Past exam of the year 2 of the earth sciences (geology) course of the University of Oxford / 2022 / Paper 1 / 6 / d

Table of contents
- i d
  - Solution i
- ii d
  - Solution ii

i

Solution

The normal-incidence acoustic impedance is

Z = ρ V_{P}

. Therefore

Z_{1} = 1500 (1500) = 2.25 \times 1 0^{6} kg m^{- 2} s^{- 1},

Z_{2} = 2000 (3500) = 7.00 \times 1 0^{6} kg m^{- 2} s^{- 1} .

These are

2.25

and

7.00

MRayl. They combine the material’s density and wave speed; using

V_{S}

would give shear rather than acoustic P impedance.

ii

Solution

Specify the amplitude convention. For normal-stress/pressure amplitudes, the seismic reflection coefficient and seismic transmission coefficient are

R_{σ} = \frac{Z _{2} - Z _{1}}{Z _{2} + Z _{1}} = \frac{19}{37} = 0.51351, T_{σ} = \frac{2 Z _{2}}{Z _{1} + Z _{2}} = \frac{56}{37} = 1.51351.

For particle displacement or velocity resolved in one fixed normal coordinate, the reflected amplitude changes sign and the transmitted ratio is instead

R_{v} = - \frac{19}{37}, T_{v} = \frac{2 Z _{1}}{Z _{1} + Z _{2}} = \frac{18}{37} = 0.48649.

These are alternative conventions, not conflicting energy predictions. The reflected and transmitted energy flux fractions are

R = \frac{361}{1369} = 0.26370, T = \frac{1008}{1369} = 0.73630,

which sum to one for the ideal lossless interface. Thus an amplitude transmission coefficient greater than one does not violate conservation of energy.

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