Built on Facts

Tuning a Guitar, Problematically

Today in my recitation we discussed several problems in acoustics. One of them involved beats. This happens when two tones which are very close in pitch are played at the same time. There’s a demonstration on the Wikipedia article. I’ll solve the problem here since if it confused people in class there’s probably people googling it. It’s an easy problem, the difficulty comes from a lack of clarity in this section of the book.

This problem is Young and Geller 12.54:

A violinist is tuning her instrument to concert A (440 Hz). She plays a note while listening to an electronically generated tone of exactly that frequency and hears a beat frequency of 3 Hz, which increases to 4 Hz when she tightens the strong slightly. What was the frequency of her violin when she heard the 3 Hz beat?

Naively, you look up the equation and solve. It’s a simple equation, what could go wrong?

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The problem is that this equation is a little confusing in that it has the possibility of a negative beat frequency. While the given equation is single-valued on its face, in reality there are two possibilities because clearly the beat frequency should be invariant under the interchange of source 1 and source 2. I think it makes much more sense to write the equation like this, and it’s what I have my students do now:

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This way it’s clear that only the difference matters, and there’s no weirdness with negative numbers. For the particular problem under consideration here, it means that there’s two possible starting frequencies for the violin that produce a 3 Hz beat frequency. It could be 443 Hz or 437 Hz. Only given the beat frequency and the reference tone, that’s all we can say.

Fortunately the problem gives us a little more to work with. She tightens the string slightly and the beat frequency increases. This breaks the symmetry of the problem and let’s up find an answer. Tightening the string increases the pitch of the string. So if making the string frequency higher makes the beat frequency higher, then the distance between the two pitches is increasing. That can only happen if we were too high to begin with. Therefore the 443 Hz answer is the correct one.

Comments

  1. #1 Asad
    November 12, 2008

    Well, it’s still ambiguous, because of the fuzziness of the word “slightly”. You could start at 437 Hz, and my definition of “slightly” increasing the tension could be different than yours and then put the string at 444 Hz.

  2. #2 Matt Springer
    November 12, 2008

    True. I take it to mean that the beat frequency increased monotonically to 4 Hz, which is I hope the intent of the question.

  3. #3 CRM-114
    November 12, 2008

    In theory, yes. In practice, I don’t think anybody can hear a frequency of a few hertz, so the tuning is matching frequencies, not zeroing out the beat frequency.

    In the Deep Space Network, the old Block 3 receivers had a beat frequency oscillator circuit which output to a loudspeaker. The operator tuned the receiver to drive the beat frequency down to about zero, then switched to Operate, and the phase-locked loop would track out the error, thus ‘locking’ the loop.

    A zillion dollars worth of equipment, and the part tourists liked best was hearing the receiver locking up. Also, the beat looked cool on an oscilloscope.

  4. #4 Anonymous
    November 12, 2008

    CRM: “so the tuning is matching frequencies, not zeroing out the beat frequency.”

    I assure you that tuning is zeroing out the beat frequency.

    I speak as someone who has been involved in music for over 30 years and whose party trick used to be tuning unplugged guitars on stage in the middle of a 100db racket by tuning out the beat between what I could feel in the neck of the guitar and the note from the tuning fork between my teeth.

    Tuning is relative, not absolute and when you tune you very, very definitely “tune out the beat”.

    That’s why you see guitarists tuning their strings in the middle of songs sometimes – they hear the beat, they know something is wrong and they know which string is out.

  5. #5 Matt Springer
    November 12, 2008

    Right. You’re not hearing a 3 Hz tone, you’re hearing the ~440 Hz tone fade in and out three times per second. Think of it like vibrato in a opera singer’s voice. She might be singing a note with pitch in the kilohertz, but the vibrato itself is modulating that pitch at only a few Hz.

  6. #6 John-Michael Caldaro
    November 12, 2008

    Matt’s ~440 brings up an interesting point. The actual tone your hear that is changing amplitude is the average of the the two frequencies producing the beat. (See Thomas Rossing’s “The Science of Sound”) In the case of 3 beats per second you would be hearing a tone of 441.5 Hz changing amplitude. If you are tuning an instrument it is therefore necessary to have a tone that is generated by a device that has the exact tone you are tuning to. Some people are said to have “perfect pitch.” Supposedly they instinctively know if a tone is not perfect. They might not have to use a device like that. Also, as was said before, the resolution of frequencies by the human ear/brain system has its limits.

  7. #7 JM
    November 13, 2008

    “It could be 443 Hz or 437 Hz. Only given the beat frequency and the reference tone, that’s all we can say.”

    Actually, Matt this isn’t right – the note can only be 443 because of the sentence in the problem “…when she tightens the strong [sic]* slightly”

    This means that the string is now 4hz away from the reference tone and since it is now sharper it can only be 444. If it had started out at 437 it would now be 438 and beating only 2x a second.

    Your analysis is correct however if the she doesn’t tighten the string and the question is asked “is the string sharp or flat in relation to the reference tone”. That you can’t tell from beats alone, what you do is tighten the string and hear which way the beats go – if faster you are sharp, if slower you are flat.

    * yippee – I’ve always wanted to use [sic].

  8. #8 JM
    November 13, 2008

    Sorry Matt – can you nix my last comment? I hadn’t read your last paragraph

  9. #9 mtc
    November 15, 2008

    #6 — I assume you’re saying one percieves the average tone of the two frequeinces. I come at this from an electrical engineering background but the math works out the same: if you add two sinusoidal signals of different frequencies together and take the fourier trasform you just get those two frequency components and nothing else. Even in a non-linear system I’m not sure how you would get the *average* of two frequencies ( I guess a subharmonic of the difference frequency could mix with the lower tone?)

    It’s a cool trick though. I find it works easier on a bass guitar when you tune on the octaves instead of the open string, I suppose because the beats are too slow at the lowest frequencies to be much help by the time it gets close enought to be useful.

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