Recent content by Nettes

  1. N

    Attenuation finding distace and amplitude

    i don't know :/ ty a lot for trying to make me understand and for your time! i think i need to take a break from this and try again later : )
  2. N

    Attenuation finding distace and amplitude

    Im sorry i don't understand this, my best guess would be x^-1 and d^-1 : /
  3. N

    Attenuation finding distace and amplitude

    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
  4. N

    Attenuation finding distace and amplitude

    d= 10 000 r= 10 000 / 2 = 5000 5000^-1 = 0,0004?
  5. N

    Attenuation finding distace and amplitude

    but i don't understand what is the attenuation :/ when u say attenuation u mean the r^-1?
  6. N

    Attenuation finding distace and amplitude

    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?
  7. N

    Attenuation finding distace and amplitude

    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.
  8. N

    Attenuation finding distace and amplitude

    in problem 2 i need to find distance so i need to change the equation so i get (A1) =(A2)/ (10^(8/20)) but I am 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 want to do when i...
  9. N

    Attenuation finding distace and amplitude

    if you ask me i don't 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!
  10. N

    Attenuation finding distace and amplitude

    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
  11. N

    Attenuation finding distace and amplitude

    because i don't know how to remove the λ from 0,5dB/λ if i now have 0,5dB/13,33. (my skills are very low)
  12. N

    Attenuation finding distace and amplitude

    Homework Statement 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)...
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