Resonant Frequency of a tuning fork

  • #1

Homework Statement



Name the two lowest resonant frequencies of a tuning fork; each tine is 18 cm long. Speed of wave is 420m/s... ignoring metal thickness.



Homework Equations



F= Velocity/Wavelength; l= wavelength/2, l= wavelength, l= 3/2wavelength, l= 2wavelength, l= 5/2wavelength



The Attempt at a Solution



fundamental = 420m/s / .72m = 583.3 Hz
first overtone = 420m/s / .36m = 1167 Hz
second overtone = 420m/s .24m = 1750 Hz

The given answer is 583 Hz
1750 Hz
2917 Hz

I would have expected to use the formulas wavelength/2, wavelength, 3/2wavelength. They have used, wavelength/2, 3/2wavelength, and 5/2wavelength. Is this because of the antinode-node boundary for each tine? I'm confused as to why they make the frequency jumps that they do. Very elementary physics and I am no student. So, I have no one to ask.
 

Answers and Replies

  • #2
Tom Mattson
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Homework Equations



F= Velocity/Wavelength; l= wavelength/2, l= wavelength, l= 3/2wavelength, l= 2wavelength, l= 5/2wavelength

These equations aren't right. If [itex]L=\lambda/2[/itex] then it means that there is a node at both ends. But one end of the tine is free. So you must have a node at one end and an antinode at the other. Does that help?
 
  • #3
These equations aren't right. If [itex]L=\lambda/2[/itex] then it means that there is a node at both ends. But one end of the tine is free. So you must have a node at one end and an antinode at the other. Does that help?

Thanks for the reply... The equations are correct in the case of an antinode at each end. I took it that the book's writer must be working it that way, even though the fork has a node at the handle and two antinodes at the ends.


If he is basing it as if it's a node and an antinode then it would be [itex]L=\lambda/4[/itex], [itex]L=\3/4lambda[/itex], etc. This would be for two different wavea, though? One in each tine? That is to say, the 'L' would be 18cm? Twice? Even so, his method ofworking the problem does not compute that way, especially in light of:

I do not understand why he worked the problem by what seems to me is 'skipping' every other overtone?

Is my question clearly stated? Thanks, again
 
  • #4
Tom Mattson
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Thanks for the reply... The equations are correct in the case of an antinode at each end.

But you don't have an antinode at each end. One end of the tine is fixed.
 
  • #5
Yup. Using the l=1/4(Lambda), l=3/4(Lambda), l=5/4(Lambda). Considering the system as a node-antinode relationship (which it clearly is), and thus employing .18m as the tine length, viola!

Thanks
 

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