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