Question

A certain electromagnetic wave traveling in seawater was observed to have an amplitude of 98.02 (V/m) at a depth of 10 m, and an amplitude of 81.87 (V/m) at a depth of 100 m. What is the attenuation constant of seawater

Answer 1

Answer:

The  value is   $\alpha = 0.002 Np/m$

Explanation:

From the question we are told that

  The first amplitude of the wave is  $E_{max}1 = 98.02 \ V/m$

  The first  depth  is  $D_1 = 10 \ m$

   The second amplitude is  $E_{max}2 = 81.87 \ (V/m)$

   The second depth is $D_2 = 100 \ m$

Generally from the spatial wave equation we have

   $v(x) = Ae^{-\alpha d}cos(\beta x + \phi_o)$

=>       $\frac{v(x)}{v(x)} =\frac{ Ae^{-\alpha d}cos(\beta x + \phi_o)}{ Ae^{-\alpha d}cos(\beta x + \phi_o)}$

So considering the ratio of the equation for the  two depth

$\frac{A}{A_S} = \frac{e^{-D_1 \alpha }}{e^{-D_2 \alpha }}$

=>   $\frac{98.02}{81.87} = \frac{e^{-10 \alpha }}{e^{-100 \alpha }}$

=>   $\alpha = \frac{0.18}{90}$

=>    $\alpha = 0.002 Np/m$