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Homework Help: Sound intensity and the hearing distance

  1. Dec 17, 2009 #1
    1. The problem statement, all variables and given/known data

    A tiny speaker emits sound with a frequency of 1000Hz evenly on all sides. On
    a distance of 1m from the loudspeaker is an amplitude variation of the particles of air 0.005 μm. On what distance we can barely hear the sound, if the limit of the audible strength or intensity is 10^-12 W/m2? The speed of sound is 340m/s, air density is 1.2 kg/m3. What is the sound power of the loudspeaker at a distance of 1m?


    2. Relevant equations

    I= P/Area

    3. The attempt at a solution

    For the second part: What is the sound power of the loudspeaker at a distance of 1m?

    I= P/ 4πr² → P= 4πr²*I
    P= 1.256 x 10^-11 W

    Can someone please give me a hint for the first part?
    Thank you for helping!
     
  2. jcsd
  3. Dec 17, 2009 #2

    cepheid

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    I don't think this is correct, because I = 10-12 W/m2 is not the sound intensity at 1 m. It is the limiting sound intensity below which humans can't hear anything, which in this case will occur at some unknown distance from the source that you are supposed to solve for. What you need is a relation that will tell you the intensity in a sound wave at a given density, sound speed, and amplitude of oscillation (all of which you have been given). I'm sure that this relation must have been derived somewhere in your notes or your book, right?
     
  4. Dec 17, 2009 #3
    You need the formula that relates intensity of sound to the wave's parameters:
    I =C f^2*rho*A^2
    Where rho is density, f is frequency, A is the amplitude (maximum displacement).
    C is a constant, I don;t remember the exact form. It shoul be in your book.
     
  5. Dec 17, 2009 #4
    So, the intensity of the tiny speaker is:
    I= C*f²*ρ*A²
    I= 340 m/s * (1000 Hz)² * 1.2 kg/m³ * (0.0000005 m) ²
    I= 1.02 x 10^-6 W/m²

    Is this correct?

    But how do I continue now? I still have to calculate the distance at which we can barely hear the sound. Help?!
     
  6. Dec 17, 2009 #5

    cepheid

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    Probably not! So you just used the formula that nasu gave you? :rolleyes: The point of what both of us were saying to you was that you should look it up in your book or notes. He didn't give you the exact form because wasn't 100% sure of it, and he specifically said that you should find out exactly what the relation is. Also, the "C" he wrote was just meant to be some sort of constant, not the speed of sound. He said as much in his post. You need to read things more carefully.

    As for the second part, calculating the distance at which the sound becomes inaudible, you already *know* how intensity varies with distance. You wrote the equations for this in your first post.
     
  7. Dec 17, 2009 #6
    Here we go again:

    I looked it up and Found the equation for the sound intensity is: I= ½ ρ*v*ω²*A², where ρ is the density of the air, v is the velocity of the sound, ω is the angular frequency, and A is the amplitude of the air oscillations

    I= ½ ρ*v*ω²*A²
    I= ½ (1.2 kg/m³)*(340 m/s)*(2π*1000 Hz)²*(0.0000005 m)²
    I= 2.01 x 10^-3 W/m²

    Is this calculation for the intensity of the tiny speaker now correct?

    And I know that the relationship between the intensity and the distance is: I → 1/d² (inverse square law relationship), but I still can’t figure out how to calculate this distance.
     
  8. Dec 17, 2009 #7
    So, if I understood the relationship correctly, this is how to calculate the distance:

    I= x/r² → r= sqrt(x/I)
    r= 22.3 m

    And for the second part: What is the sound power of the loudspeaker at a distance of 1m?

    I= P/Area
    P=I* Area → P= 4πr²*I
    P= 0.025 W

    Are my calculations correct?
    Thank you for helping!:smile:
     
  9. Dec 17, 2009 #8
    Depends what you used for x.
    Here x should be the power of the source. It is NOT the intensity that you calculated.

    You don't really need this power.
    You know the intensity at 1m and the intensity at some unknown distance r.
    The ratio between intensities is same as the inverse ratio of the distances squared.
    You coul also calculate the power of the source and go for there. Maybe this is what you have done?
     
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