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

CSIR NET- JUNE 2019

97) A seismic source at a depth of 20m generates a seismic wave which is reflected and recorded at the surface at an offset distance of 30m, Assuming that the horizontal reflector lies at a depth of 30m and the P-wave velocity of the medium is 2.5 km/s, the travel time for the reflected wave is ( in milli seconds).

A) 15

B) 20

C) 25

D) 30

(Thanks to Chandrasekhar, ANU)

Solution:

 We need to calculate the travel time from SO to OG.

Where given Offset distance (x) = 30m

Depth of the reflector(z) = 30m

P-wave velocity(V_p)= 2.5 km/s

From Pythagoras theorem, 

From triangle SOB  $SO^2 =SB^2 +BO^2---(1)$

SB is the half of the offset distance =15 m

OB = 10 m

$SO^2 =15^2 +10^2$

$SO^2 =225 +100$

$SO^2 =335$

$SO= 18.3$

Travel time from SO is $t_{SO} =\frac{Distance}{Velocity}$

$t_{SO} =\frac{18.3}{2500}$

$t_{SO} =0.0073 sec----(2)$


From triangle AOG  $OG^2 =AG^2 +OA^2$

$OG^2 =15^2 +30^2$

$OG^2 =225 +900$

$OG^2 =1125$

$OG^2 =15^2 +30^2$

$OG^2 =33.54---(3)$

Travel time from OG is $t_{OG} =\frac{Distance}{Velocity}$

 $t_{OG} =\frac{33.54}{2500}$

 $t_{OG} =0.0134--(4)$

Total travel time is $t_{SO} +t_{OG}$ =0.0073+0.0134 s

$t_{SG}= 0.0207 s$

$t_{SG}= 20.7 ms$