How does cetacean echolocation sound inside a submersible?

Question

To be more specific: assuming an inner and outer hull, with the in-between gap partially serving as a ballast tank, would the transition from water to metal to air/water to metal to air:

a) amplify or mute the sounds, and

b) affect the range of perceived frequencies.

I understand that even the "best" human ears cannot detect the full range of whale song, but would the submersible enclosure alter the frequencies perceived within the range of human hearing?

Full disclosure: I'm just as interested in how the environment within a submersible affects the perception of artificial sounds (e.g. broadband sonar "pings"). But I'm hoping/assuming the answer would be more or less the same for biologically-generated whale song and mechanically/electronically-generated sonar "chirps"; if anyone has real-world experience in this area - i.e. being on a sub and hearing whale song and/or active sonar - feel free to chime in.

Answer 1

This sounds like a class exercise, but nevertheless:

The answer to your question is in the transmission at interfaces. When arriving at an interface part of the sound will be reflected and the rest will pass the interface.

The amount of sound that transfers an interface depends on the acoustic impedance (product of density and sound speed) of the materials (water,metal,air) and the angle of arrival.

For the three materials we may for example have

• Steel: $Z=ρ_sv_s = 46.5 \times 10^6$ kg/m$^2$s
• Water: $Z=ρ_sv_s = 1.5 \times 10^6$ kg/m$^2$s
• Air: $Z=ρ_sv_s = 420$ kg/m$^2$s

For simplicity lets ignore the arrival angle and assume perpendicular incidence.

If $A_i, A_r, A_t$ are the sound aplitudes of incomming, reflected and transitted sound amplitudes, respectively, then the formulas for reflection and transmission coefficients are

• Reflection $R_{12} =A_r/A_i= (Z_2 - Z_1)/(Z_1+Z_2)$
• Transmission $T_{12} =A_t/A_i= 2Z_2/(Z_1+Z_2)$

incoming and reflected sounds are in medium 1, and transmitted sound is in medium 2

So, we have for the transmission of the sound the following cases

• water-steel: $T_{ws}=1.94$
• steel-water: $T_{sw}=0.063$
• steel-air: $T_{sa}=1.8 \times 10^{-5}$
• air-steel: $T_{as}=2 $

Ignoring multiple reflections within the double hull and assuming that frequency is high enough, i.e. steel is thick enough and gap between double hull is wide enough for sound to propagate (no tunnel effect) then we may combine all transmission factors by multiplication.

• case 1: water-steel-water-steel-air $T = T_{ws}T_{sw}T_{ws}T_{sa} = 4.2 \times 10^{-6} = -108 \mathtt{dB}$
• case 2:water-steel-air-steel-air $T = T_{ws}T_{sa}T_{as}T_{sa} = 1.3 \times 10^{-9} = -178 \mathtt{dB}$

The sound pressure within the submergible is therefore significantly lower than in the surrounding water.

Observation: in case of lower frequencies, where wavelength is much larger than the hull thickness, then effectively only the water to air transmissivity counts (hull is becomming nearly transparent).

Note: Cases that deviate from the simplifying assumption, obviously require detailed acoustic modelling.