Calculating Distance from Seismograph Arrival Times

[Student3.41]

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 8.13 km/s and the secondary, or S, wave has a speed of about 5.32 km/s. A seismograph, located some distance away, records the arrival of the P wave and then, 78.8 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?

Homework Equations

V=(wavelength)/T or f(wavelength)

The Attempt at a Solution

I solved for each wavelength for P and S

P = 641km, S= 419km

I then used the forumla L = (wavelength)/2 ... for each value...

Not sure where to go from here.. pleasse help only two i had trouble with...

 

[dacruick]

it seems like you may be over complicating this question.

 

[verty]

Start like this. Let the distance from the earthquake center to the seismograph be x. Let the earthquake happen at time t = 0. At t = t_p, the P wave arrives at the sensor, and at t = t_s the S wave arrives. Get equations for t_p and t_s. You know t_s - t_p, so...

Do these steps and if you get stuck, just holler. I've put this list here because I think by the end of this problem, you'll have a good understanding of this type of question.

-- sorry dacruick, I didn't see your post

 

[Student3.41]

Start like this. Let the distance from the earthquake center to the seismograph be x. Let the earthquake happen at time t = 0. At t = t_p, the P wave arrives at the sensor, and at t = t_s the S wave arrives. Get equations for t_p and t_s. You know t_s - t_p, so...

Do these steps and if you get stuck, just holler. I've put this list here because I think by the end of this problem, you'll have a good understanding of this type of question.

-- sorry dacruick, I didn't see your post

still unsure how to go about this problem..

So, I know 78.8s = ts-tp...

not sure how to set up the equation I am completely lost... assignments due tongiht too !

 

[gneill]

something moving at v km/sec travels a distance x. How long did it take?

 

[Student3.41]

something moving at v km/sec travels a distance x. How long did it take?

I think I am drastically complicating this question.

 

[Millacol88]

You have two unknowns: Distance, and t_p (You know that t_s = t_p + 78.8).
What do you do? Think back to some basic algebra. Hint: it has nothing to do with wavelength.

 

[Student3.41]

Can't get my head around this...

 

[Millacol88]

Can't get my head around this...

How do you express distance in terms of velocity and time?

 

[Student3.41]

How do you express distance in terms of velocity and time?

V=d/t or d=Vt

 

[Millacol88]

V=d/t or d=Vt

Correct. Can you make two equations relating the distance traveled by both sound waves to each wave's velocity and the time it took them?

 

[Student3.41]

Correct. Can you make two equations relating the distance traveled by both sound waves to each wave's velocity and the time it took them?

x= Vs-Vp/ts-tp ... ts-tp=78.8

 

[Millacol88]

x= Vs-Vp/ts-tp ... ts-tp=78.8

Try making two equations of the form d = vt.

 

[Student3.41]

Try making two equations of the form d = vt.

x=Vs(ts-tp)

x=Vp(tp+78.8)

 

[Millacol88]

Your first equation is wrong. The distance that wave s travels will be given by multiplying the velocity of that wave times the time for the movement, not the difference between the two times. Your second equation would make sense if you had vs and tp in it. Remember that the time for the s-wave is 78.8 s longer than for the p-wave.

 

[Student3.41]

Your first equation is wrong. The distance that wave s travels will be given by multiplying the velocity of that wave times the time for the movement, not the difference between the two times. Your second equation would make sense if you had vs and tp in it. Remember that the time for the s-wave is 78.8 s longer than for the p-wave.

x=vs*ts

x=vp(ts-tp)

 

[Millacol88]

x=vs*ts

x=vp(ts-tp)

This won't work if you've got 3 unknowns. And your second equation still doesn't make sense.

dp = vp (tp)

ds = vs (tp -78.8)

or:

ds = vs (ts)

dp = (vs + 78.8)

Now you've got two equations that accurately describe the situation. You need to find the values for those two variables that work in those two equations. Know how?

 

[iRaid]

You will need to find the arc length since one of the waves only travels on the surface while the other travels through the Earth (I forget which is which :p)

 

[gneill]

Forget ts - tp for now. Trying to force it into play at this point is just going to confuse things. Concentrate on the two completely separate expressions for the time of travel of the two completely separate waves. They have independent velocities. They both travel distance x. Write the two basic, separate, equations for their times of travel:

ts = ?

tp = ?

 

[Millacol88]

Forget ts - tp for now. Trying to force it into play at this point is just going to confuse things. Concentrate on the two completely separate expressions for the time of travel of the two completely separate waves. They have independent velocities. They both travel distance x. Write the two basic, separate, equations for their times of travel:

ts = ?

tp = ?

I've essentially already shown him that: any of my equations can be rearranged to the form t (s or p) = ?

 

[gneill]

I've essentially already shown him that: any of my equations can be rearranged to the form t (s or p) = ?

The problem is that by repeatedly trying to force the "78.8" number into everything every time, the same errors are made time and time again without any progress being made.

Start with the foundations: Write the very basic time vs distance expressions for the two things that are moving (S wave and P wave). Once the two *separate!* expressions for the times of travel are in hand, it will be a trivial matter to take the difference between the two and move forward from there.

 

[FearTheHump]

I know this is an old topic but I bumped into it, and wanted to know if a physics expert could tell me if my solution to this problem is correct?

d = vt

d(p) = 8.13t
d(s) = 5.32(t+78.8)

d(s) = d(p)
8.13t = 5.32(t+78.8)
8.13t = 5.32t + 419.216
8.13t -5.32t = 419.216
2.18t = 419.216
t = 419.216/2.18
= 192.3s

d(p) = vt
d(p) = 8.13 x 192.3
= 1563.406km

Edit: Hmm I just subbed in my answer with the original values and the answers don't quite match up :/
What did I do wrong?