This homework set was worth a total of 100 points. Most people did quite well on the first problem. Most people had problems on the second problem, but they were mostly the problem of not reading the assignment carefully about what was expected. Particularly the part about showing work. Read the answer for Question 2 for more details...
Figure 1 is my graph from the numbers in Table 1 of the homework. The red circles are the actual table values; the green line merely connects them to make the patterns easier to see. The dash-dot blue line is at the median number of earthquakes M 7+ per year (20), and the dashed blue lines are at the maximum (41 in 1943) and minimum (6 in 1986) number of earthquakes of that size in each year.
As you can see, while there are large fluctuations in the number of earthquakes of magnitude 7 or greater in from year to year, there seems to be no compelling evidence that the number of events is going up over time on the scale of this plot; nor is there any evidence that the number is dropping. On the other hand, on shorter time scales, many fluctuations are evident; for example, the mid 1940s had many earthquakes of this size, whereas the mid 1980s had relatively few.
I accepted any answer which could be logically drawn from the graph. Some people said there was no trend, others had their computers play with the numbers and found a verrrrry slight decrease over time (but it's insignificant), while still others decided that earthquakes were increasing since the numbers are going up from the dip in the mid 1980s. All of these were fine with me.
The best answer I got noted that this time period (roughly 100 years) is probably not long enough to give an accurate picture of what is going on. That's true. Unfortunately, the only way to get more records is probably to wait, although some people have had some success in deciphering old records of ancient quakes.
For now, the best answer is that there is no compelling evidence that the number of large earthquakes worldwide is going down or going up, except for very short-term fluctuations on 5-10 year time scales.
I was a bit surprised to find that many people had problems with this question, not because they didn't understand how to do the problem, but because they didn't read the problem completely before doing it. A number of people complained to me when I returned the homework that I had taken off points because they had not shown their work. Yes, that's true. However, I took the points off only because I told them to show their work and they didn't.
On the other hand, I'm willing to admit when I'm wrong, and when enough people complained, I accepted homeworks for re-grading. I didn't forgive the not showing work thing, but I did decide I had been a bit harsh in grading the first time.
Keep in mind, though, that when you are asked to show work, it means show all your work unless specifically told not to show some part of it. It seems trivial and picky, but it's not. If you don't show me work, I can't give partial credit for wrong answers; also, I can't tell you where you went wrong and how to fix it. That's why the homework specifically said to show work.
OK, that aside, most people did reasonably well on the homework. The only real problem was in finding the arrivals of the waves, and most people did pretty well on that. When someone did poorly on that part, he or she was usually really wrong; I didn't have too many people give me just slightly wrong picks. And I tried to give partial credit to people even when their picks were off, etc. (that show-your-work thing again)...
Here's my solution to the problem.
Let's pick an example seismogram. Figure 2 shows the seismogram for the PSP station with my picks for the arrival times noted as labeled dashed lines.
Now, I am in a position to figure out the times. I measure from the left side of the figure over to the dashed line for the P-wave and find out that the distance is 5.65 cm. All I have to do is multiply 5.65 by 2.222 and I will get seconds (since the cm cancels on the top and bottom). In this case, 5.65 × 2.222 = 12.55 seconds. So my P-wave time (relative to the time at the beginning of the bottom time scale) is 12.55 seconds.
Similarly for the S, I find that the distance is 11.42 cm. Doing a similar multiplication to that above, I find that the relative S-wave time is 25.38 seconds.
Most people failed to show the above work.
I note that the distance for 12 seconds is 94.2 km. For 13 seconds, it's 102.0 km. I can take the difference (102.0-94.2) and find it is 7.8 km. Now, I can use this and the S-P time to compute an exact distance equivalent. I take 12.83 seconds and subtract 12 seconds (the lower limit of my bracket above), and get 0.83 seconds. I can then multiply 0.83 by 7.8 and get the fraction of 7.8 km which corresponds to 0.83 seconds: it's 6.47 km. I can then add 6.47 km to 94.2 km (the lower limit again) and get the exact distance equivalent of 12.83 seconds. That number is 100.67 km.
Come see me if the above paragraphs didn't make sense; I'll be happy to go over it with you.
Since 3 cm was 50 km, all you had to do was 3 ÷ 50 to get the scale; the answer is 0.06 cm/km. Once you have that scale, you only need to multiply your distance measurements from step 3 above by the scale, and you know how big to make the radius of your circle, in cm, on the map. In our example, it's 100.67 km × 0.06 cm/km or 6.04 cm.
So I need to draw a circle on my map, centered at PSP, with a radius of 6.04 cm.
The other seismograms are dealt with in a similar fashion. Figures 3, 4, 5, and 6 show picked seismograms for EDW, GSA, GTM, and VST, respectively.
I asked you to make a table summarizing your results; here's my table:
|Station Code||P-Wave Time (sec)||P-Wave Arrival Time||S-Wave Time (sec)||S-Wave Arrival Time||S-P Time (sec)||Distance (km)||Map Circle Radius (cm)|
Figure 7 shows my map; the triangle shows the location that Caltech determined for this event.
In case you're interested, here's what Caltech has to say about this earthquake:
05 January 1998, 18:14:06.5 UTC (10:14:06.5 AM PST), magnitude 4.3, located 5 mi SSW of Chino, CA. This event was widely felt.