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

 

Q)  A wave component with a wavelength of 100m propagates through a homogeneous medium from a seismic source at the bottom of a borehole. Between two detectors, located in Boreholes at radial distances of 1 km and 2 km from the source, the wave amplitude is found to be attenuated by 10 dB. Calculate the contribution of geometrical spreading to this value of attenuation and thus determine the absorption coefficient of the medium.

(Thanks to Chandrasekhar, ANU)

  Solution:  

Given that wave length (ʎ) =100m

Amplitude attenuated by 10 dB

Radial distances $r_{1}$=1km

                        $r_{2}$=2 km

We know that $dB = 20 \log_{10}({\frac{A_{1}}{A_{2}})}$----------(1)

$dB = 20 \log_{10}({\frac{A_{1}}{A_{2}})}$

  From,

  $Energy \propto \frac{1}{radius(r)^{2}}$

   $we get  Amplitude(A) \propto \frac{1}{radius(r)}$  

    There fore

    $Amplitude(A_1) = \frac{1}{radius(r_1)}=\frac{1}{1}$

     $Amplitude(A_2) = \frac{1}{radius(r_2)}=\frac{1}{2}$

Substitute these values in equation (1) we get

$dB = 20 \log_{10}({\frac{A_{1}}{A_{2}})}$

$dB = 20 \log_{10}({\frac{1/1}{1/2})}$

$dB = 20 \log_{10}(2)$

$dB = 20 \times 0.3010$

$dB = 6$

 To get the contribution of the geometrical spreading we have to subtract this effect from the actual value

 i.e. 10dB-6dB= 4 dB is the contribution of the geometrical spreading.

Then the 1000 m ----4 dB

For 10 m ----?(let some X)

Therefore

$1000\times X =4\times 100$

$X=\frac{400}{1000}$

$X=0.4 dB/\lambda$

 

 

 

Reference:

An Introduction to Geophysical Exploration by Philip Kearey, Michel Brooks and Ian Hill.