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  • Physics 2 help? I know theres some smart ones on here

    This is a stupid question, but it's like the only one I cannot answer, and it has to do with sound waves. This is probably just a simple one to do, but for the life of me, and after reading 2 DIFFERENT physics text books, neither have helped me any.

    Anyways, here is the question asked:

    A bat hunting for food sends out a 50-khz "sound" pulse. What is roughly the smallest flying insect that can be detected by the bat? (Hint- resolution idea). Assume an air temperature of 25-degrees Celsius.

    I know the velocity of sound through air is: v=(331 + .6Tc) m/s, where Tc is temperature Celsius. So, the velocity of sound in a 25* C temperature is 346 m/s.

    So then the next thing I looked at was an Intensity problem, where I=P/A I= intesity, P= power, A= area. This kinda made sense as a useful equation since I would need to find the area taken up by the size of the insect. But.... How would you convert 50khz into an intensity? Or, am I on the wrong track here?

    Thanks.
    1992 Miata B Package
    1995 Regal 3100
    1996 Firebird LT1 T56

  • #2
    If I'm reading the question right, what they are looking for is the "size" of the sound wave being emmited, since the smallest insect detectable would have to be the same size or larger then the amplitude/durration of the soundwave otherwise the wave would be able to pass "around" the object. Sorry I don't remember how to calculate those distances anymore, haven't had physics in about five or six years. Hope that helps.

    92 GP GTP 3.4DOHC 5spd
    88 Fiero Formula CONVERTIBLE 3.4DOHC 5spd
    84 Fiero T-Top 3.4DOHC 5spd
    There seems to be a pattern there.

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    • #3
      Essentially, yeah, that is what they are looking for. A sound frequency will only travel so far. That all depends on what type of material it is passing through(air in our case), the temperature(as stated above again), and this is dry air as well, so we do not have to worry about the dampness of the air.

      Hopefully just sitting on it for tonight, I can wake up in the AM and figure this one out...
      1992 Miata B Package
      1995 Regal 3100
      1996 Firebird LT1 T56

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      • #4
        I imagine the size of the smallest detectable insect will be some multiple of the wavelength. Check yer text, they usually have answers for this sorta shizzle.
        60v6's original Jon M.

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        • #5
          The train leaving Baltimore will arrive first. A pound of feathers weighs the same as a pound of lead. And the survivors are not buried anywhere, because they are still alive.

          Marty
          '99 Z-28 - Weekend Driver
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          Originally posted by Aaron
          This is why I don't build crappy headers. I'm not sure, I don't know too much about welding.

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          • #6
            Originally posted by RacerX11
            The train leaving Baltimore will arrive first. A pound of feathers weighs the same as a pound of lead. And the survivors are not buried anywhere, because they are still alive.

            Marty
            You left one out...

            chicken or the egg?
            60v6's original Jon M.

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            • #7
              LOL to both you guys.

              Anyways, got to my class, and talked to a couple of guys in there that are also in my Statics class. We were able to figure it out. Here's how incase anyone is interested:

              f = v/2L F= frequency(which we already knew, 50khz), v= speed of sound(346 m/s), and L= length of wavelength(what we needed to solve for, as what we really needed to solve for was how big the object could be to create a wavelength, and it could only be as small as the wavelenth itself).

              So... 50khz= 346m/s /(2L)

              50khz*2= 346m/s / L

              100khz/346m/s = L

              L= 2.39
              1992 Miata B Package
              1995 Regal 3100
              1996 Firebird LT1 T56

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              • #8
                That doesn't look right... Well, you just have some of your units missed / not converted. So it ends up being ~.239mm if I did it right. Interesting.
                -Brad-
                89 Mustang : Future 60V6 Power
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                • #9
                  About one half of a 40Khz Wave length na it gotta be .239

                  346m/S at 24 degrees C. At 40KHz the wavelength is 8.65mm

                  Under 1/16th a wave length is required for bats to sense a echo, actually it closer to 1/32nd of a wave length. Bat can sense hair, even fine hair.

                  Frequency is that, intensity in other words is Amplitude.
                  I am back

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