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CSIR-NET

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 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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