Calculation of true thickness of the bed

CSIR-NET

96) The travel time differences between the P and S arrivals at two stations A and B are 3 sec and 4 sec Respectively, while the difference in the P- wave arrival time between the same stations is 1sec. What is the ratio of the P-wave velocity to that of S-wave velocity?

(Thanks to Chandra sekhar, ANU)

Solution:

Given that,

Station A travel time difference given that $(t_{s}-t_{p})_{A}= 3 seconds$

Station B travel time difference given that $(t_{s}-t_{p})_{B}= 4 seconds$ 

Difference in the P-wave arrival time between the station B to station A is given that,
$(t_{p})_{B}-(t_{p})_{A}= 1 seconds$

$\frac{V_{p}}{V_{s}}=?$

we know that,

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

(i.e $t_{p}=\frac{D}{V_{p}}$)

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

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

From the above equation,we can derive equations for station B and A respectively.

$(t_{s}-t_{p})_B= t_{p_B}(\frac{V_{p}}{V_{s}}-1)$----------------(2)

$(t_{s}-t_{p})_A= t_{p_A}(\frac{V_{p}}{V_{s}}-1)$----------------(3)

Subtract equation (3) from eq (2),

we get $(t_{s}-t_{p})_{B}-(t_{s}-t_{p})_{A}={[(t_{p})_{B}(\frac{V_{p}}{V_{s}}-1)]-[(t_{p})_{A}(\frac{V_{p}}{V_{s}}-1)]}$

$4-3=(\frac{V_{p}}{V_{s}}-1)[(t_{p})_{B}-(t_{p})_{A}]$

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

$\frac{V_{p}}{V_{s}}=1+1$

$\frac{V_{p}}{V_{s}}=2$