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

Table of contents
- Solution 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.

 Ancestors (12)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version