 Solution

ID: past-exam-of-the-year-2-of-the-earth-sciences-geology-course-of-the-university-of-oxford/2022/paper-1/6/d/i/solution

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.

 New to topics? Read the docs here!