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 one year 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...
 one year 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...

This Question is Closed

JamesJ
 one year 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
 one year 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

Cutiepo0
 one year 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

Cutiepo0
 one year 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?

luthier93
 one year ago
Best ResponseYou've already chosen the best response.0Wavelength is v/4L for closed tube. Solve that for L to find length.

JamesJ
 one year 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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