# Maths for Sound passing through different mediums

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## Summary:

What is the mathematics involved with Sound through different mediums

## Main Question or Discussion Point

What is the mathematics involved with calculating the energy lost from sound as it passes through different mediums?

If I started off with a 70dB(A) sound wave, and after 0.5m it passed through 10mm of mild steel - what would be the sound level (in dB) 1m away from the steel plate?

To clarify, there are 3 mediums: Air > Mild Steel > Air

Air:
Density of 1.225 Kg/m³

Steel:
Density of 7850 Kg/m³
Young's modulus 207 GPa
Poisson's Ratio of 0.303

Thanks

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berkeman
Mentor
Sounds like a schoolwork problem. What are the Relevant Equations?

Two hints:
1. Acoustic impedance is ##Z=\rho v## where ##\rho## is density and ##v## is sound speed.
2. Acoustic reflection coefficient is ##(\frac{Z_2-Z_1}{Z_2+Z_1})^2 ##.

sophiecentaur
Gold Member
Two hints:
1. Acoustic impedance is ##Z=\rho v## where ##\rho## is density and ##v## is sound speed.
2. Acoustic reflection coefficient is ##(\frac{Z_2-Z_1}{Z_2+Z_1})^2 ##.
For situations with three layers of similar acoustic impedance and a wide middle of the sandwich (in terms of wavelength) the sound can bounce backwards and forwards, modifying the impedance at each wavelength, according to the precise spacing. In the case of steel, the speed of the sound (and hence the wavelength) is so high that this is not a significant effect.
There is another factor and that is the metal plate would not be stationary and but vibrate as a diaphragm and you can get resonance in that mode. You would need to know the dimensions (diameter) of the plate. It could well be the main mechanism for sound transfer, I think.
Perhaps the OP should describe the problem in more detail - if it is, indeed a real life problem.

berkeman
For situations with three layers of similar acoustic impedance and a wide middle of the sandwich (in terms of wavelength) the sound can bounce backwards and forwards, modifying the impedance at each wavelength, according to the precise spacing. In the case of steel, the speed of the sound (and hence the wavelength) is so high that this is not a significant effect.
There is another factor and that is the metal plate would not be stationary and but vibrate as a diaphragm and you can get resonance in that mode. You would need to know the dimensions (diameter) of the plate. It could well be the main mechanism for sound transfer, I think.
Perhaps the OP should describe the problem in more detail - if it is, indeed a real life problem.
I suspect, as @berkeman, that this is a school work problem. In a simple analysis one would expect that the attenuation of sound in the steel is negligible and there is no excitation of resonant modes so the problem becomes computing the reflection coefficient at the first air-steel interface then computing the transmission coefficient from ##T=1-R##, then computing the refection coefficient again at the second interface using the value of T to determine transmission coefficient at second interface. Not much is getting through because the ratio ##\frac{Z_{steel}}{Z_{air}}\approx 40,000##

sophiecentaur
Hi Everyone, this is not a school work problem, I would have stated that if it was - this is a work problem; I want to know how loud my control panel is going to be. I exaggerated the thickness of the steel, so I could see and understand the maths and do the calculation myself with my actual values.

The control panel has a number of devices in it that accumulate to 70dB, this is inside a mild steel enclosure of 1800mm High x 1600mm Wide x 500mm Deep - assumed that the components producing this noise would be in the center of the enclosure.

So Acoustic Impedance is Z=ρv, but doesn't the speed of sound change as density changes, so I wouldn't use the speed of sound in a vacuum?

In regards to the Acoustic reflection, I assume that Z1 is the first medium, and Z2 would be the 2nd medium?

Where does the thickness of the material come in to this? And surely the Young's modulus and Poisson's rations would be used to calculate losses due to attenuation?

Thanks

Last edited:
sophiecentaur
Gold Member
a mild steel enclosure of 1800mm High x 1600mm Wide x 500mm Deep
So you are not dealing with the transport of sound through a bulk material. It is as I suspected, the sound being transmitted by flexing of the sides. The Zsteel is not so relevant and it is the surface density and the flexing modulus that count.
If you just need to reduce the level of sound escaping then you could brace and load the plates and possibly use the sound damping materials that are used on car body panels. There are images of what I mean in this link and this link. The whole enclosure should be sealed for a start because sound can escape from small holes, joints and slots. How many dB of reduction do you need? It's best to be aware of what you could expect from the various measures you could take, especially if it involves dismantling and rebuilding the enclosure. The enclosure needs to be totally sealed from the room, possibly venting to the outside if necessary. Also, the panel needs to be as massive as you can conveniently provide.

So you are not dealing with the transport of sound through a bulk material. It is as I suspected, the sound being transmitted by flexing of the sides. The Zsteel is not so relevant and it is the surface density and the flexing modulus that count.
If you just need to reduce the level of sound escaping then you could brace and load the plates and possibly use the sound damping materials that are used on car body panels. There are images of what I mean in this link and this link. The whole enclosure should be sealed for a start because sound can escape from small holes, joints and slots. How many dB of reduction do you need? It's best to be aware of what you could expect from the various measures you could take, especially if it involves dismantling and rebuilding the enclosure. The enclosure needs to be totally sealed from the room, possibly venting to the outside if necessary.

I do not need to achieve any level of reduction, it was just some maths that I was interested in finding out about.

In my ignorance, I thought the maths would be:

1) Converting the dB into Watts/m2.
2) Calculating the energy lost in it passing through the initial medium (air): I2 = I1 - 10 x log(d1/d2)2.
3) Calculating from the density, young's modulus and poisson's ratio how much energy reflected and dissipated as it continues through the 2nd medium at 'x' thickness.
4) Calculating how much energy is reflection, as it passes through to the 3rd medium.
5) Then calculating the energy lost in passing through the 3rd medium (air): I2 = I1 - 10 x log(d1/d2)2.
6) Converting Watts/m2 back to dB.

It's just stage 3 and 4 that I was unsure on - but I imagine this is where the Acoustic impedance comes in?

sophiecentaur
Gold Member
I do not need to achieve any level of reduction, it was just some maths that I was interested in finding out about.
OK for interest, of course but not very relevant to your system. Using the formulae above, do you get the sort of isolation that the second link in my post suggests? Acoustics is very hard!

I have attached a crude drawing of the system indicating the reflected and transmitted acoustic power using the equations for the reflection coefficient ##R=(\frac{Z_{steel}-Z_{air}}{Z_{steel}+Z_{air}})^2## and the transmission coefficient ##T=1-R##. I didn't account for attenuation due to distance from the plate.

sophiecentaur