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

94. The primary (P) and secondary (S) waves from a shallow focus earthquake reached a seismological observatory at 8:30:04 Hrs and 8:31:16 Hrs respectively. If the velocities of the P and S waves are in the ratio 1.6:1, then the time of occurrence of the earthquake would be

A) 8:27:16 Hrs

B) 8:27:32 Hrs

C) 8:28:04 Hrs

D) 8:28:52 Hrs

Solution:

Travel time of P-wave  from earthquake to seismometer  $ t_p =8:30:04hr$

                              =30,604sec

  

Travel time of S-wave from earthquake to seismometer $ t_s = 8:31:16 hrs$

                            = 30,676sec

 

${v_p}:{v_s}= {1.6}:{1}$

$v_p$ =1.6 velocity of the P-wave

$v_s $= 1 velocity of the S-wave

Distance can be calculated by using below this formula

$t_{s}-t_{p}=D(\frac{1}{v_s}-\frac{1}{v_p})$

$D=\frac{t_s-t_p}{(\frac{1}{v_s}-\frac{1}{v_p})}$

$D=\frac{72}{(\frac{1}{1}-\frac{1}{1.6})}$

$D=\frac{72}{(\frac{1.6-1}{1.6})}$

$D=\frac{72*1.6}{0.6}$

D=192 km

The relation between the distance, velocity of the P-wave and origin time of earthquake$ (t_0)$

$D=V_p (t_p-t_0)$

$t_0= (t_p-\frac{D}{V_p})$

$t_0= (30,604-\frac{192}{1.6})$

$t_0= (30,604-120)$

$t_0= 30,484 sec$

Convert 30,484 sec we will get 8:28:04 Hrs

Reference :http://mjgeophysics.blogspot.com/