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

CSIR-NET 2014

108. A seismic survey over a horizontal reflector brought out a two-way travel time of 1.0 sec at the shot point, while the refraction survey brought out the intercept time of 0.8 sec. A detector placed at 1.8km distance from the shot point received the refracted and reflected waves simultaneously. The depth to the reflector is...

A) 0.8km          B) 1.2km           C) 1.6km               D) 2.0

Solution:

Horizontal reflector is given in above problem

So, using NMO (normal move out)

Formula:

Δt_n = X² / (2*V²*ti) -----------------(1)

Δt_n = reflector travel time – intercept time

      = 1.0-0.8

Δt_n = 0.2 s

Offset distance (X)= 1800m

t_i = intercept time = 0.8 s

Substitute these values in eq (1)

above formula velocity (V) = X / (2*t_i* Δt_n)^1/2 ----------------------(2)

                                             = 1800/ ( 2*0.8*0.2)^1/2

                                            = 3185 m/s

Therefore the depth to the reflector h = (V*t_i) / 2

                                                              = (3185 *0.8) / 2

                                                              = 1274 m

                                                             = 1.274 km