Which depth corresponds to the depth at which surface waves energy is reduced to approximately 1/25 of the original power?

Which depth corresponds to the depth at which surface waves energy is reduced to approximately 1/25 of the original power?

• A. 1 inch
• B. 4 inches
• C. 1 wavelength
• D. 4 wavelengths

Answer: (C)

Surface waves lose energy quickly with depth, concentrating near the surface. The decay is close to exponential with depth, so the amplitude at depth z can be described as A(z) ≈ A0 e^{-kz}, where k = 2π/λ and λ is the surface wavelength. Since power is proportional to amplitude squared, the power at depth z follows P(z) ≈ P0 e^{-2kz} = P0 e^{-4πz/λ}. To get about 1/25 of the original power, set e^{-4πz/λ} ≈ 1/25. Solving gives z/λ ≈ (ln 25)/(4π) ≈ 0.26, i.e., roughly a quarter of a wavelength. So the depth at which the surface-wave energy is reduced to about 1/25 of its surface value is about one-quarter of the wavelength. In practical terms, if your wavelength in the material is four inches, that depth is about one inch. The deeper depths (a full wavelength or more) would reduce energy far more than 1/25, while shallower depths would not reach that level of attenuation.

Surface waves lose energy quickly with depth, concentrating near the surface. The decay is close to exponential with depth, so the amplitude at depth z can be described as A(z) ≈ A0 e^{-kz}, where k = 2π/λ and λ is the surface wavelength. Since power is proportional to amplitude squared, the power at depth z follows P(z) ≈ P0 e^{-2kz} = P0 e^{-4πz/λ}.

To get about 1/25 of the original power, set e^{-4πz/λ} ≈ 1/25. Solving gives z/λ ≈ (ln 25)/(4π) ≈ 0.26, i.e., roughly a quarter of a wavelength. So the depth at which the surface-wave energy is reduced to about 1/25 of its surface value is about one-quarter of the wavelength.

In practical terms, if your wavelength in the material is four inches, that depth is about one inch. The deeper depths (a full wavelength or more) would reduce energy far more than 1/25, while shallower depths would not reach that level of attenuation.