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

GATE-2007

Problem:-A P-wave generated from a surface source is incident at an angle of 15° on the horizontal interface between two 100m thick layers with velocities v1=2km/s and v2=4km/s for the first and second layers respectively.

71)The crossover distance (m) for a head wave from the interface between the two layers is

Solution:-

In Distance-Travel time(x-t) plot ,direct and doubly refracted rays cross each other at this distance, which is accordingly called the “crossover distance,X_cr”

Cross- Over distance (X_cr)= 2Z (√((V2 + V1) / (V2-V1)) -----------(1)

Thickness Z= 100m

1st layer velocity V₁= 2 km/s = 2000m/s

2nd layer velocity V₂= 4 km/s = 4000m/s

X_cr= 2* 100 (√((4000 + 2000) / (4000- 2000))

= 200 √ (6000 / 2000 )

= 200 √3

= 200 * 1.73

X_cr= 346.4 m

72) A reflection from the base of the second layer is recorded at an offset (Source-receiver) distance(m) of

Solution:

Source- Receiver distance is also called “offset distance (X)”

In given figure AO + OO' is the “half offset distance”

So, AO'+O'E is the offset distance

Then,

tan15˚= AO / OB

AO = 100 * tan15˚

AO= 26.79 m

From Snell's law

sini / sinr = V1/ V2

sin15˚/ sinr = 2/ 4

sinr = 2 * sin 15˚

       = 2 * 0.2588

       = 0.517

refraction angle (r)= sinˉ1( 0.517)

                                = 31. 17˚

tan31.17˚= FC/BF ( BF= 100m)

FC = 100 * tan 31.17˚

FC = 100 * 0.6049

FC = 60.49 m

AO' = AO+OO' = 26.79 + 60.49

                          = 87.2806 m

Source- Receiver distance (X) = AO'+ O'E

                                                   = 87.2806 + 87.2806

                                                   = 174.56 m

73. The total travel time (ms) taken for the P-wave generated at the surface to reach the detector after reflection from the base of the second layer is?

Solution:

ΔABO cos15˚ = OB/ AB

AB= OB / cos15˚

AB= 100 / 0.965

AB= 103.6

ΔBFC cos31.17˚ = BF/ BC

BC= BF / cos31.17˚

     = 100/ 0.855 

     = 116.95 m

Travel time for the ray to move AB in first layer is = AB/ V₁

             = Distance / velocity

             = (103.5 / 2000)

             = 0.05185 s

Travel time for the ray to move BC in second layer is = BC / V₂

           =(116.95 / 4000)

             = 0.02923 s

Total travel time =( ((2 * (AB/ V₁)) + (2 * (BC/ V₂)) )

                            = 2* 0.0518 + 2* 0.02923

                            = 0.1036 + 0.058475

                            =0.162 s

Total travel time = 162 ms