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B Sound Waves -- propagating, dispersing, volume effects?

  1. Jul 25, 2017 #1
    Generally sound waves are depicted as simple sine waves, where volume is related to amplitude, and there is periodic motion. Realistically sound waves aren’t as simple. I attached a picture of a dissipating sound wave. I would appreciate if you guys could answer a couple questions I have about it.

    1.) Aren’t waves defined by periodic motion? Can real sound waves really be described as waves if they change in amplitude over time?

    2.) How loud would the sound wave be at each of the three points? Would they each have the same volume as perfect non-dissipating sine waves would if they were drawn out to go through those points?
     

    Attached Files:

    Last edited by a moderator: Jul 25, 2017
  2. jcsd
  3. Jul 25, 2017 #2
    Yes, a decaying wave is still a wave.
    You would need to measure the amplitude at these points and use the formula for converting sound pressure into decibels.
    $$L_{p}=20log_{10}\left(\frac{p_{rms}}{p_{ref}}\right)$$
    where ##p_{ref}## is the standard reference sound pressure of 20 micropascals in air.
     
  4. Jul 25, 2017 #3
    i am not expert in sound wave field, but how is the wave amplitude being measured? connecting the peaks with a line will show you the type of decay function.

    is a wave periodic? in general no. continuous waves that form patterns typically have a mathematical function to define it, but how would you describe a single ripple that travels?

    model it like pebble in infinite pond. thus take your XY graph and now rotate it 360 degree to get your spatial representation.

    and following NFuller post, db is that log scale, 10db is 10x, 20db is 100x, so if the point is -10db from origin the intensity will be -10x, if its -20db the intensity will be -100x.
     
  5. Jul 25, 2017 #4
    But what is the amplitude at point B? The amplitude is generally taught to be the highest point of a wave, but if the wave is decaying this wouldn’t be as easy.
     
  6. Jul 25, 2017 #5
    i mentioned decay function. to find this function the sound pressure(x) has to be recorded at multiple spots along wave path. it would become a simple factor in the equation.

    $$L_{p}=20log_{10}\left(\frac{p_{rms}}{p_{ref}}\right)*{decay(x)}$$
     
  7. Jul 25, 2017 #6
    Notice the value ##p_{rms}## is a root-mean-square value meaning it is a type of average of the sound pressure at a given point in time. So ##p_{rms}## is not zero at point B.
    You should not multiply a decay function with this equation. Use the equation as is.
     
    Last edited: Jul 25, 2017
  8. Jul 25, 2017 #7
    Wouldn’t the root-mean-square value be constant, and therefore indicate a constant volume at any point on the wave?
     

    Attached Files:

  9. Jul 25, 2017 #8
    The rms value is defined over an interval from ##t_{1}## to ##t_{2}##.
    $$p_{rms}=\sqrt{\frac{1}{t_{2}-t_{1}}\int_{t_{1}}^{t_{2}}\left[p(t)\right]^{2}dt}$$
    Generally speaking, the time interval should be short enough to respond to changes in amplitude but not shorter than several periods of the signal. So if the signal is slowly attenuated over many rms measurement intervals, the rms value will steadily decrease. If the signal very rapidly decays such that the decay is captured in one rms measurement interval, which is roughly the case in your drawing, there will be a sudden drop in the rms value.
     
  10. Jul 25, 2017 #9

    JBA

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    Air has mass and the propagation of sound waves depends upon moving that mass which requires energy drawn from the sound wave and therefore the reduction of the amplitude of the wave form with its distance from the source point.
     
  11. Jul 25, 2017 #10
    Would this just approximate the volume of the wave? I don’t have the knowledge to fully grasp what you’re saying, but the times chosen for the time interval seem somewhat arbitrary.
     
  12. Jul 25, 2017 #11
    The whole idea of volume in sound is sort of an approximation. The decibel system isn't perfect; it is meant to help quantify loudness perceived by the human ear. Our ears will not pick up a rapidly decaying waveform as a decrease in loudness but rather as a unique sound. The waveform in your drawings is similar to the sound of a drum being hit: a load sound which rapidly falls off in amplitude.
    They are, I think a lot of rms devices like multimeters use an interval from 0.1-0.5 seconds.
     
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