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									    General Properties of Waves
        Reflection Seismology
              Geol 4068
Questions and answers to first lecture

         September 8, 2005
      Homework 1-due September 8,
           2005 at 9.30 a.m.
Q. 1 What is the P-wave velocity of the following
earth materials measured near the surface of the

basalt, granite, peridotite, gabbro and iron

             Answer. +/- 10%
             basalt-- 5 km/s
             granite --- 6.5 km/s
             peridotite -- 8.1 km/s
             gabbro---7.2 km/s SOURCES must
             be referenced!!!
Q. 2 If following a surface explosion, a “ray” of sound
enters the blue synform, what will the angle of refraction
at point A for the following interface geometry? Apply
Snell’s Law with the values provided. Explain your work
clearly and succinctly.

Hint: simplify the geometry of the geology

 We can assume that the explosion is at the center of a circle and
 that the synform describes, approximately the shape of a circle.
 With this assumption, we start by noting that lines of radius
 intersect the circumference of a circle at right angles. If so then
 the incident angle of refraction is 90 degrees. At this normal
 incidence, reflected rays will return along the same path as the
 incident ray. Not all the energy of the ray is is returned entirely
 to the place where of the explosion took place. Some energy may
 be transmitted through the boundary.
Q. 2
Q. 3 If the lowest frequency your body size can register is
about 8 times your greatest dimension, does this value
change whether you are in water or in air? What are these
values? Will your body “feel” or register these waves? Or,
is your body too small? Assume your are 2 m tall !
Assume sound travels at (I) 1500 m/s in water and (II)
330 m/s in air. Assume you are 2 m high. Consider
that the dominant frequency in your signal is (a) 10 Hz,
and (b) 400 Hz ,
Let us start with the following equation:
               VP  frequency ( f )  wavelength ( )
               VP  f     (1)
If we rearrange (1) we obtain the wavelength as a function
of velocity and frequency we are better ready to answer the question:
                          VP
Note that based on the information provided in the question
 you would have to grasp that if you are smaller than 8 times
the dominant wavelength your body will not "register" or
"feel" the wave. Or, in other words you will be too small to
 be "seen" by the wave.
STEP 1: Substitute values for part a into (1)
1500  10  150
m 1
  m
s s
The wavelength is 150m. BUT, you are looking to see if
you will be able to feel the wave through your body. You
would then have to be no shorter than 1/8 of 150 m, i.e.
~18.75 m

We perform similar calculations for a velocity of 330 m/s,
330 m/s = 10 Hz x 33 m, and 1/8 of 33 is about 4 m which
is taller than the tallest human and therefore undetectable.

(b) Similarly, if the frequency is 400 Hz, the minimum
height you would have to be is inversely proportional to
the frequency. From the calculations you should obtain

1500 m/s = 400 Hz x 3.75 m, and 1/8 of 3.75 m divided by
8 is ~0.5 m, which is much smaller than anyone.

We perform similar calculations for a velocity of 330 m/s,
330 m/s = 400 Hz x 0.825 m, and 1/8 of 0.825 is about 10 cm.

It is best sometimes to graph the results and see what we
can infer. I have done this in MATLAB. You could also do
it by hand.
Matlab code
Q. 4 What is the critical angle between water and basalt?
This is a typical scenario in oceanic spreading ridges.
Assume a P-wave velocity of 1500 m/s

First, let’s study Snell’s Law for this case at critical conditions, i.e.,
when the angle of refraction 90 degrees and Snell’s Law reduces
         sin(incident angle) / V in first medium              =
         sin(90deg.) / V in second medium          (1)
     sin(critical angle)/V in first medium = 1/V in second medium    (2)
            sin(critical angle) = V in first medium/V in second medium

           critical angle = angle whose sine is:
                    V in first medium/V in second medium            (3)

             In other words, critical angle = arcsin(V1/V2)

         Note carefully, that your calculator may either return
arcsin in radians or degrees. Always consider you units,
Now, taking V1=1500 m/s and V2=5000 m/s we obtain that
arcsin(1500/5000) = 17.45 degrees
We can also plot the data out for the general case to
understand the behavior better:

                                                         Matlab code

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