The acoustic impedance or acoustic impedance specific is the resistance that the material means when sound waves pass. It is constant for a given medium, which varies from a rock layer inside the Earth to biological tissue.
Denoting Z as acoustic impedance, mathematically we have to:
Z = ρ.v
Where ρ is the density and v is the speed of sound in the medium. This expression is valid for a plane wave moving in a fluid.
In SI International System units, density comes in kg / m 3 and velocity in m / s. Therefore, the acoustic impedance units are kg / m 2 .s.
Likewise, acoustic impedance is defined as the ratio of pressure and velocity:
Z = p / v
Expressed in this way, Z is analogous to the electrical resistance R = V / I, where pressure plays the role of voltage and speed than current. Other units of Z in the SI would be Pa.s / m or Ns / m 3 , completely equivalent to those given above.
Sound wave transmission and reflection
When you have two different impedances, Z 1 and Z 2 , part of a sound wave that hits the interface of both can be transmitted and another part can be reflected. This reflected wave, or echo, is the one that contains important information about the second medium.
How the energy carried by the wave is distributed depends on the reflection coefficients R and transmission T, two very useful quantities for studying the propagation of the sound wave. For the reflection coefficient, it is the quotient:
R = I r / I or
Where I o is the intensity of the incident wave and r is the intensity of the reflected wave. Analogously, we have the transmission coefficient:
T = I t / I or
Now, it can be shown that the intensity of a plane wave is proportional to its amplitude A:
I = (1/2) Z.ω 2 .A 2
Where Z is the acoustic impedance of the medium and ω is the wave frequency. On the other hand, the quotient between the transmitted amplitude and the incident amplitude is:
The T / A a = 2 Z 1 / (Z 1 + Z 2 )
This allows the I t / I o ratio to be expressed in terms of incident and transmitted wave amplitudes as:
I t / I o = Z 2 A t 2 / Z 1 A or 2
Using these expressions, R and T are obtained in terms of the acoustic impedance Z.
Transmission and reflection coefficients
The above quotient is precisely the transmission coefficient:
T = (Z 2 / Z 1 ) [2.Z 1 / (Z 1 + Z 2 )] 2 = 4Z 1 Z 2 / (Z 1 + Z 2 ) 2
As losses are not covered, it is true that the incident intensity is the sum of the transmitted intensity and the reflected intensity:
I o = I r + I t → (I r / I o ) + (I t / I o ) = 1
This allows us to find an expression for the reflection coefficient in terms of the impedances of the two media:
R + T = 1 → R = 1 – T
Doing some algebra to rearrange the terms, the reflection coefficient is:
R = 1-4Z 1 Z 2 / (Z 1 + Z 2 ) 2 = (Z 1 – Z 2 ) 2 / (Z 1 + Z 2 ) 2
And since information related to the second medium is found in the reflected pulse, the reflection coefficient is of great interest.
Thus, when the two media have a large impedance difference, the numerator of the previous expression becomes larger. Therefore, the intensity of the reflected wave is high and contains good information about the medium.
As for the part of the wave transmitted to the second medium, it is gradually attenuated and the energy is dissipating as heat.
Applications and exercises
Transmission and reflection phenomena give rise to several very important applications, for example, sonar developed during World War II and used to detect objects. In fact, some mammals like bats and dolphins have a built-in sonar system.
These properties are also widely used to study the Earth’s interior in seismic prospecting methods, obtaining medical ultrasound images, measuring bone density and capturing images of different structures in search of flaws and defects.
Acoustic impedance is also an important parameter in evaluating the sound response of a musical instrument.
– Exercise solved 1
The ultrasound technique to obtain images of biological tissue uses high frequency sound pulses. Echoes contain information about the organs and tissues they pass through, translated by the software into an image.
An ultrasound pulse aimed at the muscle-fat interface is incised. With the data provided, locate:
a) The acoustic impedance of each tissue.
b) The percentage of ultrasound reflected in the interface between fat and muscle.
Grease
- Density: 952 kg / m 3
- Sound speed: 1450 m / s
Muscle
- Density: 1075 kg / m 3
- Sound speed: 1590 m / s
Solution for
The acoustic impedance of each tissue is found by substituting in the formula:
Z = ρ.v
In this way:
Fat Z = 952 kg / m 3 x 1450 m / s = 1.38 x 10 6 kg / m 2 .s
Muscle Z = 1075 kg / m 3 x 1590 m / s = 1.71 x 10 6 kg / m 2 .s
Solution b
To find the percentage of intensity reflected at the interface of the two tissues, the reflection coefficient given by:
R = (Z 1 – Z 2 ) 2 / (Z 1 + Z 2 ) 2
Here, fat Z = Z 1 and muscle Z = Z 2. The reflection coefficient is a positive quantity, guaranteed by the squares in the equation.
Replacing and evaluating:
R = (1.38 x 10 6 – 1.71 x 10 6 ) 2 / (1.38 x 10 6 + 1.71 x 10 6 ) 2 = 0.0114.
By multiplying by 100, we have the percentage reflected: 1.14% of the incident’s intensity.
– Exercise solved 2
A sound wave has an intensity level of 100 decibels and normally falls on the surface of water. Determine the intensity level of the transmitted wave and that of the reflected wave.
Data:
Water
- Density: 1000 kg / m 3
- Sound speed: 1430 m / s
Air
- Density: 1.3 kg / m 3
- Sound speed: 330 m / s
Solution
The intensity level in decibels of a sound wave, indicated as L, is dimensionless and is given by the formula:
L = 10 log (I / 10-12 )
Increasing to 10 on both sides:
10 L / 10 = I / 10-12
Since L = 100, it results in:
I / 10 -12 = 10 10
Intensity units are given in terms of power per unit area. In the international system, they are Watt / m 2 . Therefore, the intensity of the incident wave is:
I o = 10 10 . 10 -12 = 0.01 W / m 2 .
To find the transmitted wave intensity, the transmission coefficient is calculated and multiplied by the incident intensity.
The respective impedances are:
Z water = 1000 kg / m 3 x 1430 m / s = 1.43 x 10 6 kg / m 2 .s
Z ar = 1.3 kg / m 3 x 330 m / s = 429 kg / m 2 .s
Replacing and evaluating in:
T = 4Z 1 Z 2 / (Z 1 + Z 2 ) 2 = 4 × 1.43 x 10 6 x 429 / (1.43 x 10 6 + 429) 2 = 1.12 x 10 -3
Therefore, the intensity of the transmitted wave is:
I t = 1.12 x 10 -3 x 0.01 W / m 2 = 1.12 x 10-5 W / m 2
Your intensity level in decibels is calculated by:
L t = 10 log (I t / 10-12 ) = 10 log (1.12 x 10-5 / 10-12 ) = 70.3 dB
In turn, the reflection coefficient is:
R = 1 – T = 0.99888
Thus, the intensity of the reflected wave is:
I r = 0.99888 x 0.01 W / m 2 = 9.99 x 10 -3 W / m 2
And its intensity level is:
L t = 10 log (I r / 10-12 ) = 10 log (9.99 x 10 -3 / 10 -12 ) = 100 dB
