Frequency of a Tuning Fork

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Homework Statement



A 40-cm-long tube has a 40-cm-long insert that can be pulled in and out. A vibrating tuning fork is held next to the tube. As the insert is slowly pulled out, the sound from the tuning fork creates standing waves in the tube when the total length L is 42.5 cm, 56.7 cm, and 70.9 cm.

Homework Equations



The wavelength= 2L/m where m=1,2,3,4...
frequency: mv/2L where m=1,2,3,4

The Attempt at a Solution



This is an open ended-opened tube. Since I know the values of L, I can find the wavelengths (but I still don't know m). The frequency of the tuning fork would not change, but I am confused about what to do next. How do I approach this question? Thank you for all the help.



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Answers and Replies

  • #2
G01
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Each value of m gives you a standing wave with a different amount of nodes, correct? Notice that you have three L's which give you standing waves. How would these relate to the values of m?
 
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I still don't really understand how the value of m connects to the three values of L. And how can that help me to find the frequency of the tuning fork?

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  • #4
G01
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The frequency of the tuning fork will have to be equal to the frequency of the wave created by it.

What does the value of m correspond to physically? Notice that the shortest standing wave is set up with an L of 42.5cm. This will then correspond to a standing wave with one antinode. The second distance would then have to correspond to a standing wave with two antinodes, etc.
 
  • #5
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I assumed that this system is an open-open system. Is that right? If it is accurate, then wouldn't m represent the number of nodes? Would the second length represent the second harmonic then? Wouldn't changing the length change the fundamental frequency? How do you know that the shortest standing wave has one anti-node?
 
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  • #6
G01
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I assumed that this system is an open-open system. Is that right? If it is accurate, then wouldn't m represent the number of nodes? Would the second length represent the second harmonic then? Wouldn't changing the length change the fundamental frequency? How do you know that the shortest standing wave has one anti-node?

An open-open system? Do you mean open-closed? Remember there is a movable barrier at one end, so that end isn't open.

I mis-spoke before. The m value for an open-closed system does not correspond to the number of antinodes or nodes. Still, m=1 corresponds to the standing wave with the least number of anti nodes, and the number of anti-nodes increases as m increases, but they are not equal. Also, remember for a open-closed system, m can only be odd and the formulas are different from the ones you have above. Make sure you use the correct ones.

Remember the frequency of the wave can't change, since all that depends on is the tuning fork itself. Changing the barrier will change the amount of anti-nodes and nodes of the standing wave that can fit in the tube, that is all.

Since, we have an open-closed system, the fundamental standing wave will be one that has a node at the barrier and an antinode at the end of the tube. This has to correspond to the L=42.5 case, since there are no standing waves for any shorter distances. So, m=1 corresponds to L=42.5m.

Now, what would m=3 and m=5 correspond to? (Remember m can only take odd values in this case.)
 
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  • #7
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I am doing the same problem as well. So from what I have read, since the frequency of the tubes does not change, the wavelength of the standing wave would be 56.7-42.5=14.2cm or 0.142m. Then I would take the speed of sound divided by the wavelength to get the frequency, am I correct? Also, is the speed of sound for the question 343m/s or 340m/s? Thanks in advanced.

Edit: wait is the above mentioned wavelength only half the value? Do I need to multiply 0.142m by 2 to get 0.284m, the actual wavelength?


Nvm yea it's half the wavelength, the answer is 1210 Hz for future references.
 
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