# Wavelength - how far away did earthquake occur

1. Dec 4, 2007

### mathcrzy

1. The problem statement, all variables and given/known data

Assuming typical speeds of 8.9 km/s and 5.8 km/s for P and S waves, respectively, how far away did the earthquake occur if a particular seismic station detects the arrival of these two types of waves 1.0 min apart?
_____km

2. Relevant equations

v=(lamba)(f)
1/T=frequncy
lamba=one wave length
T=period

X=same

Vpwave=deltaX/T

Vswave=dletaX/T+60

3. The attempt at a solution

8.9=

5.8=

2. Dec 4, 2007

### dynamicsolo

This is basically what you need. Solve the first equation for an expression for T and substitute it into the second equation, so you can eliminate the unknown T. You now have an equation you can solve for deltaX.

[As a check, consider this. Both types of waves started from the same "source". About how much do the P-waves gain on the S-waves each second? How much longer do the S-waves need to make up the difference? How long will it take for the S-waves to be a full minute behind? How far will the P-waves have traveled in that time? The algebraic solution described above is equivalent to answering these questions.]

Last edited: Dec 4, 2007
3. Dec 5, 2007

### mathcrzy

Vp=deltaX/T
VpT=deltaX
T=deltaX/Vp

Vs=deltaX/T+60
Vs(T+60)=deltaX
VsT+Vs60=deltaX
T=deltaX-Vs60/Vp

deltaX-Vs60/Vp=deltaX/Vp
deltaX-Vs60=deltaX
deltaX=Vs60

Is this how you solve for deltaX

4. Dec 5, 2007

### dynamicsolo

You were OK to here. The next line would then be

T = (deltaX - Vs·60)/Vs ,

so from there,

(deltaX - Vs·60)/Vs=deltaX/Vp

(deltaX/Vs) - (deltaX/Vp) = 60

(deltaX) · [ (1/Vs) - (1/Vp) ] = 60

delta X = [ (Vs·Vp) / (Vp - Vs) ] · 60

5. Dec 5, 2007

### mathcrzy

i got the answer but just wondering how you got from:

(deltaX - Vs·60)/Vs=deltaX/Vp

to

(deltaX/Vs) - (deltaX/Vp) = 60

6. Dec 5, 2007

### dynamicsolo

(deltaX - Vs·60)/Vs = deltaX/Vp

(deltaX)/Vs - (Vs·60)/Vs = deltaX/Vp

(deltaX)/Vs - (60) = deltaX/Vp

(deltaX/Vs) - (deltaX/Vp) = 60

In fact, you could really start right from here because this just says that the difference between the time it takes the S-waves to reach the station and the time it takes for the P-waves to do so is 60 seconds.