GATE-2018 (24) :Wyllie – Time average equation

Gate_2019

78) A split-spread reflection survey is carried out along a profile in the direction of the dipping interface. The difference in arrival times of the reflected waves from the interface at two geophones with an offset distance of 1000 m from the shot-point on both sides is 20 msec. If the velocity of the layer above the dipping interface is 3000 m/s, then the dip of the bed is ________degrees. (Round off to 1 decimal place). (Assumption 2d >> X, where‘d’ is depth below the shot-point normal to the interface and X is the source-geophone spacing)

(Thanks to Chandra Sekhar, ANU)

Solution:

Given that $\triangle t_{d}$ = 20 msec=0.02 seconds

offset distance X= 1000 m

V = 3000m/sec

Dip angle $\theta$= ?

We know that 

 

$\triangle t_{d} = \frac{2x\sin\theta}{V}$-----------(1)

By substituting the given values in equation (1) we get 

 

$\triangle t_{d} = \frac{2x\sin\theta}{V}$

$0.02 = \frac{2\times1000\times\sin\theta}{3000}$

$\sin\theta =\frac{(0.02)(3000)}{2000}$

$\sin\theta = 0.03$

$\theta=\arcsin(0.03)$

Dip of the bed is $(\theta) = 1.719$

Reference :http://appliedgeophysics.berkeley.edu/seismic/seismic_23.pdf