# How to derive a formula for and calculate the specific acoustic impedance of air and water at STP, knowing that: The density of air at STP is 1.29kg/m³ and the speed of sound is 332m/s; The...

How to derive a formula for and calculate the specific acoustic impedance of air and water at STP, knowing that:

• The density of air at STP is 1.29kg/m³ and the speed of sound is 332m/s;
• The density of water at STP is 1003kg/m³ and the speed of sound is 1500m/s.

theconjecture | Certified Educator

Specific acoustic impedance is defined as the ratio of acoustic pressure (p) to specific flow (u) (also known as acoustic flow velocity).

`z = p/u `

We know that for sound waves traveling in a medium, the displacement of the particles in the y direction at a position x and at a time t is given by:

`y(x,t) = A_0sin(kx - wt) `

Where Ao is the maximum displacement (the amplitude), k is the wave number (defined as k=2`pi`/`lambda`) and w is the angular frequency of the oscillation. It is also known that the velocity u of a particle in this sound wave and the acoustic pressure p are given by the following two equations:

`u(x,t) =(del y)/(del t) = -A_0wcos(kx - wt)`

`p(x,t) = -kappa (del y)/(del x) = -kappa A_0kcos(kx - wt) `

Here, `kappa` is defined as the adiabatic bulk modulus (it basically measures the resistance of a substance to uniform compression). Now, substituting this results in our definition of the specific acoustic impedance, we get:

`z= (kappa k )/w `

But it can also be shown, when studying sound waves, that

`kappa = rho v^2 `

Where v is the speed at which the sound wave travels in this medium and `rho` is the density of the substance. Finally, we have for the specific acoustic impedance, using the definition of k:

`z = rho v^2 k / w = rho v `

Now we see that the acoustic impedance of a substance is a function of the density of the medium and the velocity at which the sound wave travels in it. But both of this variables depend on the temperature, so the impedance is really a function of the temperature.

Using the values given by you, we have for air:

`z_(air)=(1.29 kg m^(-3))(332 m s^(-1)) = 428.28 (kg) / (m^(2) s)`

And for water:

`z_(water) = (1003 (kg)/(m^3))(1500 m/s)=1.5045xx10^6 (kg)/(m^2s)`

What this show us is that sound is less intense (~3500 times less) in water than in air for a given pressure amplitude! Things sound louder in air than in water!