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anonymous
 3 years ago
What is the shortest air column, closed at one end, that will resonate at a frequency of 440 Hz when the speed of sound is 344 m/s?
I tried using the wavelength= v/f formula, but I keep getting the wrong answer...
anonymous
 3 years ago
What is the shortest air column, closed at one end, that will resonate at a frequency of 440 Hz when the speed of sound is 344 m/s? I tried using the wavelength= v/f formula, but I keep getting the wrong answer...

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JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1well you can calculate the wavelength. But how does that wavelength relate to the column of air?

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1...and it depends if it is open or closed. Yours is closed. This might help you: http://hyperphysics.phyastr.gsu.edu/hbase/waves/opecol.html

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I still don't really know what to do :( I don't know how the length of the column relates to anything

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0SO because the wavelength is .8, and in a closed cylinder, the length is 1/4 the wavelength, the answer is .2?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Wavelength is v/4L for closed tube. Solve that for L to find length.

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.1The wave doesn't oscillate at the closed end; it can't. That's why there's a node at that end. At the open end, that's the natural place for an antinode. So if you drew out one entire wave form 0 to max to 0 to min to 0, you can see that that first antinode is at the max, and that is 1/4 of the total wave length. So, given f and v, you have solved for wavelength, lambda. Now the length of that tube is L = lambda/4.
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