Resonance, find the lengths

1. May 2, 2006

endeavor

"A tuning fork with a frequency of 440 Hz is held above a resonance tube partially filled with water. Assuming that the speed of sound in air is 342 m/s, for what heights of the air column will resonances occur?"

f1 = 440 Hz
v = 342m/s
I'm not sure if this is refering to an open pipe, or a closed pipe.
For an open pipe,
fn = (nv)/(2L) = nf1
v/(2L) = f1
L = v/(2f1)
L = 0.3886 m
But this is the length for the sound traveling up and down... i think. So the length is 0.194m. However, the answer gives 3 values, 0.194m, 0.583m, 0.972m. Even for a closed pipe, I there is only 1 value...
I must be doing something wrong. I don't think the speed of sound in water is a factor, because my first answer seems to be correct...

2. May 2, 2006

Andrew Mason

This is a closed pipe. There is a node at the bottom and an anti-node at the open end. So resonances occur where the length of the pipe is $L = \lambda/4, 3\lambda/4, 5\lambda/4,... (2n+1)\lambda/4$

Using the universal wave equation:

$$\lambda = v/f$$

Substituting the resonance criterion: $\lambda = 4L/(2n+1) = v/f$

resonance occurs at:

$$L = (2n+1)v/4f = (2n+1)*342/4*440 = (2n+1).194 m.$$

AM

3. May 3, 2006

endeavor

That makes sense, except for why f is kept as the initial frequency. The formula I have is:
L = (mv)/(4fm) where m = 1,3,5,.....
and
fm = mf1
thus
L = v/(4f1)
why do I use only the initial frequency here?

4. May 3, 2006

Andrew Mason

In this case, there is a tuning fork which provides only the fundamental frequency - a pure sine wave. There are no higher frequencies to resonate in the air column. So you are dealing with the fundamental frequency only.

AM

5. May 3, 2006

endeavor

Oh, Ok. Thanks!