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Sound as compression or rarefaction of air

  1. Jun 6, 2015 #1
    When a loudspeaker moves back and forth, sound is generated. The moving of the loudspeaker diaphragm causes condensation of air when it moves outward, and rarefaction of air when it moves inward. When this movement is happening at a certain frequency, we hear a sound. See for example this illustration:

    https://www.physicsforums.com/attachments/sound-gif.84528/?temp_hash=012fb293b7e650bf392ecc98c949c953 [Broken]​

    I was thinking about this and I understand we perceive the condensation/rarefaction as sound. But what happens if we have only one or the other? I.e. if we have a loudspeaker only moving from it's initial position outward ONCE, only condensation of air happens. In the illustration above, this would correspond to only one crest of the wave, but no trough.

    Same question goes if the loudspeaker only moves backwards from it initial position once, so there's only rarefaction (one trough, no crest).

    So my questions are:
    1. If ONLY ONE condensation (one "crest") is happening, will we still perceive it as sound?
    2. If ONLY ONE rarefaction (one "trough") is happening, will we still perceive it as sound?
    3. If the answer is yes to any or both of question 1 and 2, are the sounds in 1 and 2 different from each other? If so, how?
    4. If the answer is yes to any or both of question 1 and 2, are the sounds different from a sound that contains both condensation and rarefaction? If so, how?
     
    Last edited by a moderator: May 7, 2017
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  3. Jun 6, 2015 #2

    Vanadium 50

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    If you have a crest without a trough, where did the extra material come from?

    Your question is based on an impossible premise, I'm afraid.
     
  4. Jun 6, 2015 #3
    So what happens if the diaphragm of the loudspeaker only moves outward once from it's initial position? Wouldn't that create only condensation - i.e. only a crest?
     
  5. Jun 6, 2015 #4

    Vanadium 50

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    The answer is still no. If you have an overpressure and not an underpressure, where does the extra material come from?
     
  6. Jun 6, 2015 #5

    jbriggs444

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    From half a wavelength (or less) over?

    You move the speaker membrane in toward the room by some increment. The air displaced by this motion is a local excess. The pressure associated with this local excess causes material to move further into the room causing a local excess farther in. And so no. The phenomenon is called a wave. Ignoring dispersion, many wave shapes are possible. Including a wave form that consists of a single pulse above or below the baseline.

    Of course, if one has a single pulse in the middle of an otherwise flat baseline, one might argue that the baseline amounts to a single infinitesimally deep trough.

    Or am I failing to take your point?
     
  7. Jun 7, 2015 #6

    CWatters

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    How about the pressure wave/pulse from an explosion?
     
  8. Jun 7, 2015 #7

    Vanadium 50

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    Makes for a poor loudspeaker. Too much harmonic distortion.
     
  9. Jun 7, 2015 #8

    CWatters

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    I see no reason why you can't also create a negative pulse. For example by evacuating a cylinder slowly until it fails (or a diaphragm in the wall fails).
     
  10. Jun 7, 2015 #9

    boneh3ad

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    The premise almost works. You would have a pulse that quickly drops to ambient after the speaker finishes its movement. It would overshoot on its way to ambient, I'd wager, thus conserving mass. So in that sense it's still one wavelength, not half a wavelength as originally proposed.
     
  11. Jun 7, 2015 #10

    DaveC426913

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    Regardless of half versus full wavelength arguments, a single pulse is tantamount to an infinite wavelength, which is a frequency of zero. We do not perceive sounds below 20Hz. Elephants can perceive sound down to 10Hz.

    But 0Hz is not a sound. It is simply a shock wave. And I don't think it's physically possible to have one alone.
     
  12. Jun 7, 2015 #11

    jbriggs444

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    Nonsense. A single pulse has Fourier components at infinitely many wavelengths.
     
  13. Jun 7, 2015 #12

    DaveC426913

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    How does that negate anything I said?
     
  14. Jun 7, 2015 #13

    jbriggs444

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    Because it is a compression wave in air with a component at audible frequencies.
     
  15. Jun 7, 2015 #14

    DaveC426913

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    (EDIT: changed the question)
    What frequency would that be? A single overpressure?
     
  16. Jun 7, 2015 #15

    jbriggs444

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    Infinitely many, as already stated.

    Edit: The question is ill-posed. You have to specify the wave form for the single pulse before its components are determined.
     
  17. Jun 7, 2015 #16

    DaveC426913

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    Are you asserting that a single overpressure would result in a person hearing a sound at all frequencies?
     
  18. Jun 7, 2015 #17

    jbriggs444

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    Have you ever heard of Fourier analysis?
     
  19. Jun 7, 2015 #18

    DaveC426913

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    I have. I think you are caught up in the mathematics of the problem, and have lost sight of the question being asked, which is about perception.

    Ears do not hear Fourier transformations. Ears require periodic pulses to interpret them as sounds.
     
  20. Jun 7, 2015 #19

    jbriggs444

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    Reference, please.

    You have falsely claimed that a shock wave is 0 hz. I'd like a reference for that claim as well.
     
  21. Jun 7, 2015 #20

    DaveC426913

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    The onus is on you to demonstrate the relevance of post #11 - Fourier transformation in the perception of sound. I think you've diverged onto a tangent. (My post 12 essentially says: "relevance?")

    (Dang. Gotta step out. Won't be online for a few hours. But I'll be back.)
     
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