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What does this graph say?

  1. Sep 8, 2005 #1

    honestrosewater

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    I read the explanation several times, but I don't know how to interpret what I'm looking at. It's the third graph on this page, under the heading Spectrum (the paragraph above the graph explains it). Did they take a section of the recording and average the frequencies and amplitudes? Could someone please walk me through how they constructed the graph and maybe tell me what one point says?

    BTW, I'm studying this for phonetics. At the moment, I know very little about acoustics or physics, so any extra explanation about sound waves and their properties that are being measured here would help. For instance, I only have a fuzzy idea of what frequency, amplitude, and phase are. :frown:
     
    Last edited: Sep 8, 2005
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  3. Sep 8, 2005 #2

    G01

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    Go to the topic Sound Wave Calculations. I tied explaining some stuff about waves in general, especially sound waves there.
     
  4. Sep 8, 2005 #3

    LURCH

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    I don't think it's averaged out; if I read the description correctly, this is a snapshot that comes as close as possible to isolating a "single moment" in time, so there's nothing to average. It looks to me like, at the moment this snapshot was taken, the speaker was producing a large number of different frequencies of soundwave. Most of the waves he was producing were in the mid-range (between 2000 and 5000) and of medium volume, but he was also making a few waves in a narrow band of bass (below one thousand) and these waves, though fewer, were louder than the rest. I don't think that means that the bass was louder over-all, because these few louder waves are not loud enough to dominate over the great majority of waves at the mid-range at medium volume.

    Sidenote:
    When I came back to reply, I was going to ask you what kind of sound you thought this set of waves would produce. I haven't ever studied phonetics or accustics, but it looked to me as though they caught the speaker in the middle of pronouncing a vowel. I figured you could tell me if I was right or wrong about that, but then I saw that they tell you right above the graph that they took this image in the beginning of an "O" sound.
     
  5. Sep 8, 2005 #4

    FredGarvin

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    It is a FFT (fast fourier transform) of the source at the instant they are referring to. The FFT breaks a signal down into it's constituaent frequencies. The peaks represent the major frequencies that make up the overall sound. In other words, you are looking at a signal in the frequency domain, not in the time domain as you may be used to (plot #1). This is a numerical technique.
     
  6. Sep 8, 2005 #5

    honestrosewater

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    Okay, so pretend that this graph is labelled as the first one. Mostly loud and low pitch. ?? This might be a man grunting, perhaps as he's punched in the stomach (just my thoughts). I don't know what to make of the shape of the graph though - the sharpness of the curves, for instance. The frequencies being 'concentrated' or not varying much - would that indicate a purer tone? :yuck: Ugh, I think I'll read around and learn the basics of sound waves before going further.
     
  7. Sep 8, 2005 #6

    FredGarvin

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    The large peak indicates that the signal is predominantly made up of a pure wave at the frequency that it corresponds to, in this case about 30 hz. The "pureness" of the signal is representative of how many peaks you see. All of the peaks beyond that are overtones that give a sound it's richness. A pure tone would have one single, sharp peak. If you have repeating peaks at integer multiples of the dominant frequency, i.e. 60, 90, 120 hz...then those would be harmonics.
     
  8. Sep 8, 2005 #7

    Gokul43201

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    That's right (about the sharpness indicating a purer tone) ! But don't say that the frequencies are not varying - that seems to suggest a time-dependence.
     
  9. Sep 8, 2005 #8

    Gokul43201

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    Also, it might be useful to make a comparizon with a light spectrum. The (white) light from an incandescent bulb (or the sun) would have a broad spectrum, indicating that it is a mixture of all frequencies. Light (the emission spectrum, really) from a laser will be sharply peaked about a single frequency.
     
  10. Sep 9, 2005 #9

    honestrosewater

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    Wow, our nervous systems are amazing: 20 to 20,000 Hz and 0 to 120+ dB?! Maybe that's not impressive to you guys, but it sure surprised me. :surprised And the precision! :biggrin: Anywho...

    I think I've got most of it. I just can't visualize the actual propagation of the sound wave. Here's my source s, receptor r, and medium, say, air molecules, o:
    Code (Text):
    s o o r
    The source moves, hitting its neighbor:
    Code (Text):
     so o r
    Now what?
    Code (Text):
    s  oo r
    ?? Why did the source move back to its original position? Solely because of the collision with its neighbor? (Is there any 'vacuum forcey stuff' going on there?) Would it proceed like so:
    Code (Text):
    s o  or
    s o o  r
    s o o r
    Erm, ideally. ;)
     
    Last edited: Sep 9, 2005
  11. Sep 9, 2005 #10

    Gokul43201

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    The source moves back (not because of collisions with air molecules or vacuum thingies but) because it is designed to vibrate - it is a drum-head or a string in tension. The elastic restoring forces in the material of the source are what cause it to move back and forth (vibrate). Damping forces in the material cause the motion to die down (decay) with time unless externally "pumped".
     
  12. Sep 10, 2005 #11

    LURCH

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    I think she may not be referring to the original source, but the substance within the medium. As a wave propogates, any part of the medium through which it vibrates can be said to be the source of the subsequent vibration further down the line.

    But, even if that is what you're asking, Rose, the above post holds the answer to that question, as well. As a speaker or drumhead bounces back from its initial motion, it sucks air in behind it, forming a low-pressure area imediately behind the high-pressure compression wave it sends out. Of course, with vocal chords I don't think this occurs, because the compression wave is being created by the movement of air through the larynx. So the compression waves in speach are riding on a moving current of air.
     
  13. Sep 10, 2005 #12

    honestrosewater

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    Okay, cool. Thanks. :biggrin: I guess I'm having trouble understanding how the wave keeps moving in an orderly way instead of just kind of scattering or falling apart almost immediately. Say the source has collided with the first o and moved back into its original position:
    Code (Text):
    s  oo o r
    The first o moves to the right, creating an area of (relatively) low pressure to its left. The first o collides with the second o, creating an area of high pressure. Due to this collision (and the differences in pressure?), the first o moves back ~into its original position, and the second o moves to the right:
    Code (Text):
    s o  oo r
    And the process keeps repeating. ?? That seems pretty clear, but for some reason, I can't get a mental model of everything that's happening in more complex situations. What, if any, are some of the other significant things going on when a wave travels through the air?
     
  14. Sep 11, 2005 #13

    SpaceTiger

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    Looks like there are some excellent answers so far. I just wanted to say a few things that may clarify the picture a bit.

    We should be careful to specify what we mean by "oscillation" or "wave" when we're analyzing this picture. If we take the source to be an oscillating drumhead, then we're talking about a harmonic oscillator, while with the propagation of sounds waves, one would use the wave equation. Both produce an oscillation, but the qualitative picture is quite different.

    The harmonic oscillator is somewhat easier to understand, I think. Basically, you have a system that can move in either direction about some eqilibrium point, but with a force that will always try to pull it back to center. If you displace it a little bit in one direction, then the force will pull it back, but once it reaches center, it has momentum, so it keeps going past the equilibrium point until it's displaced by nearly the same amount in the opposite direction. If it weren't for friction and drag, this process would repeat indefinitely, but in practice the oscillation will die out eventually. This is basically how a drumhead or a guitar string would work.

    However, the propagation of sound waves through the air is in both time and space. The wave equation basically takes a disturbance (of any shape) and propagates it forward. The only reason it takes the form of a steady up-and-down oscillation in the example being discussed is that the source is undergoing such an oscillation. One could just as easily propagate the disturbance created by a clap of the hands as by the vibration of a guitar string. In fact, it's fortunate for us that the medium doesn't continue oscillating after the disturbance passes because that would create an awful lot of background noise!


    The pictures you've drawn are a good representation, but I would caution against taking them too literally. It's probably safer to think of the propagation in terms of the properties of space at each location. For example, let's say that I have a wave moving to the right. I might represent the initial density of the gas as:

    0 1 0 0 0 0 0

    where "0" means the ambient density and "1" is an overdensity created by a disturbance of some kind. At the next moment, it might look like:

    0 0 1 0 0 0 0

    and then the next:

    0 0 0 1 0 0 0

    ...and so on. The disturbance doesn't have to be so simple, however, I could just as easily propagate the following:

    0 1 -2 4 0 0 0 0

    0 0 1 -2 4 0 0 0

    0 0 0 1 -2 4 0 0

    The point is that the wave equation just "moves" the disturbance along, maintaining its shape. The mechanism for this propagation is much like your picture demonstrates, but it's tempting to interpret the motion of the "o" in your picture as the actual motion of the molecules in the gas. This is a little deceptive, since the sound wave is an "average" effect (i.e. the combination of the complicated motions of many molecules).
     
    Last edited: Sep 11, 2005
  15. Sep 12, 2005 #14

    honestrosewater

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    Which wave equation did you mean? The only 'wave equation' I understand is [itex]v = \lambda f[/itex]. I understand some of its concepts, but I don't speak calculus, so I can't really read the others.

    Could "0 1 0 0 0 0 0" actually represent a wave? That looks like a compression without a rarefaction. I thought that the 'cycle', a compression and rarefaction, was a necessary characteristic of a sound wave. Can you have just a compression or just a rarefaction? What are the minimum requirements that a disturbance must meet in order to count as a sound wave? I mean, I get the difference between a hand clap and a vibrating string, but doesn't the clap still produce at least one cycle?
    Sheesh, sorry, I have a lot of questions, but I'll try to make the rest easy to answer.
    1) In

    0 1 -2 4 0 0 0 0

    the numbers represent amplitude?

    2a) Is amplitude measured from ambient density (or pressure*) to the peak of the compression (or rarefaction)? So that in

    0 -1 3 0 0 0

    the amplitude of the compression would be 3, and the amplitude of the rarefaction would be 1. Or is amplitude just half of the intensity (intensity being the difference between the peak of the compression and the peak of the rarefaction)? So that, in the above wave, the amplitude of both the compression and rarefaction would be 2.

    *2b) Can I use density and pressure interchangeably?

    3) Could

    0 -1 3 0 0 0

    be a wave? That is, assuming amplitude is measured from ambient pressure, must the amplitudes of the compression and rarefaction be equal?

    4) Could

    0 -2 2 0 0 0

    be travelling in either direction, i.e., must rarefaction follow compression, or can rarefaction precede compression?

    5) How fine-grained are the changes in pressure? Could I safely expand

    0 -2 2 0 0 0

    into, for instance,

    0 0 0 0 -.5 -1 -1.5 -2 -1.5 -1 -.5 0 .5 1 1.5 2 1.5 1 .5 0 0 0 0

    6) Does the pressure change, increasing or decreasing, in a predictable way? Perhaps at a certain rate depending on frequency and wavelength?

    7) I may be able to figure this out, but I guess I may as well ask. If you know the speed of the wave and the ambient pressure, do you need to hear an entire cycle in order to determine the frequency, wavelength, and amplitude, or can you determine them from only a segment of a cycle? What is the smallest segment that you need? And so on - just the general "if I know x, I can determine y" kind of questions. I'm not so much asking for an answer - it's just something I'm thinking about.
    The amplitude seems easy. If it's measured from ambient pressure, you just need to hear the segment where the pressure changes from increasing to decreasing or vice versa, i.e, one of the peaks. But if the pressure changes in a predictable way, or at a certain rate, perhaps any segment could tell you the amplitude, perhaps given the frequency and wavelength.

    Eh, I realize I'm asking a lot. I was hoping that this would be a simple little foray, but the more I learn, the more questions I have. And I think that my book is oversimplifying things and not giving me the whole story. Maybe someone can recommend a book or website that could answer my questions?

    And as always, thank you so much - you guys rock! :biggrin:
     
    Last edited: Sep 12, 2005
  16. Sep 12, 2005 #15

    SpaceTiger

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    Well, I would certainly call it a wave, but let's avoid any potential terminological debates and say the following:

    - The above disturbance could be created in air.
    - The disturbance would be propagated and described by the wave equation.

    I used that simple example because I wanted to demonstrate the behavior of the wave equation. In practice, both a compression and rarefaction will usually occur because total matter has to be conserved. If I want to compress the gas in one place, then I must take particles from elsewhere.


    Unfortunately, the wave equation I'm referring to does involve calculus, but let's see if we can step through it. For simplicity, we'll continue to work in one dimension:

    [tex]\frac{\partial^2f}{\partial^2x}=\frac{1}{v^2}\frac{\partial^2f}{\partial^2t}[/tex]

    What this equation basically says is that the shape of the function in space is equal to the shape of the function in time multiplied by a normalization constant (it's a little more complicated than that, but that's the basic idea). To visualize this, imagine you're in a boat on a flat lake and there's a wave coming at you. If the wave were governed by the above equation, then the following two things would be equal:

    1) Sitting motionless and letting the wave pass under you (it's moving at speed v).
    2) Freezing the wave and passing over it at speed v.

    Notice that I didn't say anything about the character of the wave. It could be a single hump, a wiggle, or a sine function. The point is that the shape is maintained as it moves forward. In practice, water waves are not well described by the above equation, but sound waves are, so let's move on to the issue at hand...


    I'm sure it depends on who you ask. I would say that any disturbance that satisfies the wave equation could be called a wave, but these things aren't written in stone. Some people would probably require, as you say, at least a single cycle.


    The numbers are meant to be representations of the density relative to the ambient ("normal") value. However, they could just as easily represent pressure or temperature.


    Amplitude is traditionally used to refer to a property of simple waves, like sine waves, where the symmetry makes the definition is unambiguous. In the case of sound, they usually talk about the "root mean square" amplitude, which is the square root of the average of the square of the deviation from the ambient value. A mouthful, I know. In your case:

    0 -1 3 0 0 0

    this would be given by:

    [tex]A_{rms}=\sqrt{\bar{A^2}}=\sqrt{\frac{(-1)^2+3^2}{2}}=\sqrt{5}[/tex]


    Well, they're not numerically equal, but they both are governed by the wave equation.


    I would give the same answer as above -- it depends on who you ask. I would say it could be a wave (and so would scienceworld), but I've seen a variety of definitions.


    It can go either way. Both are solutions to the wave equation, but are distinguished by the initial conditions.


    In theory, it could show any amount of small scale variation (down to intermolecular distances). This goes back to the graph with which you started the thread, because it's the spectrum that tells you how fine-grained it is. If the spectrum shows a lot of power at high frequencies, the expansion will likely be more like:

    0 0 0 -0.5 -1.5 -2.5 -1.5 -2.5 -1.5 -1 0.5 -0.5 1 1.5 2.5 1.5 2.5 1.5 2.5 0 0 0 0


    It depends on the circumstance. Musical notes are very predictable, while a talking crowd would likely produce a very erratic sound.


    That's an excellent question. I don't know how sophisticated the software in our ear is (you should ask in the biology forum), but in astronomical systems, we regularly infer periodic motions from only a small portion of a cycle. Making this inference does, of course, require assumptions, so we could easily interpret something as having a certain frequency when it really doesn't!
     
  17. Sep 12, 2005 #16

    honestrosewater

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    :rofl: Yes, that makes sense.
    Okay, I'm easy. I just wonder what frequency and wavelength would mean for a disturbance that doesn't have a cycle.

    Anyzok, (:tongue2:) thanks for taking the time to answer. I think I'm comfortable enough now that I can move on and just learn more as needed.
     
  18. Sep 12, 2005 #17

    SpaceTiger

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    You don't assign a single frequency or wavelength to it, but you can construct it from a combination of many sine waves that each have their own frequency and wavelength. The relative contributions of these simple waves is what one is looking at when they view the spectrum.



    You just made my list of favorite people. :biggrin:
     
  19. Sep 12, 2005 #18

    honestrosewater

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    Oh, cool, it's all starting to click.
    It was delicious. :!!) I want more - please say there's more. :puppydogeyes:
     
  20. Sep 12, 2005 #19

    SpaceTiger

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    Gunther was based on something I wrote for a high school writing evaluation, so there is more, but I have no idea where it is. :frown:

    As for that particular story, I add to it whenever I'm in the mood. Perhaps it's time for another update...
     
  21. Sep 13, 2005 #20
    the graph says, that at lower frequencies, the amplitude ( ie. volume) ( ie. loudness) of the sound is higher, but the higher frequencies are not as loud as the lower frequencies..... if this were all caused by the same expenditure of energy, i would say this is a case of the higher frequencies using up more power to produce the oscillation that the lower ones...
     
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