96. A Thick Sedimentary formation, in Which the seismic wave Velocity V Increases with depth Z from $V_0$ at the surface to $V_1$ at the bottom according to the relation $V_1=V_0e^{\lambda z}$ . the Two way reflection travel time at a point closest to the short point is …..
(Special Thanks to Pawan Singh)
(Thanks to Ashok, AKNU)
Solution:
Given that $V=V_0e^{\lambda z}$
At Z = 0 $V
= V_0$ ( surface )
At Z
= Z $V =
V_1$ ( bottom )
We know that two Way Trave Time is
$dt = \frac{2dz}{v}$
By integrating this equation we get
$t = 2\int_{0}^{z} \frac{1}{V_0e^{\lambda z}}dz$
$=\frac{2}{V_0}\int_{0}^{z}{e^{-\lambda z}}dz$
$= \frac{2}{V_0}[\frac{e^{-\lambda z}}{-\lambda}]^z_0$
$= \frac{2}{V_0\lambda}[{-e^{-\lambda
z}}]^z_0$
since
${e^{-\lambda z }}= \frac{V_0}{V_1}$
$= \frac{2}{V_0\lambda}[-\frac{V_0}{V}]^z_0$
$= \frac{2}{\lambda}[-\frac{1}{V(z)}+\frac{1}{V_0}]$
$= \frac{2}{\lambda}[-\frac{1}{V_1}+\frac{1}{V_0}]$
$t = \frac{2}{\lambda}[\frac{1}{V_0}-\frac{1}{V_1}]$
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