Sound pressure

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The movement of the molecules of the medium generate variations in the density and pressure in the medium, which can be detected.

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Mechanisms

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Code
Concept
Formation of pressure
Sound Pressure

Mechanisms

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Sound Pressure

Description

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As sound propagates, it causes displacement of molecules at the boundary of the system, leading to impacts against the wall. These impacts transfer momentum to the wall, resulting in a force. Because the force is generated by a high number of particles, its effect depends on the surface area of the system, giving rise to a pressure.

It's important to understand that sound pressure is not the same as ambient pressure. In air, the latter is on the order of $10^5,Pa$, whereas sound pressure is typically much lower than $1,Pa$.

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Formation of pressure

Concept

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If we displace a cube's face, we generate an increase or decrease in concentration, which leads to a decrease or increase in collisions of molecules with the face of the volume:

Since pressure is the transfer of momentum due to collisions of molecules with the wall, the change in volume leads to an increase or decrease in pressure.

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Model

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Calculations

Variables

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$F$
F
Force of the medium
N
$Z$
Z
Impedance
kg/m^2s
$\rho$
rho
Mean density
kg/m^3
$u$
u
Molecule speed
m/s
$p$
p
Momento de la partícula
$L$
L
Noise level, air
dB
$p$
p
Pressure
Pa
$p_{ref}$
p_ref
Reference pressure
Pa
$S$
S
Section or Area
m^2
$p$
p
Sound pressure
Pa
$c$
c
Speed of sound
m/s
$t$
t
Time
s
$\Delta V$
DV
Volume with molecules
m^3
$\lambda$
lambda
Wavelength of Sound
m

Parameters

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units


First, select the equation: to , then, select the variable: to

Calculations

Symbol
Equation
Solved
Translated

Calculations

Symbol
Equation
Solved
Translated

Variable Given Calculate Target : Equation To be used


Equation

#
Equation

$ \Delta V = S \lambda $

DV = S * lambda


$ F =\displaystyle\frac{ d p }{ d t }$

F = dp / dt


$ L = 20 \log_{10}\left(\displaystyle\frac{ p }{ p_{ref} }\right)$

L = 20* log10( p / p_ref )


$ p \equiv\displaystyle\frac{ F }{ S }$

p = F / S


$ p = \rho c u $

p = rho * c * u


$ Z =\displaystyle\frac{ p }{ u }$

Z = p / u


$ Z = \rho c $

Z = rho * c

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Definition of pressure

Equation

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The water column pressure ($p$) is calculated from the column force ($F$) and the column Section ($S$) as follows:

$ p \equiv\displaystyle\frac{ F }{ S }$

$F$
Force of the medium
$N$
5815
$p$
Pressure
$Pa$
5224
$S$
Section or Area
$m^2$
5405

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Force exerted by molecules

Equation

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According to Newton, force can be expressed as the change in momento de la partícula ($p$). This momentum is generated by the rebound of particles, which transfer momentum to the wall. Since momentum is conserved and the particle's momentum changes from $p_{particle}$ to $-p_{particle}$ upon rebound, by momentum conservation we then have:

$p_{particle} = p_{wall} - p_{particle}$



which implies

$p_{wall} = 2p_{particle}$



Thus, the change in the momento de la partícula ($p$) on the wall is the time ($t$), generating force exerted by the Molecules ($F$), which is

$ F =\displaystyle\frac{ d p }{ d t }$

$F$
Force of the medium
$N$
5815
$p$
Momento de la partícula
$kg m/s$
9047
$t$
Time
$s$
5264

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Volume with molecules

Equation

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When a sound wave travels through ($$), it expands and contracts over a distance on the order of ($$), resulting in a volume variation depending on the section or Area ($S$) perpendicular to the direction of propagation.

Therefore, the volume variation is equal to:

$ \Delta V = S \lambda $

$S$
Section or Area
$m^2$
5405
$\Delta V$
Volume with molecules
$m^3$
5080
$\lambda$
Wavelength of Sound
$m$
5079

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Variation of the moment by molecules

Equation

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The sound pressure ($p$) can be understood as the momentum density calculated from the mean density ($\rho$) and the molecule speed ($u$), which is then multiplied by the speed of sound ($c$) via

$ p = \rho c u $

$\rho$
Mean density
$kg/m^3$
5088
$u$
Molecule speed
$m/s$
5072
$p$
Sound pressure
$Pa$
5084
$c$
Speed of sound
$m/s$
5073

The variation in momentum $dp$ is associated with the mass of the molecules $m$ and the velocity of sound $u$ through:

$dp = 2mu \approx mu$



Thus, in a time interval equal to the period $dt \approx T$, we have:

$F=\displaystyle\frac{dp}{dt}=\displaystyle\frac{mu}{T}$



Therefore, the sound pressure ($p$) can be calculated using the pressure



the speed of sound ($c$) is

$ c = \displaystyle\frac{ \lambda }{ T }$



and the volume with molecules ($\Delta V$) which varies

$ \Delta V = S \lambda $



as follows:

$p=\displaystyle\frac{1}{S} \displaystyle\frac{dp}{dt}=\displaystyle\frac{1}{S}\displaystyle\frac{mu}{T}=\displaystyle\frac{muc}{ScT}=\displaystyle\frac{muc}{S\lambda}=\displaystyle\frac{muc}{\Delta V}=\rho u c$



In the last term, both numerator and denominator are multiplied by $c$. The expression in the denominator represents the volume of gas displaced by the sound in $T$, so we can replace the mass divided by this volume with density, yielding:

$ p = \rho c u $

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Noise level as function of the sound pressure, air

Equation

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The noise level, air ($L$) encompasses a wide range of the sound pressure ($p$), making it useful to define a scale that mitigates this difficulty. To do so, we can work with the logarithm of the pressure normalized by a value corresponding to zero on this scale. If we take the minimum pressure that a person can detect, defined as the reference pressure ($p_{ref}$), we can define a scale using:

$ L = 20 \log_{10}\left(\displaystyle\frac{ p }{ p_{ref} }\right)$

$L$
Noise level, air
$dB$
5119
$p_{ref}$
Reference pressure
3.65e+10
$Pa$
5121
$p$
Sound pressure
$Pa$
5084



which starts at 0 for the audible range. In the case of air, the reference pressure ($p_{ref}$) is $20 \mu Pa$.

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Acoustic impedance

Equation

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The concept of impedance ($Z$) provides a measure of the system's resistance to transmit the sound wave. It considers a pressure acting and establishes a measure in which the exposed medium is displaced. In this way, the sound pressure ($p$) is compared to the molecule speed ($u$).

Therefore, impedance ($Z$) is defined as:

$ Z =\displaystyle\frac{ p }{ u }$

$Z$
Impedance
$kg/m^2s$
5104
$u$
Molecule speed
$m/s$
5072
$p$
Sound pressure
$Pa$
5084

ID:(3414, 0)



Impedance in waves

Equation

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To calculate impedance ($Z$) from the mean density ($\rho$) and the speed of sound ($c$), the formula used is:

$ Z = \rho c $

$Z$
Impedance
$kg/m^2s$
5104
$\rho$
Mean density
$kg/m^3$
5088
$c$
Speed of sound
$m/s$
5073

Since impedance ($Z$) is calculated from the sound pressure ($p$) and the molecule speed ($u$) using

$ Z =\displaystyle\frac{ p }{ u }$



along with the expression for the sound pressure ($p$) in terms of the mean density ($\rho$) and the speed of sound ($c$),

$ p = \rho c u $



we obtain

$ Z = \rho c $

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