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Resonance, find the lengths

by endeavor
Tags: lengths, resonance
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endeavor
#1
May2-06, 10:00 PM
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...
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Andrew Mason
#2
May2-06, 11:37 PM
Sci Advisor
HW Helper
P: 6,654
Quote 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 [itex]L = \lambda/4, 3\lambda/4, 5\lambda/4,... (2n+1)\lambda/4[/itex]

Using the universal wave equation:

[tex]\lambda = v/f [/tex]

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

resonance occurs at:

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

AM
endeavor
#3
May3-06, 12:43 AM
P: 176
Quote 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 [itex]L = \lambda/4, 3\lambda/4, 5\lambda/4,... (2n+1)\lambda/4[/itex]

Using the universal wave equation:

[tex]\lambda = v/f [/tex]

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

resonance occurs at:

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

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?

Andrew Mason
#4
May3-06, 07:26 AM
Sci Advisor
HW Helper
P: 6,654
Resonance, find the lengths

Quote 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
endeavor
#5
May3-06, 12:58 PM
P: 176
Quote 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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