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}]
Post a Comment
Post a Comment