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

Gate_2019

80) In a seismogram of a shallow focus (h = 5 km) earthquake, the difference between the arrival times of the S and P phases is 1.34 s. Assuming the average P wave velocity of the crust to be 6.0 km/s and the Poisson’s ratio to be 0.27, the epicentral distance is _______kilometer’s. (Round off to 1 decimal place)

(Thanks to Chandra Sekhar, ANU)

Solution:

Given that focal depth(h)= 5 km 

 

$t_{s} - t_{p} $ = 1.34 sec

$V_{p}$ = 6 km/sec

Poissions ratio $\nu$ = 0.27

Epicentral distance ($\triangle$) = ?

Formula’s

1. ) $\frac{V_{p}}{V_{s}}=\sqrt{\frac{2(1-\nu )}{(1-2\nu)}}$------------(1)
   
   
2. ) $t_{s} - t_{p}= D(\frac{1}{V_{s}}-\frac{1}{V_{p}})$-----------------(2)
   
   

From equation (1) 

 

$\frac{V_{p}}{V_{s}}=\sqrt{\frac{2(1-\nu )}{(1-2\nu)}}$

$\frac{6}{V_{s}}=\sqrt{\frac{2(1-0.27 )}{(1-2\times0.27)}}$

$\frac{6}{V_{s}}=\sqrt{\frac{1.46}{0.46}}$

$\frac{6}{V_{s}}=\sqrt{0.317}$

$\frac{6}{V_{s}}=1.781$

$V_{s}=\frac{6}{1.781}$

$V_{s}= 3.37 km/sec$

From equation (2)

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

$1.34= D(\frac{1}{3.37}-\frac{1}{6})$

1.34= D(0.296-0.166)

1.34= D(0.1294)

$D=\frac{1.34}{0.1294}$

D= 10.35 km

The epicentral distance can be calculated from the above figure

 From Pythagorean equation,

$ D^{2}=\triangle ^{2} + h^{2}$

$\triangle ^{2}=D^{2} -h^{2}$

$\triangle ^{2}=(10.35)^{2} -(5)^{2}$

$\triangle ^{2}=107.1225-25$

$\triangle ^{2}=82.1225$

$\triangle =9.06 km$

The epicentral distance is 9.06 km 

Reference: Fundamentals of Geophysics William Lowrie