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

In summary: 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).
  • #1
lola1227
25
7
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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  • #2
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?
 
  • #3
Steve4Physics said:
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?
 
  • #4
lola1227 said:
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:
  • #5
Steve4Physics said:
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?
 
  • #6
lola1227 said:
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.
 
  • #7
Steve4Physics said:
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
 
  • #8
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).
 
  • #9
Steve4Physics said:
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!
 

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

1. What is the relationship between beats and resonance?

The phenomenon of beats occurs when two sound waves with slightly different frequencies interfere with each other, resulting in a periodic variation in the amplitude of the resulting wave. Resonance, on the other hand, is the tendency of a system to vibrate at its natural frequency when excited by an external force. In the context of sound, beats and resonance are related in that resonance can be created by the interference of sound waves, resulting in beats.

2. How can beats and resonance be used to find the length of a string?

In order to find the length of a string using beats and resonance, one must first know the frequency of the sound produced by the string. This can be achieved by plucking the string and using a frequency meter or by calculating the frequency based on the tension and mass of the string. Then, by adjusting the length of the string and creating beats with a known frequency, the length of the string can be determined using the formula L = v/2f, where L is the length of the string, v is the speed of sound, and f is the frequency of the beats.

3. Can beats and resonance be used to find the length of any object?

While beats and resonance can be used to find the length of a string, they cannot be used to find the length of any object. This method relies on the ability to adjust the length of the object and create beats with a known frequency. This is not possible with all objects, such as solid objects with fixed lengths.

4. How accurate is the method of using beats and resonance to find length?

The accuracy of using beats and resonance to find length depends on the precision of the equipment used and the skill of the experimenter. With careful measurements and accurate frequency readings, this method can yield precise results. However, external factors such as ambient noise and variations in the tension of the string can also affect the accuracy of the measurements.

5. Are there any limitations to using beats and resonance to find length?

There are some limitations to using beats and resonance to find length. As mentioned before, this method may not be applicable to all objects. Additionally, the accuracy of the results may be affected by external factors, and the process can be time-consuming. It is also important to note that this method assumes a linear relationship between frequency and length, which may not always be the case.

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