Recent content by Nettes
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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 : )- Nettes
- Post #26
- Forum: Introductory Physics Homework Help
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Attenuation finding distace and amplitude
A^-1 / d^1- Nettes
- Post #24
- Forum: Introductory Physics Homework Help
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Attenuation finding distace and amplitude
x^-1 / d^-1?- Nettes
- Post #22
- Forum: Introductory Physics Homework Help
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Attenuation finding distace and amplitude
Im sorry i don't understand this, my best guess would be x^-1 and d^-1 : /- Nettes
- Post #20
- Forum: Introductory Physics Homework Help
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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- Nettes
- Post #18
- Forum: Introductory Physics Homework Help
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Attenuation finding distace and amplitude
d= 10 000 r= 10 000 / 2 = 5000 5000^-1 = 0,0004?- Nettes
- Post #16
- Forum: Introductory Physics Homework Help
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Attenuation finding distace and amplitude
but i don't understand what is the attenuation :/ when u say attenuation u mean the r^-1?- Nettes
- Post #14
- Forum: Introductory Physics Homework Help
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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?- Nettes
- Post #12
- Forum: Introductory Physics Homework Help
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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.- Nettes
- Post #10
- Forum: Introductory Physics Homework Help
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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...- Nettes
- Post #8
- Forum: Introductory Physics Homework Help
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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!- Nettes
- Post #7
- Forum: Introductory Physics Homework Help
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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- Nettes
- Post #5
- Forum: Introductory Physics Homework Help
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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)- Nettes
- Post #3
- Forum: Introductory Physics Homework Help
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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)...- Nettes
- Thread
- Amplitude Attenuation
- Replies: 25
- Forum: Introductory Physics Homework Help