# Attenuation finding distace and amplitude

• Nettes
In summary: I want you to solve the equation for attenuation at B in terms of x.Then you can use that information to solve the equation for attenuation at C in terms of d.
Nettes

## 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)?

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.

## Homework Equations

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

## 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 really tryed to understand this for 3 days now and I am going mad, if someone could show me how to solve this i will be verry grateful!

Why did you stop in problem 1?

1 person
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)

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?

1 person
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

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?

1 person
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!

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 see this problem now(dont know why :p), but i also don't understand how this equation can give me the location of the nearest point.

Well, on part 2, you should explain what you know about "geometric spreading" and how that results in attenuation.

1 person
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.

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.

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?

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.

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

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?

1 person
d= 10 000
r= 10 000 / 2 = 5000
5000^-1 = 0,0004?

Where do all those numbers come from? And why divide by 2?

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

I also said "do not use numbers just yet". And no, r is not "radius" here, it simply means "distance".

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

So if amplitude is A1 at the source, and A2 at distance d, what is A1/A2?

x^-1 / d^-1?

There was no x in the question.

A^-1 / d^1

That question was what is A1/A2?

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 : )

## What is attenuation?

Attenuation is the reduction in strength or intensity of a signal as it travels through a medium.

## How is attenuation measured?

Attenuation is typically measured in decibels (dB), which represents the ratio of the initial signal strength to the final signal strength.

## What factors affect attenuation?

The main factors that affect attenuation are distance and the properties of the medium through which the signal is traveling, such as its density and composition.

## What is the relationship between attenuation and distance?

The further a signal travels, the more it will be attenuated. This is because the signal has to pass through more of the medium, which results in a greater reduction in signal strength.

## How does amplitude relate to attenuation?

Amplitude is a measure of the strength of a signal, and it is directly affected by attenuation. As a signal travels through a medium and is attenuated, its amplitude decreases.

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