Solve using these equations only K$=\frac{E}{3(1-2v)}$

$\frac{v_p}{v_s}=\sqrt[\phantom{2}]{\frac{2-2v}{1-2v}}$

$v_s=\sqrt[\phantom{2}]{\frac{G}{}}$

$v_p=\sqrt[\phantom{2}]{\frac{E}{}}=\sqrt[\phantom{2}]{\frac{\frac{4}{3}+K}{}}$

$\frac{sin1}{V{p}_1}=\frac{sin3}{V{s}_1}=$

$\frac{sin2}{V{s}_2}=\frac{sin4}{V{p}_2}$

$M_L=lo{g}_{10}{A}_o+3lo{g}_{10}[8t]-2\mathrm{.}92$

$d=\frac{t_{p-s}}{\frac{1}{v_s}-\frac{1}{v_p}}$

For the following soil profile, if the P$-$wave entered the middle layer at $20$ from the vertical then refracted within the layer by $35.$ If the thickness of the middle layer $=25\mathrm{.}5$ meters

$1-$ Sketch the time$-$distance diagram, illustrating all the necessary information on it$.$

$2-$ Find $V_1$ and $V_2$

$3-$ What is the cross$-$over distance

$=2039\mathrm{.}736k\frac{g}{m^3}$

$4-$ What is the time intercept $T_i$

$5-G$ and $E$ for the middle layer

$(\frac{E}{G}{)}^{0\mathrm{.}5}=1\mathrm{.}54,$ bulck Modulus $=35000$kPa

$=25k\frac{N}{m^3},T_{i2}=0\mathrm{.}5sec$

$E=360000$kPa,$=20k\frac{N}{m^3}$