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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 

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