions? explanation could help thanks Well, the question directly from the book: "What aspect of the standing wave in a closed pipe makes it impossible for it to have the same wavelength as the standing wave in an open pipe of the same length?" Boundary conditions would be "any restriction on the behavior at specific points" i.e. knowing you can set some variable to a specific value. So do you see that the answer has been given already? Link to comment Share on other sites More sharing options...
thedarkshade Posted March 29, 2008 Share Posted March 29, 2008 "What aspect of the standing wave in a closed pipe makes it impossible for it to have the samewavelength as the standing wave in an open pipe of the same length?" Well then the source of the wave in the closed pipe and the source the wave in the opened pipe produce waves with different frequencies. I mean, you can't the same wavelength in an opened pipe and in a closed pipe if the source of the wave produces waves with the same frequency, and this is due to reflection. Link to comment Share on other sites More sharing options...
jinsan Posted March 30, 2008 Author Share Posted March 30, 2008 yeah i got it now, thanks for all the help. The closed pipe has a node at the end which means it can only have odd harmonics (odd number of 1/4 \lambda) while the open ended pipe can have odd and even.. Link to comment Share on other sites More sharing options...
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