Calculating standing waves, given a fruequency? :O

[SFSKabam]

Homework Statement

"Design a tube closed on one end so that it's lowest frequency standing wave is 440Hz. (i.e. determine the length of the tube."

Homework Equations

Um...I was unsure.
I thought maybe...I might use w=2(pi)f...and the function 2A sin (kx) cos (wt), but I epically failed on it.

The Attempt at a Solution

My solution was rather embarrassing, and it acquired me 0/10 points on my exam, thus I went online trying to get help (from Wiki and the sorts), then I found this site, thought I'd try it. :)
The attempt was as follows...
w = 2(pi)f = sqrt(g/L)

I then squared both sides to get (2(pi)f)^2 = g/L

Then simplified to L = g/(2(pi)f)^2) --Hence epic fail, for, come to find out, sqrt(g/l) is for pendulums. --

And now I'm lost. -.-

 

[Kurdt]

Its really rather more simple than that. A tube closed at one end will have a fundamental frequency, where the wavelength is 4 times the tubes length.

 

[Moetasim]

I would like to first clarify that the equation w=2(pi)f is not applicable to this problem as it is the angular frequency for simple harmonic motion, not standing waves.
To calculate the standing wave frequency of a tube closed on one end, we can use the equation f=nv/4L, where n is the harmonic number, v is the speed of sound, and L is the length of the tube.
Since we are given the frequency of 440Hz and we want to find the length of the tube, we can rearrange the equation to solve for L.
L = nv/4f
Plugging in the values, we get L = (1)(343 m/s)/(4*440 Hz) = 0.195 m or 19.5 cm.
Therefore, the length of the tube must be 19.5 cm for its lowest frequency standing wave to be 440Hz.
I hope this helps and I encourage you to continue practicing and seeking help when needed.