Solving Sound Reduction Index Problem: Air to Water

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SUMMARY

The normal incidence sound reduction index (SRI) for sound traveling from air to water is calculated using the formula R_n = 10 log_{10}(1/τ), where τ is the transmission coefficient. The transmission coefficient τ is derived from the acoustic impedances of air and water, specifically τ = 4c₁ρ₁c₂ρ₂ / (c₁ρ₁ + c₂ρ₂)². Given the speed of sound in air (337 m/s) and water (1470 m/s), along with their respective densities (1.1 kg/m³ for air and 996 kg/m³ for water), the resulting SRI is 29.9 dB.

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tyco05
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I hate acoustics, but unfortunately I have to get through this subject.
Most of the time I don't even know where to start with the problems, and this is one of those times.
I'm sure there's some easy formula for it, but I just can't seem to find it.

What is the normal incidence sound reduction index for sound traveling from air to water in dB? Assume that the speed of sound is 337m/s in air and 1470m/s in water. Assume that the density of air is 1.1kg/m^3 and that the density of water is 996 kg/m^3

Everything I try I get the wrong answer. Please can anyone give me a hint of how to start this question? It's got to be easy!
 
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Can you calculate the transmission coefficient? Hint: Think acoustic impedance!

The sound reduction index (probably in your textbook) is SRI = -10 log T = 10 log 1/T
 
Got it, thanks.

So I use

[tex]R_n = 10log_{10}(\frac{1}{\tau})[/tex]

Where

[tex]\tau=\frac{4c_1\rho_1c_2\rho_2}{(c_1\rho_1+c_2\rho_2)^2}[/tex]

and I ended up with 29.9 dB

cheers
 

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