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Are you in tune with this beat?

  1. Nov 28, 2006 #1
    Two-out-of-tune flutes play the same note. One produces a tone that has a frequency of 258 Hz, while the other produces 266 Hz. When a tuning fork is sounded together with the 258 Hz tone, a beat frequency of 3 Hz is produced. When the same tuning fork is sounded together with the 266 Hz tone, a beat frequency of 5 Hz is produced. What is the frequency of the tuning fork?
     
  2. jcsd
  3. Nov 28, 2006 #2
    What I was thinking that I have to set 2 equations up ...

    258 - X = 3
    266 - X = 5

    ... but, that doesn't make sense, am I looking at this wrong?
     
  4. Nov 28, 2006 #3

    turbo

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    You're looking at this wrong, but not too far wrong. You should Google on "beat frequency" and see how they arise, then look at your frequencies and beat values. You'll go "Aha!"
     
  5. Nov 28, 2006 #4
    I figured it out in a really weird way ... nevertheless, I set the two equations to each other and found out the frequency of the tuning fork is 261 Hz.
     
    Last edited: Nov 28, 2006
  6. Nov 28, 2006 #5

    turbo

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    If you understand this at the level that I think you've "gotten it", let me pose you another question. Let's say that your reference (tuning fork) is 440Hz and you have to notes (x and z) to compare it with and those notes are 10Hz apart. Note x gives you 15Hz beats when played with the reference tone and note z gives you 20Hz beats when played with the reference tone. What can you tell me about the frequencies of note x and note z?
     
  7. Nov 28, 2006 #6
    Note "Z" has a lower frequency from "A" (440 Hz) and note "X" has a higher frequency ... am I right?
     
  8. Nov 28, 2006 #7

    turbo

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    No, since the frequency differential between x and z is smaller than the beat frequency of either with respect to the reference 440Hz, they are both on the same side of the reference note (higher or lower Hz). Can you solve the frequencies of x and z?

    Edit: Your teacher gave you a special case.
     
    Last edited: Nov 28, 2006
  9. Nov 28, 2006 #8
    Note "X" has a frequency of 450 Hz and note "Z" has a frequency of 420 Hz ... that's my final answer ... [scratches noggin]
     
  10. Nov 28, 2006 #9

    turbo

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    Since notes x and z are 10Hz apart, they cannot be 450Hz and 420Hz. Consider that x and z can be either at higher or lower frequencies than the reference note, and re-compute. You may not get a single set of frequencies.

    Your teacher gave you a special case in which the beat frequencies were both smaller in magnitude than the spread between the two notes
     
  11. Nov 28, 2006 #10
    Now, I'm lost ... [cries]
     
  12. Nov 28, 2006 #11

    AlephZero

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    Turbo-1, are you sure you haven't made a typo in your question? I think I have a pretty good understanding of beat frequencies (I've done a lot of research into early tuning systems for musical instruments, tune my own piano and harpsichord, etc) but I can't make sense of the numbers in your question, if we are talking about pure (sine-wave) tones.
     
  13. Nov 28, 2006 #12

    AlephZero

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    Sailordragonball, I think you "got it" - if you have a tone at 258 and a beat frequency of 3, the other tone can be at either 258+3 or 258-3. So your original equation "258-X=3" is only telling half of the the story.
     
  14. Nov 28, 2006 #13

    turbo

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    The notes can either both be at higher OR lower frequencies than the reference note. You will end up with two sets of possible answers. I told you that your teacher gave you a special circumstance.
     
  15. Nov 28, 2006 #14

    turbo

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    You are absolutely right - I was trying to edit the problem to make it easy and ended up with unworkable beats/frequency differentials between the notes. sailordragonball, I apologize. Let me set this up properly to explain the problem.

    Suppose your tuning fork (reference) is 440Hz and you have beat frequencies of 20Hz with tone x and 30Hz with tone z. These two tones must be 10Hz apart, and there are two sets of frequencies that can produce these beat frequencies. You can have x at 420Hz and z at 410Hz OR you can have have x at 460Hz and z at 470Hz. In the first case, z is at the lower frequency(x/z), and in the second case, x is at the lower frequency(x/z).

    I apologize for the error in the problem that I set up for you, sailordragonball. I was trying to simplify it to demonstrate that the beat frequency cannot necessarily give you a real fix on the tone that you are comparing to the reference tone and I did not proofread after editing.
     
  16. Nov 29, 2006 #15
    It's OK, Turbo-1 ... I still don't get it ... LOL - but, I got the answer ... but, I would like a mathematical explanation though.
     
  17. Nov 29, 2006 #16

    turbo

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    Your teacher gave you an example in which the beat frequencies were so small that the reference frequency just had to fall in between the two flute tones. In real life, tones above and below your reference frequency can cause the same beat frequency. A tone 10Hz above the reference tone will cause a beat frequency of 10Hz, but so will a tone 10Hz below the reference frequency. That's why if you have a reference frequency and a beat frequency produced when a note is played along with the reference, there are always two possible frequencies for the played note.
     
  18. Nov 29, 2006 #17
    OIC ... I understand now ... something was wrong when you gave your example ... LOL - thanks.
     
  19. Nov 29, 2006 #18

    turbo

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    Yeah, I wrote out the problem, then decided to simplify it to rounder numbers, but failed to proofread it and didn't revise the beat frequency of one of the played notes. I'm sorry for the confusion. Beat frequency is the absolute difference (in Hz) between any two notes, meaning that if you have a reference (like a 440Hz A tuning fork) there are two notes (one higher, one lower) that can produce any given beat frequency with the tuning fork.

    As a guitarist, I always detune a string, then tune up to the reference note. This minimizes mechanical problems with gear backlash, friction at the nut and bridge etc, but it is also a quick and dirty way to know which way to tune to reduce beat frequency.
     
    Last edited: Nov 29, 2006
  20. Nov 29, 2006 #19

    OlderDan

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    It all comes down to setting up the correct equations. You only know the absolute difference of the frequencies

    |258 - X| = 3 » X = 255 OR 261
    |266 - X| = 5 » X = 261 OR 271

    X = 261

    Your beat frequencies both happen to be less than the difference between the known frequencies, so the unknown is between the two known frequencies. If one or both of the beat frequencies had been greater than the difference, the unknown would be above the higher or below the lower and closer to the one with the lower beat frequency.
     
  21. Nov 29, 2006 #20
    Yeah, I read in the book that it was the ABSOLUTE VALUE ... I'm just a nerd. Thanks.
     
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