Question

A compressional acoustic wave takes $55~\mu s$ to travel $0.3048~\text{m}$ through a rock formation having bulk modulus of $37.5~\text{GPa}$ and shear modulus of $31~\text{GPa}$. The bulk density of the rock is ............. kg/m$^{3}$ (rounded to two decimal places).

Hint:

For $V_p$ in solids, use $V_p=\sqrt{(K+\tfrac{4}{3}G)/\rho}$. With travel time over $1$ ft ($0.3048$ m), $V_p\,[\text{m/s}]\approx \dfrac{0.3048}{t\,(\text{s})}$.

Answer 1

Step 1: Compute $V_p$ from travel time.
$V_p=\dfrac{\text{distance}}{\text{time}}=\dfrac{0.3048}{55\times10^{-6}}=5541.818~\text{m/s}\ (\text{approx}).$
Step 2: Use isotropic elastic relation for $V_p$.
$V_p=\sqrt{\dfrac{K+\tfrac{4}{3}G}{\rho}} \Rightarrow \rho=\dfrac{K+\tfrac{4}{3}G}{V_p^{2}}.$
Step 3: Substitute $K=37.5~\text{GPa$, $G=31~\text{GPa}$.}
$K+\tfrac{4}{3}G=37.5\times10^{9}+\tfrac{4}{3}(31\times10^{9})=78.833\!\times\!10^{9}\ \text{Pa}.$
$\Rightarrow\ \rho=\dfrac{78.833\times10^{9}}{(5541.818)^{2}}=2566.88~\text{kg/m}^{3}$ (to two decimals).