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/
Reference :http://mjgeophysics.blogspot.com/
Post a Comment
Post a Comment