Earthquake distance to seismograph

  • Thread starter Hypnos_16
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  • #1
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Homework Statement



When an earthquake occurs, two types of sound waves are generated and travel through the earth. The primary, or P, wave has a speed of about 7.87 km/s and the secondary, or S, wave has a speed of about 4.86 km/s. A seismograph, located some distance away, records the arrival of the P wave and then, 76.8 s later, records the arrival of the S wave. Assuming that the waves travel in a straight line, how far is the seismograph from the earthquake?

Pwave = 7.87km/s OR 7870m/s
Swave = 4.86km/s OR 4860m/s


Homework Equations



I don't think there's an equation specifically for this, if there is, i haven't got it yet


The Attempt at a Solution


I've figured out that when the pwave reaches the station it takes 76.8s for the swave to get there. So the distance at that point is 373.248km or 373248m. But i can't get the distance that it took before then.

Pwave--------------------------------------------| Station
Swave------------|_______________________| Station
x 373,248m

I tried to make it easier to understand, i can't find the distance "x", i know it traveled some distance before getting to the station i just don't know how to find that.
Someone help.
 
Last edited:

Answers and Replies

  • #2
gneill
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Suppose an earthquake occurs at some distance d from the station. How long before the P wave arrives? Write an expression for it. Call that tp. Do the same for the S wave, and call that time ts. Now do something clever with the time difference.
 
  • #3
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Suppose an earthquake occurs at some distance d from the station. How long before the P wave arrives? Write an expression for it. Call that tp. Do the same for the S wave, and call that time ts. Now do something clever with the time difference.
soooo, something like that? but what can i do with that?

tp = d / 7.87m/s
ts = d / 4.86m/s
 
  • #4
gneill
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It's probably easier to work symbolically so that there aren't a lot of decimal places cluttering up the thought processes. You've written:

tp = d/vp
ts = d/vs

These are the arrival times for the P and S waves from an even that occurs at an assumed t0 = 0 seconds at distance d.

Now, the information that the seismograph station has to work with is the difference in arrival times. Can you write that equation?
 
  • #5
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tp = d/vp
ts + 76.8 = d/vs

something like that?
 
  • #6
gneill
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Don't plug in any values yet. Work symbolically. Write an expression for the time difference between the arrivals, Δt.
 
  • #7
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Don't plug in any values yet. Work symbolically. Write an expression for the time difference between the arrivals, Δt.
okay,

Tp = d/vp
Ts + ΔT = d/vs

so something like that then?
 
  • #8
gneill
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Why didn't you use the expression you obtained for ts? You're losing ground!

Δt = ???
 
  • #9
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Why didn't you use the expression you obtained for ts? You're losing ground!

Δt = ???
Δt = 76.8 seconds, the time between when the p wave arrives at the station and then when the s wave arrives.
 
  • #10
gneill
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You're plugging in numbers again. Write the expression for Δt in terms of the expressions for ts and tp.
 
  • #11
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Δt =Ts - Tp

Something more like that?
 
  • #12
gneill
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Δt =Ts - Tp

Something more like that?
Yes, now substitute your expressions for ts and tp.
 
  • #13
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Yes, now substitute your expressions for ts and tp.

Δt =Ts - Tp
Δt =d/vs - d/vp
Δt =d(vp - vs) / vs - vp

and come up with something like that i imagine.
 
  • #14
gneill
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Something like that. But your last line is not correct.

Δt = d/vs - d/vp
Δt = d*(1/vs - 1/vp)
Δt = d*(vs*vp)/(vp - vs)

So now you can solve for d, right?
 
  • #15
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Something like that. But your last line is not correct.

Δt = d/vs - d/vp
Δt = d*(1/vs - 1/vp)
Δt = d*(vs*vp)/(vp - vs)

So now you can solve for d, right?
wouldn't you just go

Δt = d/vs - d/vp
Δt = d*(1/vs - 1/vp)
Δt = d*(vs*vp)/(vp - vs)
∆t * (vp - vs) / (vs * vp) = d
???

but that only gives like 6km, so, i'm thinking that isn't how you do it...
 
  • #16
gneill
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wouldn't you just go

Δt = d/vs - d/vp
Δt = d*(1/vs - 1/vp)
Δt = d*(vs*vp)/(vp - vs)
∆t * (vp - vs) / (vs * vp) = d
???

but that only gives like 6km, so, i'm thinking that isn't how you do it...
Sorry, my mistake... sometimes I type faster than I think...:redface:

Δt = d/vs - d/vp
Δt = d*(1/vs - 1/vp)
Δt = d*(vp - vs)/(vs*vp)
d = Δt*(vs*vp)/(vp - vs)
 
  • #17
153
1
Happens to me all the time. No worries

So i got the whole thing to be

Tp = d/vp
Ts = d/vs
∆T = Ts - Tp
∆T = (d/vs) - (d/vp)
∆T = d*(1 / vs - 1 / vp)
∆T = d*(vp - vs) / (vs*vp)
d = ∆T*(vs*vp) / (vp - vs)
d = (76.8)(4.86*7.87) / (7.87 - 4.86)
d = 975.9km

which seems like a more likely answer than 6.
 

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