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S and P waves.

  • Thread starter tristan_fc
  • Start date
  • #1
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It seems to me that this is just a simple algebra problem, that doesn't have much to do with waves, but it's the only problem on the homework that I can't figure out. :-/ I know there's an easy solution to it, but I keep getting the wrong answer. Anyway, here it goes:

A seismographic station receives S and P waves from an earthquake, 18.2 s apart. Suppose that the waves have traveled over the same path, at speeds of 4.50 km/s and 7.00 km/s respectively. Find the distance from the seismometer to the epicenter of the quake.

Anyone have any help? I know the solution must be blindingly obvious. [?]
 

Answers and Replies

  • #2
Doc Al
Mentor
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Both S and P waves travel the same distance. Use the venerable formula, Distance = Speed x Time, to calculate D based on the difference in T.
 
  • #3
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Thanks I knew it had something to do with that formula, and I figured it out.

change in time = d/v1 - d/v2. :)
 
  • #4
chroot
Staff Emeritus
Science Advisor
Gold Member
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As you know,

[tex]
\begin{equation*}
\begin{split}
distance &= velocity \times time\\
s &= v t
\end{split}
\end{equation*}
[/tex]

The distances are the same in each case, so you have

[tex]
s = v_s t_s = v_p t_p
[/tex]

where s,p denote the two kinds of waves.

The s wave takes 18.2 seconds to reach the detector than the p wave. This means

[tex]
t_s = t_p + 18.2
[/tex]

Substitute this into the previous equation:

[tex]
v_s (t_p + 18.2) = v_p t_p
[/tex]

Solve for [tex]t_p[/tex]. You then know the time taken by the p-wave, and the speed of the p-wave, so the distance is easily found.

- Warren
 
  • #5
chroot
Staff Emeritus
Science Advisor
Gold Member
10,226
34
Originally posted by tristan_fc
Thanks I knew it had something to do with that formula, and I figured it out.

change in time = d/v1 - d/v2. :)
Yup. :smile:

- Warren
 

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