Q) A 10Hz seismic wave travelling at 5 km/sec propagates for 1000m through a medium with an absorption coefficient of 0.2 dB/ ʎ. What is the wave attenuation in decibels due solely to absorption?
(Thanks to Chandrasekhar, ANU)
Solution: Given that velocity(C) of the seismic wave = 5 km/sec
= 5000 m/sec
Frequency of the seismic wave (f) = 10 Hz
The relation between frequency, velocity and wavelength is given by,
C=f\lambda
\lambda=\frac{C}{f}
\lambda=\frac{5000}{10}
\lambda=500 m
Then for the wavelength 500m the absorption coefficient is 0.2 dB/ ʎ Then what is the wave attenuation for the wave length 1000m
500 ------ 0.2 dB/ ʎ
Then 1000 -------? ( let be some X)
500 \times X=0.2\times 1000
X=\frac{200}{500}
X= 0.4 dB/\lambda
Therefore, the wave attenuation in decibels due solely to absorption is0.4 dB/\lambda
Reference:
An Introduction to Geophysical Exploration by Philip Kearey, Michel Brooks and Ian Hill.
Post a Comment
Post a Comment