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Two More Sound Wave Problems

  1. Dec 13, 2004 #1
    >> A jackhammer, operated continuously at a construction site, behaves as a point source of spherical sound waves. A construction supervisor stands 52.0 m due north of this sound source and begins to walk due west. How far does she have to walk in order for the amplitude of the wave function to drop by a factor of 1.90?

    I'm not really sure how to set this problem up. The only thing I have so far is that 1/1.90 Amplitude1 = Amplitude2...

    _________________________________________________________________
    >> A loudspeaker is placed between two observers who are 125 m apart, along the line connecting them. If one observer (observer A) records a sound level of 70 dB and the other (observer B) records a sound level of 100 dB, how far is the speaker from each observer?

    Again, I'm not sure how to start this. I'm fairly certain I have to use intensity levels, though.
     
  2. jcsd
  3. Dec 13, 2004 #2

    Pyrrhus

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    For the second problem i would get both intensities then use

    [tex] I_{1}r_{1}^2 = I_{2}r_{2}^2 [/tex]

    [tex] r_{1} + r_{2} = 125 [/tex]
     
    Last edited: Dec 13, 2004
  4. Dec 13, 2004 #3

    Andrew Mason

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    This is just an application of the 1/s^2 rule for sound intensity and a little trigonometry.You also have to know the relationship between amplitude and intensity (hint: it is not linear). How far from the source does the supervisor have to be to have the intensity drop to that level (ie. the intensity where amplitude is 1/1.9 of the original)?

    Am
     
  5. Dec 13, 2004 #4

    Okay, I think I have this set up right now but the algebra is tripping me up. I found the intensity of observer A to be 1.0x10^-5 and the intensity of observer B to be .01. Then I used the formulas I(a)r(a)^2=I(b)r(b)^2 and r(a)+r(b)=125.

    So then plugging things in, I get r(a)+((I(b)r(b)^2)/(I(a)))^1/2 = 125. This should be the easy step, but I don't know how to solve for r(a). When I try it I get an answer of around 2700m... which can't be right since the two observers are only 125 meters apart to begin with!
     
  6. Dec 13, 2004 #5

    Pyrrhus

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    What do you mean?

    i got good answers, check your algebra.

    [tex] r_{1}= \sqrt{\frac{I_{2}}{I_{1}}} r_{2} [/tex]

    Substituing

    [tex] r_{2} + \sqrt{\frac{I_{2}}{I_{1}}} r_{2} = 125[/tex]

    [tex] r_{2} = \frac{125}{1+\sqrt{\frac{I_{2}}{I_{1}}}}[/tex]

    Solutions

    [tex] r_{1} = 121.16 m [/tex]

    [tex] r_{2} = 3.84 m [/tex]
     
  7. Dec 14, 2004 #6
    Okay I've gotten all the problems for this homework except this one about the jackhammer. I'm still not sure how to set this up... I know intensity is equal to power over amplitude?
     
  8. Dec 14, 2004 #7

    Andrew Mason

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    No. Intensity = Power/Area (eg. units of watts/m^2). Intensity varies as the square of the amplitude.

    AM
     
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