Calculation of true thickness of the bed

17. A seismograph receivers a S-wave 60s after it receives the P-wave. If the velocities of P- and S-waves are 7km/s and 6km/s respectively, then the distance of the seismic focus from the seismograph is

A) 2520 km

B) 42 km

C) 7070 km

D) 72km

Solution:

Formula:

$t_s-t_p = D( \frac{1}{V_s} -\frac{1}{V_p})$ ----(1)


The difference between S and P –wave is $(t_s- t_p)$ = 60s


Velocity of the P-wave $(V_p)$ = 7 km/s


Velocity of the S-wave $(V_s)$ = 6 km/s


D is the distance from focus to seismograph


Substitute above values in eq(1)


$60 = D*(\frac{1}{6}–\frac{1}{7})$


$60= D*\frac{(7-6)}{42}$


$60= \frac{D}{42}$


D= 60*42


Therefore, (D) = 2520 km