# resonance, find the lengths

by endeavor
Tags: lengths, resonance
 P: 176 "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...
HW Helper
P: 6,559
 Quote by 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.
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
P: 176
 Quote by 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
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?

HW Helper
P: 6,559

## resonance, find the lengths

 Quote by 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?
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
P: 176
 Quote by 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
Oh, Ok. Thanks!

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