Calculation of true thickness of the bed

30)  A 10 Hz seismic wave Propagates for 40 km through a material with a P-wave velocity of 5 km /sec and Quality factor (Q) of 100. The percentage of the initial amplitude retained in the attenuated wave________ (Round off to 1 decimal place)

 (Thanks to Chandrasekhar, ANU)

Solution:

 Given that Frequency (f) = 10 Hz

 

Distance from the source (r) =40 km = 40,000 meters

 

Velocity of the P-wave= 5 km/sec= 5000 m/sec

 

Quality factor (Q) = 100

 

The Percentage of initial amplitude retained in the attenuated wave  $\frac{A}{A_{0}}$=?

 

We know that the Damped amplitude of a seismic wave at a distance of r from its source is

 

$A = A_{0}e^{-\frac{\pi}{Q}\frac{r}{\lambda}}$

 

$\frac{A}{A_{0}} = e^{-\frac{\pi}{Q}\frac{r}{\lambda}}$-------------------------------(1)

 

 Here $\lambda$ can be calculated from the equation $\lambda=\frac{c}{f}$

 

$\lambda=\frac{5000}{10}$

 

$\lambda=500m$

 

 Substitute all the given values in the equation (1)  

 

$\frac{A}{A_{0}} = e^{-\frac{\pi}{Q}\frac{r}{\lambda}}$

 

$\frac{A}{A_{0}} = e^{(-\frac{3.14}{100})(\frac{40000}{500})}$

 

$\frac{A}{A_{0}} = e^{-\frac{3.14\times4}{5}}$

 

$\frac{A}{A_{0}} = e^{-2.512}$

 

$\frac{A}{A_{0}} = 0.0811$

 

$\frac{A}{A_{0}}$=8.11%