Finding epicenter distance from P and S wave arrival times

[tristan_fc]

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. [?]

 

[Doc Al]

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.

 

[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. :)

 

[chroot]

As you know,

$$\begin{equation*}<br /> \begin{split}<br /> distance &= velocity \times time\\<br /> s &= v t<br /> \end{split}<br /> \end{equation*}$$

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

$$s = v_s t_s = v_p t_p$$

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

$$t_s = t_p + 18.2$$

Substitute this into the previous equation:

$$v_s (t_p + 18.2) = v_p t_p$$

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

- Warren

 

[chroot]

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.

- Warren