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
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.
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...
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!
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
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)...