Built on Facts

Earthquake!

Every section of Physics 218 I’ve taught this semester has asked me about this question. Really it’s less of a physics question than it is a math question, but either way it gives people fits. It’s not all that surprising. While it seems like it should be simple, to most beginning students it’s not at all intuitive how all the given information gits together. Without further ado, the problem:

Earthquakes produce several types of shock waves. The most well-known are the P-waves and the S-waves. In the earth’s crust, the P-waves travel around 8.9 km/s while the S-waves move at about 2.7 km/s. The time delay between the arrival of these two waves at a seismic recording station tells geologists how far away the earthquake occurred. If the time delay is 14 s, how far away from the seismic station did the earthquake occur?

Numbers are for suckers. Call the faster speed v1 and the slower one v2. Call the time delay T. Call the unknown distance d. Call the time between the earthquake and the first wave t1, and the time between the earthquake and the second wave t2.

We know v1, v2, and T. We don’t know d, t1, or t2.

Why did we do all that defining? If you don’t know where to start, the first thing to do is to think systematically about what you know and what you don’t. That might give you an idea as to how you might proceed. Distance is velocity times time, so this is true:

i-d4bb9fe09d6b78eb832985821d8a307f-1.png

Which is nice, but not enough. There are three unknown quantities and two equations.

As a rule, if you have a certain number of unknown quantities you need that same number of equations to uniquely specify all of them – or to guarantee that there is no consistent answer. To be fair the mathematicians can tell you a whole stack of caveats to that statement, but they essentially boil down to requirements that your equations also not be mere restatements of each other.

So we need a third equation. We happen to have one, hidden in the fact that we know the time delay:

i-88516ca505d84f969b7ab15a965c3fb3-2.png

It doesn’t matter which order we solve these three equations in, so the following is one of several paths leading to the answer. Take this T equation, solve it for t2, and plug it into the second of our d equations. You’ll get:

i-d18063683dcc0d42b9be45451a84d1e3-3.png

Take that, solve it for t1, and plug it into our first d equation. You’ll get:

i-80f9713f6d11461837a9f9b540684e36-6.png

We’re almost done because that equation only involves d and the things we’re given. So solve it for d.

i-bb5c2c6b0452df43a61e3974bd9b473f-5.png

Now we’re done. The longer the time delay, the farther the quake is. For very large v1, the distance is very close to v2 t2 as we expect. Nice to see that the limiting cases check out.

Is it very difficult, once you know how to do it? Of course not. But it is great practice in recognizing equations you may not immediately know you have.

Comments

  1. #1 rob
    February 11, 2009

    solving physics problems is often the first time for students to be forced to *really* solve problems with a consistent technique. there are a lot of physics principles and corresponding derived equations included in texts. there is no guaranty that the “correct” equation will be in the text. since this equation is not one i have ever seen derived explicitly in a text, the students will quickly hit a brick wall by just plugging the values into random equations.

    they need to use a consistent problem solving method like you outline in your solution.

    the faster they learn that technique, the faster they can switch from doing math to actually learning physics.

  2. #2 Uncle Al
    February 11, 2009

    Do the two rays take the same path? The compressional wave doesn’t much care where it goes but the transvers wave will be averse to water.

  3. #3 Peter
    February 11, 2009

    Or realize the time taken to travel a unit distance is the reciprocal of velocity so the difference (1/2.7-1/8.9) is the time delay per kilometer distance. So the total distance to the quake is the observed time delay divided by this ie 14/(1/2.7-1/8.9).

    Using your symbols, time delay per kilometer is (1/v2-1/v1) so T = d*(1/v2-1/v1) and d = T/(1/v2-1/v1)

    I find it sometimes helps to go back to basic principles (in this case for velocity) rather than start by reaching for equations that may or may not be applicable.

    In reality the P and S waves do not take the same path over very long distances since the S wave does not pass through the molten part of the core and there is refraction of the P wave as well. The two types of waves observed at a specific location may also be coming from different areas of the fault since the source may be kilometers long and the events are of signifiant duration.

  4. #4 Nemo
    February 11, 2009

    Your formula bothered me at first because it lacks obvious symmetry… If you exchange v1 and v2 and negate T, you should get the same answer. So no reason for v2 to be all by itself there in the numerator.

    T*v1*v2 / (v1-v2)

    Ah, much better. :-)

    (Peter’s formula is good too.)

  5. #5 Carl Brannen
    February 11, 2009

    What I find interesting about P and S waves is that they follow massless Klein-Gordon wave equations, with the two waves, P and S, using different wave speeds.

  6. #6 chat
    February 12, 2009

    very good sites

  7. #7 Maranda Murphy
    January 19, 2010

    Silly question, but I am not seeing how you solved for d in your second to last step. How did you get 1-v2/v1 in the numerator?

    Thanks!
    Maranda