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GATE 2017

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GATE-2017

22) The travel time difference between arrival times of a shear wave(s) and primary wave(p) observed on a seismogram recorded at an epicenter distance of 100 km from a near surface earthquake is _______ seconds(Assume the average P and S wave velocities to be 6 km/sec and 3.5 km/sec respectively)


(Thanks to Chandrasekhar, ANU)

Solution:

Given that the epicentral distance(D) = 100 km

Velocity of P- wave(V_{p})= 6 km/sec

Velocity of S- Wave(V_{s})= 3.5 km/sec

t_{s}-t_{p}=?

The travel time difference between arrival time of shear wave and Primary wave is

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

Substitute these values in the equation (1) we get


t_{s}-t_{p}=100(\frac{1}{3.5}-\frac{1}{6})


t_{s}-t_{p}=100(0.285-0.166)


t_{s}-t_{p}=100(0.1184)


t_{s}-t_{p}=11.84




Extra information:

If D is the distance travelled from the focus to the receiver then


t_{p}= \frac{D}{V_{p}}


t_{s}= \frac{D}{V_{s}} then


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


The intercept to the horizontal axis (in wadati diagram ) gives the time of occurrence of the earthquake (t_{0})


i.e D=V_{p}(t_{p}-t_{0})


if \triangledown_{km} is the epicentral distance and D is the distance travelled by the wave , d is the focal depth then


d = \sqrt{D^{2}-\triangledown_{km}^{2}}




Reference : Fundamentals of Geophysics by William Lowrie.


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