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Attenuation finding distace and amplitude

  1. Feb 13, 2014 #1
    1. The problem statement, all variables and given/known data
    problem 1
    A 20 Hz seismic wave traveling at 3 km/s propagates for 2000 m from point X to point
    Y through a medium with an attenuation rate of 0.5 dB/λ. What is the ratio between
    the amplitude of the wave measured at X and the amplitude of the wave measured at Y (i.e., AX/AY)?

    and
    problem 2
    A seismic wave travels from the source to two separate recording points in space following a single, straight ray path. The farhest recording point is located at 10km from the source and the signal is attenuated by 8dB between the two recording points. Assuming that all the attenuation is due to geometrical spreading, give the distance from the source to the nearest recoring point.


    2. Relevant equations
    LdB = 20log10(A1/A2), where LdB is a value of either attenuation or dynamic range that is given in decibels (dB).




    3. The attempt at a solution
    problem 1
    i was thinking i somehow need to get rid of the λ behind 0,5dB.
    λ=3000/20 = 150m and
    2000/150 =13,33 λ from Ax to Ay....

    problem 2
    no idea.. tryed 8log = 20log10 (A0/A1)
    (A1/A0)= 10^(8/20) = 2,512.....

    i realy tryed to understand this for 3 days now and im going mad, if someone could show me how to solve this i will be verry grateful!
     
  2. jcsd
  3. Feb 13, 2014 #2
    Why did you stop in problem 1?
     
  4. Feb 13, 2014 #3
    because i dont know how to remove the λ from 0,5dB/λ if i now have 0,5dB/13,33. (my skills are very low)
     
  5. Feb 13, 2014 #4
    You have found that λ = 150 m. 0.5 dB/λ means that the wave attenuates at 0.5 dB when it passes the distance of one λ. Then how much does it attenuate when it passes 2000 m?
     
  6. Feb 13, 2014 #5
    so one λ is 150m and i got 2000m, 2000/150=13,333 λ, meaning the there is 13,33 λ in those 2000m.
    then i do 0,5*13,333=6,665 ?
    and then i can put this into the equation ?
    doing (A2/A1)= 10^(6,665/20)=2,15
     
  7. Feb 13, 2014 #6
    That would be correct, but: I think you also have to include the attenuation due to the geometrical spreading like in problem 2. Or no?
     
  8. Feb 13, 2014 #7
    if you ask me i dont know : ) my subteacher said ''i should not think about that part so much''. Ty very much for the help, was not so hard when u explained it slowly! if you got any thoughts on problem 2 and have time let me know!
     
  9. Feb 13, 2014 #8
    in problem 2 i need to find distance so i need to change the equation so i get
    (A1) =(A2)/ (10^(8/20)) but im not certain if the (A2) is in the right position now.
    Then i would get that the nearest point (A1) = 10000 / 10^(8/20)= 3981,07m from the source.
    This is what i wanna do when i see this problem now(dont know why :p), but i also dont understand how this equation can give me the location of the nearest point.
     
  10. Feb 13, 2014 #9
    Well, on part 2, you should explain what you know about "geometric spreading" and how that results in attenuation.
     
  11. Feb 13, 2014 #10
    hmm i found in my book that wave amplitude falls off as r^-1 due to geometrical spreading. not sure if this is what u mean. we did not learn so much theory exept that attenuation is due to 1) spreading of energy at interfaces. 2) geometric spreading and 3) absorption due to imperfect elasticity.
     
  12. Feb 13, 2014 #11
    Very well. So you have three points: A, B, C. A is the source of the wave. B and C are recording stations. You know the distance between A and C. What is the attenuation of the signal between A and C due to geometrical spreading? Then you have the unknown distance between A and B, say x. You can still write an equation for attenuation at B. Then you know the difference between B and C. Write all that down, and solve the equation.
     
  13. Feb 13, 2014 #12
    dont think i follow
    if r is half the distance = 5000m
    wave amplitude falles off as 5000^-1 = 0,0002?
    and the 8dB gives A2/A1 = 2,51
    2,51/0,002=1255?
     
  14. Feb 13, 2014 #13
    Don't use numbers just yet. Let the AB distance be x, and the AC distance be d.

    Write down attenuation at B in terms of x.

    Write down attenuation at C in terms of d.

    Finally, you have a relationship between the signal levels at B and C.
     
  15. Feb 13, 2014 #14
    but i dont understand what is the attenuation :/
    when u say attenuation u mean the r^-1?
     
  16. Feb 13, 2014 #15
    If a wave starts with amplitude A, and travels distance d, what is its amplitude there, taking the geometric spreading into account? What is the attenuation in this case?
     
  17. Feb 13, 2014 #16
    d= 10 000
    r= 10 000 / 2 = 5000
    5000^-1 = 0,0004?
     
  18. Feb 13, 2014 #17
    Where do all those numbers come from? And why divide by 2?
     
  19. Feb 13, 2014 #18
    you said to let the distance AC be d, and is say ''The farhest recording point is located at 10km from the source'', so i thought AC was 10000m from each other. and if the r^-1 meant radius i took half of those 10km xD
     
  20. Feb 13, 2014 #19
    I also said "do not use numbers just yet". And no, r is not "radius" here, it simply means "distance".
     
  21. Feb 13, 2014 #20
    Im sorry i dont understand this, my best guess would be x^-1 and d^-1 : /
     
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