HOMEWORK NUMBER 3
“Real problems from the underworld”
Due Date: Tuesday, October 4th at 9.30 a.m.
1. A human rights organization wants to know whether seismology can be used to detect old
buried land mines. A small land mine can be 20-30 cm in diameter and 10 cm thick.
The top of these mines can be buried 10 cm.
Calculate what dominant frequency you will need to use in order to resolve its (1) thickness
and (2) width.
--For the limit of vertical resolution use a lambda/2 as the minimum thickness resolvable.
--For the horizontal resolution use a Fresnel zone width no more than ½ the body’s width.
Velocity = lambda * frequency (units of m/s)
Velocity/ lambda = frequency (units of 1/s)
If lambda/2 = 0.1 m, then lambda = 0.2 m
Assuming Velocity of ~1000 m/s:
frequency = 1000 /0.2
= 5000 Hz is the dominant frequency I will need in order to
resolve the thickness
If the lateral resolution is 0.25/2 m (Fresnel zone 1) and the depth of burial is 0.1 m and the
assumed velocity is 1000 m/s then using equation 3.36 (Liner, 2004; p. 74) we start with:
Fresnel Zone 1 = square root( 2 * dominant lambda * depth)
FZ1 * FZ1*
= 2 * dom_lambda * depth
FZ1* FZ1/2/depth = dom_lambda
But, we know that for any lambda, frequency = Velocity/lambda so…
Dominant frequency = Velocity/ dom_lambda
Dominant frequency = 1000/( FZ1* FZ1/2/depth)
= 1000/( 0.25/2 )* (0.25/2) / 2 / depth)
= 1000/(0.125 * 0.125 / 2 / 0.1)
= 1000/(1/8 *1/8 * 1/2 * 10)
= 1000 * 128 / 10
= 12800 Hz
2. A military research laboratory wants to know whether they can use seismology to locate
caves. For 10-100 Hz, plot out the minimum dimensions and maximum depth of the cave for
which you will be able to resolve its size. If it is smaller than those dimensions will you still be
able to see it???
Let's use 5000 m/s for the cave wall velocity.
For 10-100 Hz, I have lambdas of 500 - 50 m, respectively. I can therefore resolve caves,
250 to 25 m in height. (There are other complications you can also argue).
For the first Fresnel zone, I have:
Resolvable widths = 1/2 of square root( 2 * lambda * depth to top of cave)
(other arguments can be made for the resolvable width, but here I am just following from
the suggestion given in question 1)
For the above values assumed, then
Resolvable widths = 1/2 of square root[ 2 * (50 to 500 m) * depth of cave]
= 1/2 of square root[(100 to 1000m) * depth of cave]
3. At mid-oceanic ridges, the depth to the top of the magma chamber can be about 1 km.
Calculate what the relative strength of the reflection (amplitude) from the top of a magma
chamber will be at the just below the surface of the ocean directly above the magma
chamber. Assume a normal water velocity and density. Assume that the magma displays an
acoustic velocity of 4 km/s, the basalt and gabbro above the magma chamber 5 km/s.
Assume that the amplitude is calculated relative to 1. (Note that the values given here have
yet to be verified with the refereed literature)
V in water = 1500 m/s rho water = 1000 kg/m3
------------------------------------------------V basalt = 5000 m/s rho_basalt= 2800 kg/m3
---------------------------------------------------V magma = 4000 m/s rho magma=2700 kg/m3
Final Amplitude = 1.* TC (water-to-basalt) * RC(basalt-to-magma) * TC(basalt-to-water)
Units for I (Acoustic Impedance) = kg /m^3 * m / s^2
I(basalt) = 5000 * 2800 = 1.4 * 10 ^7 ;
I(water) = 1500 * 1000 = 1.5 * 10^6
I(magma)= 4000 * 2700 = 1.08 * 10^7
[I(basalt) + I (water)] = [I (water) + I(basalt) ] = 1.55 * 10^7;
I(basalt) + I (magma)] = 2.48 * 10^7
TC(water-to-basalt)
= 2 * I (water) / [I(basalt) + I (water)]
= 2 * 1.5 * 10^6/[1.55 * 10^7]
= 3.0 * 10^6/1.55 * 10 ^7
~ 0.1935
RC(basalt-to-magma) = [I (magma)] - I (basalt) ]/ [I(basalt) + I (magma)]
=[1.08 - 1.4] * 10^7 / [2.48 * 10^7]
= -0.32 * 10^7 / 2.48 * 10^7
~ -0.129
TC(basalt-to-water)
= 2 * I(basalt) / [I (water) + I(basalt) ]
= 2 * 1.4 * 10 ^7/[1.55 * 10^7]
= 2.8 * 10 ^7 /1.55 * 10 ^7
~ 1.858
By substiution of values we obtain that the
Final Amplitude = 1.* TC (water-to-basalt) * RC(basalt-to-magma) * TC(basalt-to-water)
~ 1 * 0.1935 * -0.129 * 1.858
~ -0.0464 or, nearly 5% of the original amplitude with opposite polarity
because it bounces from the top of the magma chamber, the top of a low-velocity layer.
Please e-mail me suggested improvements to this exercise to juan@geol.lsu.edu
October 6, 2005 Juan Lorenzo