# Elastic modulus, earthquake waves?

• Feb 1st 2009, 11:29 AM
realintegerz
Elastic modulus, earthquake waves?
A longitudinal earthquake wave strikes a boundary between two types of rock at a 35 degree angle. As it crosses the boundary, the specific gravity (SG = d_rock / d_water ) of the rock changes from 3.7 to 2.8. Assuming that the elastic modulus is the same for both types of rock, determine the angle of refraction.

To solve this I have two equations.

1) (velocity 1/ velocity 2) = (sine angle 1 / sine angle 2), a ratio

2) velocity of an earthquake wave = sq. root ( elastic modulus / density )

The problem I'm having is, I don't know what the elastic modulus is and what the density or (rho) is, and also density is supposed to = m/v

Otherwise, I could solve this easily...

Can anyone help? (Worried)
• Feb 1st 2009, 11:57 AM
skeeter
Quote:

Originally Posted by realintegerz
A longitudinal earthquake wave strikes a boundary between two types of rock at a 35 degree angle. As it crosses the boundary, the specific gravity (SG = d_rock / d_water ) of the rock changes from 3.7 to 2.8. Assuming that the elastic modulus is the same for both types of rock, determine the angle of refraction.

To solve this I have two equations.

1) (velocity 1/ velocity 2) = (sine angle 1 / sine angle 2), a ratio

2) velocity of an earthquake wave = sq. root ( elastic modulus / density )

The problem I'm having is, I don't know what the elastic modulus is and what the density or (rho) is, and also density is supposed to = m/v

Otherwise, I could solve this easily...

Can anyone help? (Worried)

the problem says that the elastic modulus is the same for both types of rock, so it will cancel in the equation.

$\displaystyle \frac{v_1}{v_2} = \frac{\sqrt{\frac{e}{\rho_1}}}{\sqrt{\frac{e}{\rho _2}}}$

$\displaystyle \frac{v_1}{v_2} = \sqrt{\frac{\rho_2}{\rho_1}}$

you also know the respective SG's ...

$\displaystyle \frac{\rho_1}{\rho_w} = 3.7$

$\displaystyle \frac{\rho_2}{\rho_w} = 2.8$

$\displaystyle \frac{\frac{\rho_1}{\rho_w}}{\frac{\rho_2}{\rho_w} } = \frac{3.7}{2.8}$

note that the value for water density also cancels ...

$\displaystyle \frac{\rho_1}{\rho_2} = \frac{3.7}{2.8}$

you should be able to finish up