Question

If \( \alpha \) and \( \beta \) are P- and S-wave velocities, respectively, then \( \alpha^2 - \frac{4}{3} \beta^2 \) is equal to: \[ (\kappa \text{ is the bulk modulus, } \mu \text{ is shear modulus, and } \rho \text{ is density}) \]

• A. \( \kappa / \rho \)
• B. \( \mu / \rho \)
• C. \( \kappa + \mu / \rho \)
• D. \( \kappa - \mu / \rho \)

Hint:

The relationship between the P-wave and S-wave velocities, and the moduli of elasticity, allows us to derive the expression for \( \alpha^2 - \frac{4}{3} \beta^2 \) as \( \kappa / \rho \).

Answer 1

The Correct Option is A

The relationship between the P-wave velocity \( \alpha \), the S-wave velocity \( \beta \), the bulk modulus \( \kappa \), the shear modulus \( \mu \), and the density \( \rho \) is given by: \[ \alpha^2 = \frac{\kappa + \frac{4}{3} \mu}{\rho} \text{and} \beta^2 = \frac{\mu}{\rho}. \] Thus, we can derive: \[ \alpha^2 - \frac{4}{3} \beta^2 = \frac{\kappa}{\rho}. \] Therefore, the correct answer is (A). Final Answer: (A) \( \kappa / \rho \)