GATE-2018 (24) :Wyllie – Time average equation

Question : A Plane seismic wave traveling vertically downwards, the upper layer seismic velocity and densities are given by 2500 m/s and 2500 kg/m³ respectively, the lower layer seismic velocity and densities are 3500 m/sec and 2800 kg/m3 respectively. Then what are the amplitude ratios of transmitted and reflected waves?

 (Thanks to Chandrasekhar, ANU)

Solution:

The amplitude ratio for the reflected waves are (R)= $\frac{A_{1}}{A_{0}}$

The amplitude ratio for the transmitted waves are (T)= $\frac{A_{2}}{A_{0}}$

 Where,

$A_{0}$ is the amplitude of the incident wave

 $A_{1}$ is the amplitude of the reflected wave

 $A_{2}$ is the amplitude of the transmitted  wave

 Given that,

 $V_{1}$= 2500 m/sec                           $V_{2}$= 3500 m/sec

 $\rho_{1}$=2500 kg/m³                         $\rho_{2}$= 2800 kg/m³

 

Then Reflection coefficient (R) =$\frac{\rho_{2}V_{2}-\rho_{1}V_{1}}{\rho_{2}V_{2}+\rho_{1}V_{1}}$

By substituting the given values in the above formulae, we get

$=\frac{\rho_{2}V_{2}-\rho_{1}V_{1}}{\rho_{2}V_{2}+\rho_{1}V_{1}}$

$=\frac{(2800)(3500)-(2500)(2500)}{(2800)(3500)+(2500)(2500)}$

$=0.2212$

 We know that R +T=1

Then T=1-R

              = 1-0.2212

              =0.78.

Then from the amplitude ratio of reflected wave $A_{1}=R \times A_{0}=0.2212 \times A_{0}$

Then from the amplitude ratio of transmitted  wave $A_{2}=T\times A_{0}=0.78 \times A_{0}$

 

Then the amplitude ratio of the transmitted wave and the reflected wave is as follows

$\frac{A_{2}}{A_{1}}=\frac{0.78 \times A_{0}}{0.2212 \times A_{0}}$

               =$\frac{0.78}{0.2212}$

               =3.52