Reinforcement depth in marine recording

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Problem 3.5a

For a source at a depth $ h $, show that the maximum amplitude of a downgoing incident wave and its reflection at the surface of the sea occurs at the depth $ \lambda /\left(4\cos \theta \right) $, where $ \theta $ is the angle of incidence, by expressing the pressure $ P $ in the form used in equations (3.1b,d) and applying appropriate boundary conditions.

Solution

Since the interface is liquid/vacuum, only two waves exist, the incident and reflected P-waves. Taking the z-axis positive downward, we take $ {\mathcal {P}} $ in the form

$ {\begin{aligned}{\mathcal {P}}=A_{0}e^{\mathrm {j} \omega p\left(x+z\cot \theta \right)}+A_{1}e^{\mathrm {j} \omega p\left(x-z\cot \theta \right)}.\end{aligned}} $

There is only one boundary condition, namely that $ {\mathcal {P}}=0 $ at $ z=0 $. This gives $ A_{1}=-A_{0} $ Using Euler’s formulas (see Sheriff and Geldart, 1995, problem 15.12a), we get

$ {\begin{aligned}{\mathcal {P}}&=A_{0}e^{\mathrm {j} \omega px}\left(e^{\mathrm {j} \omega pz\cot \theta }-e^{-\mathrm {j} \omega pz\cot \theta }\right)\\&=2\mathrm {j} A_{0}e^{\mathrm {j} \omega \left(px-t\right)}\sin \left(\omega pz\cot \theta \right)\end{aligned}} $

upon inserting the time factor. The amplitude of the combined incident and reflected waves is

$ {\begin{aligned}2A_{0}\sin[\left(\omega pz\cot \theta \right]=2A_{0}\sin \left[\left(\omega z/\alpha \right)\cos \theta \right].\end{aligned}} $

It is a maximum when $ \left(\omega z/\alpha \right)\cos \theta =\pi /2 $, that is, when

$ {\begin{aligned}z=\left(\pi /2\right)\alpha /\left(\omega \cos \theta \right)=\lambda /\left(4\cos \theta \right).\end{aligned}} $

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