Answer:
The value is [tex]\alpha = 0.002 Np/m[/tex]
Explanation:
From the question we are told that
The first amplitude of the wave is [tex]E_{max}1 = 98.02 \ V/m[/tex]
The first depth is [tex]D_1 = 10 \ m[/tex]
The second amplitude is [tex]E_{max}2 = 81.87 \ (V/m)[/tex]
The second depth is [tex]D_2 = 100 \ m[/tex]
Generally from the spatial wave equation we have
[tex]v(x) = Ae^{-\alpha d}cos(\beta x + \phi_o)[/tex]
=> [tex]\frac{v(x)}{v(x)} =\frac{ Ae^{-\alpha d}cos(\beta x + \phi_o)}{ Ae^{-\alpha d}cos(\beta x + \phi_o)}[/tex]
So considering the ratio of the equation for the two depth
[tex]\frac{A}{A_S} = \frac{e^{-D_1 \alpha }}{e^{-D_2 \alpha }}[/tex]
=> [tex]\frac{98.02}{81.87} = \frac{e^{-10 \alpha }}{e^{-100 \alpha }}[/tex]
=> [tex]\alpha = \frac{0.18}{90}[/tex]
=> [tex]\alpha = 0.002 Np/m[/tex]
