# Beats and Resonance - How to Find the Length With only Frequency?

lola1227
Homework Statement:
An air column, closed at one end has a fundamental (lowest frequency) of 400Hz. What are two other frequencies that produce resonance with this air column? (This is a difficult problem when no air column length is given. Try to find a way to get around the problem.)
Relevant Equations:
V=f(lambda)
Closed Air Column Fractions - 1/4, 3/4, 5/4
So, my thinking was that we use the formula

V=f(lambda)

and substitute the f so,

V = 440(lambda)

but then i don't have another number to cancel or rearrange by.

And since closed air columns have the fractions of 1/4, 3/4, and 1 1/4 (5/4), we could divide by those?

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You should be given a value for the speed of sound. What is it?

You have fractions:1/4, 3/4 and 5/4. What do these tell you?

lola1227
You should be given a value for the speed of sound. What is it?

You have fractions:1/4, 3/4 and 5/4. What do these tell you?
Hi, we were just given this formula and another formula that we have is that the speed of sound is V=331+0.59T

this is why i was confused with the question because there was not enough information given

the fractions tell me that the frequency will be divided by 1/4?

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Hi, we were just given this formula and another formula that we have is that the speed of sound is V=331+0.59T

this is why i was confused with the question because there was not enough information given

the fractions tell me that the frequency will be divided by 1/4?
EDIT. Sorry for any confusion. You don't need temperature.

Look at this diagram very carefully:
http://hyperphysics.phy-astr.gsu.edu/hbase/Waves/imgwav/ccyl.gif
Can you see how the frequencies of the different resonances are related?

Last edited:
lola1227
You would usually be given a value for the speed of sound in this sort of question. Since you only have your formula, assume a convenient/realistic value for temperature, state your assumption at the start of your answer, then use your formula to find the speed (v).

The fractions tell you how many wavelengths fit in a pipe closed at one end when the air inside resonates. Look at this diagram very carefully:
http://hyperphysics.phy-astr.gsu.edu/hbase/Waves/imgwav/ccyl.gif

The top picture shows the fundamental resonance for the pipe. Note 1/4 of a wavelength fits in the pipe, so the length of the pipe is 1/4 of a wavelength. You should now be able to find the pipe's length.

The next picture shows the next possible resonance. Note this corresponds to 3/4 of a wavelength fitting in the pipe. Since you know the pipe's length, you should be able to find the new wavelength and hence the new frequency.

The bottom picture shows the next possible resonance. Note this corresponds to 5/4 of a wavelength fitting in the pipe. So you should be able to complete the problem.
Just to see if I understand properly,

I would use a number such as 343, or 331 as my speed of sound and then I would divide the 400 by this number?

After that I would use L=1/4(lambda) and multiply the number that I got previously by the different frecations?

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Just to see if I understand properly,

I would use a number such as 343, or 331 as my speed of sound and then I would divide the 400 by this number?

After that I would use L=1/4(lambda) and multiply the number that I got previously by the different frecations?
Many apologies, I made it much more complicated than necessary. See my updated post #4.

You don't need temperature. And you don't need wavelength. You just need to understand how the different resonant frequencies are related. Check my updated post #4.

lola1227
Many apologies, I made it much more complicated than necessary. See my updated post #4.

You don't need temperature. And you don't need wavelength. You just need to understand how the different resonant frequencies are related. Check my updated post #4.
That's Okay!

So the relation is that you have more of that number every time? So like at first you have a 1/4 wavelength which is how you got 400 frequency

then to get the 3/4 frequency you would multiply by 3 so you would get 1200Hz

and then for 5/4 multiply by 5 so you would get 2000Hz?

The relation is that the frequency that was received for the first quarter is multiplied by the numerator of the next fraction

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Yes - those are the right answers. Well done!

It's best to think of the fractions in terms of wavelength. For the first resonance, 1/4 of a wavelength fits in the pipe. (Because you must have a node at the closed end and an antinode at the open end. You can see this on the diagram.)

For the second resonance, 3/4 of a wavelength fits in the pipe (node at closed end, antinode at open end). So the second wavelength must be 3 times smaller than the first one.

This means the second frequency must be 3 times bigger than the first one (since f = v/λ and v is constant).

Same argument when 5/4 wavelengths fit the pipe (node at closed end, antinode at open end).

lola1227
Yes - those are the right answers. Well done!

It's best to think of the fractions in terms of wavelength. For the first resonance, 1/4 of a wavelength fits in the pipe. (Because you must have a node at the closed end and an antinode at the open end. You can see this on the diagram.)

For the second resonance, 3/4 of a wavelength fits in the pipe (node at closed end, antinode at open end). So the second wavelength must be 3 times smaller than the first one.

This means the second frequency must be 3 times bigger than the first one (since f = v/λ and v is constant).

Same argument when 5/4 wavelengths fit the pipe (node at closed end, antinode at open end).
Thank you so much for your help!