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Home , Acoustic wave, Acoustic wave equation, Particle displacement, Particle velocity, Wave equation

Chapter 5 – The Acoustic Equation and Simple Solutions

(5.1) In this chapter we are going to develop a simple linear for propagation in fluids (1D). In reality the equation is nonlinear and therefore more complicated than what we will look at in this chapter. However, in most common applications, the linear approximation to the wave equation is a good model. Only when sound have high enough do nonlinear effects show themselves.

What is an acoustic wave? It is essentially a change. A local pressure change causes immediate fluid to compress which in turn causes additional pressure changes. This leads to the propagation of an acoustic wave.

Before we derive the wave equation, let’s cover a few definitions and concepts.

Generation ® Transducer (piston for example) creates a particle (which in turn has an associated pressure and density change). This change affects the immediately adjacent region, etc., so that the disturbance (wave) propagates.

Uniform plane wave ® common phase and amplitude in a plane ^ direction of propagation.

Particle We will talk a lot about . It is defined in terms of . A particle contains many molecules (large enough to be considered as a continuous medium) but its dimensions are small compared to the distances for significant changes in the acoustic parameters (for example, small compared to l or small enough to consider acoustic variables constant throughout particle volume).

Parameters acoustic pressure particle displacement density changes velocity potential

! particle position rxxyyzz=++ˆˆˆ

! particle displacement xx=++xyzxyzˆˆ x xˆ

! ! ¶x particle velocity u = ¶t density r

Oelze ECE/TAM 373 Notes - Chapter 5 pg 1 rr– condensation s = 0 r0 undisturbed (equilibrium) density

acoustic pressure pPP= – 0

instantaneous undisturbed (ambient pressure)

Three laws used to develop wave equation for fluids

1) Equation of State – determined by thermodynamic properties Relates changes in Pressure (P) and density (r). Dependent upon material (for example, an ideal gas is different from a liquid). We can expand the Equation of state into linear and nonlinear terms, however, we will only be looking at the linear terms (for now).

2) Equation of Continuity – Essentially this is conservation of mass. Relative Motion of fluid in a volume causes change in density.

3) Equation of Motion – Force Equation – Newton’s 2nd law Pressure variations generate a force (F = P ´ Area) that causes particle motion

(5.2) Let’s first look at the Equation of State:

An equation of state must relate three physical quantities describing the thermodynamic behavior of the fluid. For example, the equation of state for a perfect gas is

P= r rTK

3 where P is the pressure in Pascals, r is the density (kg/m ), r is the , and TK is the temperature in Kelvin. If the gas has high thermal conductivity then any slow compressions of the gas will be isothermic and P r = . P00r If the thermal conductivity is sufficiently low, the heat conduction during a cycle of the acoustic disturbance becomes negligible. In this case the condition is considered adiabatic and the relation between the pressure and density for the perfect gas are: g P æör = ç÷ P00èør

C where g is the ratio of specific heats, p . Cv

Oelze ECE/TAM 373 Notes - Chapter 5 pg 2 The perfect gas is a simple case for an adiabat. To model this relationship, let expand P as a function of r (about r0) in a Taylor series:

2 æö¶¶PP1 æö 2 P = P00+ç÷(rr-) +ç÷2 ( rr- 0) +... ¶¶rr2 èør0 èør0

The Taylor series shows the fluctuations of P with r. We can rewrite this as: 11 P=+ P As + Bs23 + Cs +... 0 2 3! where æö¶P A = r0 ç÷, ¶r èør0 2 2 æö¶ P B = r0 ç÷2 . ¶r èør0

2 The Taylor series is nonlinear due to the quadratic term (r-r0) . The nonlinearity of a propagation medium is usually characterized by the ratio of B/A.

We linearize the Taylor series by assuming small fluctuations so that only the lowest order term in

(rr- 0 ) need be retained. This gives:

p= P- P0 = As where we define the coefficient A as the adiabatic , B: æö¶P A ==B r0 ç÷ ¶r èør0 æö¶P Let’s examine the unit for B = r0 ç÷ ¶r èør0 P : Pascal where 1Pa = 1 N/m2 = 1kg /s2/m r : kg/m3 P m2 ® r s2 P ® c2 (speed) (experimentally determined) r so 2 pcss==r0 B is the Equation of State for linear acoustic waves in fluids (small changes in density |s| << 1).

Oelze ECE/TAM 373 Notes - Chapter 5 pg 3 *************************** Example 5.1 *************************** Compare the condensation for water and air at 20˚C if the pressure change is 1 atm: Sol: 2 p = Bs = r0c s

3 From tables: for fresh water @ 20˚C: r0 = 998 kg/m and c = 1481 m/s

For p = 1 atm = 1.0133 bar = 1.0133 ´ 105 Pa

5 rr– 0 p 1.0133´ 10 -5 s ===22 =4.63x10 rr0 oc (998)(1481)

3 for air @ 20˚C: r0 = 1.21 kg/m and c = 343 m/s

1.0133´ 105 s ==0.712 (1.21)(343)2

1 atm is a huge density fluctuation in air.

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(5.3) The Equation of Continuity

This is essentially a statement of conservation of mass.

The book looks at a 3D representation of a fluid particle. We are going to look at a 1D representation and then generalize to 3D.

cross sectional area A

x x +x (x,t) x + Dx x x + Dx + xx (x + Dx,t)

Let’s first consider an undisturbed volume of mass = r0 AxD . Then, the mass is disturbed (expands) so that the new volume is: Axéù+D x+xx x +D- xt,, x+ xt = AD x+ xx x +D- xt ,, xt ëûxx( ) ( ( )) ëûéù xx( ) ( ) . For small Dx ¶x xx( xxtxt+D-@D,,) ( ) x x (approximately) xx¶x

Oelze ECE/TAM 373 Notes - Chapter 5 pg 4 giving the new volume

æö¶xx AxDç÷1+ . èø¶x Our disturbed density will then be mass r = volume

r0 AxD æö¶xx r = ®rr0 =ç÷1 + . æö¶xx è ¶x ø AxDç1+ ÷ è ¶x ø

Now, recall that the condensation is: rr– s = 0 . r0 Solving for the total density gives

rrr=+s 00. Substituting into our equation above yields

æö¶xx rrr000=+(s )ç÷1 + èø¶x or rearranging ¶¶xx ssrr=--xx r . 00¶¶xx 0 ¶x If we assume that s and x are small so that the second term is negligible (most often the case). ¶x ¶x s =- x ¶x Notice that the condensation, s , is the fractional change in density. This equation tells us that if the displacement, xx , varies with x then there is a density change. The minus sign is significant. If the displacement increases with x what happens to the density? Why?

If we generalize the equation to 3-D we get ¶x ¶x ¶x s =---x y z ¶¶¶xyz or æö¶¶¶ sxyzxyz=-ç÷ˆˆ++ˆˆ×(xxxxyz ˆˆ+ + ) èø¶¶¶xyz ! s =-Ñ×x If we differentiate with respect to ¶ ! s =-Ñ×x ¶t ( ) ! ! ¶¶! x ¶x ! we note that -Ñ×x =-Ñ× and = u so ¶¶tt( ) ¶t Oelze ECE/TAM 373 Notes - Chapter 5 pg 5

¶s ! =-Ñ×u . (the linear equation of continuity) ¶t or alternatively if we substitute in for the density we have: ¶r ! =-r Ñ×u . ¶t 0

(5.4) The Simple Force Equation: Euler’s Equation

Again, let’s consider a small volume of fluid (dV = dx dy dz) that is undergoing a force in the x- direction. The fluid element is moving with the surrounding fluid. The fluid element has a mass, dm. We will consider the 1D case and then generalize to the 3D case.

The force on the fluid from Newton’s 2nd law implies: df= adm . force Now, the pressure, P = , and in the figure we see the force is exerted on the cross sectional area area, dA = dy dz. Looking at the differential force or pressure across the dm element in the x- direction. é¶ùæöP dfx =êú P-ç÷ P+ dx dA ëûèø¶x

¶P df=- dv x ¶x Generalizing to 3 dimensions: ! df=++ dfxyz xˆˆ df y df zˆ ! æö¶¶¶PPP df=-ç÷ xˆˆ+ y + zˆ dv èø¶¶¶xyz ! df=-Ñ Pdv Relating to Newton’s 2nd Law (3D)

Oelze ECE/TAM 373 Notes - Chapter 5 pg 6 ! ! df= adm where ! ! du a = . dt Recall, there is a difference between ! ! du ¶u and . dt ¶t

!! ! ! ! du¶¶ u x ¶¶ u y ¶¶ u z ¶ u =+++ dt¶¶ x t ¶¶ y t ¶¶ z t ¶ t or simplifying !! ! ! ! du¶¶¶¶ u u u u =+++uuu dt¶¶¶¶ xxyz y z t !! !!!du¶ u auu==( ×Ñ) +. dt¶ t Using the fact that dm= r dv then relating the forces gives ! éù!!¶u -ÑPdv=r ( u×Ñ) u + dv ëûêú¶t

! éù!!¶u -ÑPuu=r ( ×Ñ) + ëûêú¶t This is the nonlinear, inviscid force equation (viscosity introduced later as a loss mechanism).

To get the linear Euler’s equation we will retain only the 1st order terms. st 1 : pPP=-0 ® Ñ p=Ñ P. 2nd: The particle velocity is assumed small so terms of 2nd order are negligible giving: ! ¶u -Ñp = r ¶t rd 3 : If we assume that |s| << 1 (small) then rr= 0 giving finally:

! ¶u -Ñp = r (linear Euler’s equation, Newtons’s Law) 0 ¶t

(5.5) The Linearized Wave Equation

(1) Equation of State (Linearized): 2 ps==B r0 cs æö¶P Where B = r0 ç÷ ¶r èør0 Oelze ECE/TAM 373 Notes - Chapter 5 pg 7 (2) (Linearized): ¶r ! =-r Ñ×u ¶t 0 (3) Euler’s Equation (Linearized): ! ¶u -Ñp = r 0 ¶t To derive the Linear Wave Equation we first take the derivative of s rr- Recall s = 0 r0 ds1 dr dr ds So that = or = r0 . dtr0 dt dt dt Substituting this into the Continuity Equation (2) gives the relative change in density as the divergence of the particle velocity. ¶s ! rr=- Ñ×u 00¶t Substituting in the Equation of State (1) yields ¶ éùp ! =-Ñ×u ¶t ëûêúB

¶p ! =-B Ñ×u (4) ¶t Taking the derivative of (4) wrt time gives: 2 ! ¶¶pu! ¶ =-BB Ñ×u =- Ñ× . (5) ¶¶tt2 ¶ t If we take the divergence of Euler’s Equation (3): ! éù¶u Ñ× -Ñp = r0 ëûêú¶t ! 1 ¶u -Ñ2 p =Ñ×. r0 ¶t Substituting into (5) gives the result: 2 ¶ p B 2 2 =Ñp ¶t r0 2 Using B = r0c gives us the Linearized Wave Equation: ¶2 p =cp22Ñ ¶t 2 This is also (in form) the Classical Wave Equation!

The is given by: c2

We can also define a velocity potential (similar to EM).

Oelze ECE/TAM 373 Notes - Chapter 5 pg 8 From eq (3), Euler’s equation, we note that the curl of a gradient is zero (Ñ´Ñf =0 ) so: ! ìü¶u Ñ´íý -Ñp = r0 îþ¶t ! ¶Ñ´u -Ñ´Ñp =0 = r 0 ¶t which implies ! Ñ´u =0 . This means that the particle velocity can be expressed as the gradient of a function: ! u =ÑF . Substituting back into the Euler’s equation gives: ¶ÑF -Ñp = r 0 ¶t or the relation between the pressure and the velocity potential is: ¶F p =-r . 0 ¶t The velocity potential, F , can also be shown to satisfy the wave equation.

(5.6) Speed of Sound in Fluids

Let’s evaluate the Equation of State:

æö¶P 2 2 æö¶P B ==rr00ç÷ c so that c = ç÷. ¶r ¶r èør0 èør0

GASES Now apply thermodynamic principles: PV = nRT (). To derive speed of sound in air for example, Newton applied Boyle’s Law: PV = constant where T is held constant. This means that heat is conducted from one region to another so that temperature does not change. This is known as an ISOTHERMAL PROCESS, that is, PV = constant at constant P T or = constant at constant T. r Consider: For an isothermal process, PPP r oo=®P = . rroo r Relating this back to the speed of sound,

2 æö¶¶P æörPPoo c ==ç÷ ç÷ = ¶¶rrrr èøo èøooo P Therefore, c = o ro

For air at STP where 5 2 Po ==1atm 1.013´ 10 Pa= 101.3 kPa (1 Pa =1 N / m ) Oelze ECE/TAM 373 Notes - Chapter 5 pg 9 3 ro =1.293 kg/m The propagation speed assuming an isothermal process is 5 Po 1.013x 10 Pa c ==3 =279.9 m / s (Newton’s first estimate of sound speed) ro 1.293 kg / m

But, at STP, the actual propagation speed in air is 331.6 m/s! Let’s examine the assumptions. Sound in the audio and ultrasonic frequency ranges is an adiabatic process. What does an adiabatic process mean in terms of ?

If heat has sufficient time to flow between compressed and rarefied regions, then the process maintains approximately the same temperature, and is therefore an isothermal process. Acoustic waves propagating in air do not follow an isothermal process (this is born out by the experimental measurements of the speed of sound in air).

P Consider the ADIABATIC GAS LAW, that is, PV g = constant, or = constant where g is the r g c ratio of specific heats, that is, g = p cv Consider: g PPooP r gg=®P = g (Eq. A9.23) rroo r Relating this back to the speed of sound: g 2 æö¶¶P æörgPPoo c ==ç÷ ç÷g = ¶¶rrrr èøo èøooo g P P Therefore, c = o , which is corrected with respect to c = o ro ro

*************************** Example 5.2 ***************************

Determine sound speed in air assuming an adiabatic process. Sol: For air at STP where 5 2 Po ==1atm 1.013´ 10 Pa= 101.3 kPa (1 Pa =1 N / m ) 3 ro =1.293 kg/m g =1.402 Thus, the propagation speed assuming an adiabatic process is 1.402 1.013x 105 Pa g Po ( )( ) c ==3 =331.4 m / s (what we actually measure) ro 1.293 kg / m *******************************************************************************

Oelze ECE/TAM 373 Notes - Chapter 5 pg 10 Combining the Equation of State (perfect gas law or ideal gas equation - see Eq 5.2.1 or Eq A9.17)

g Po P= r rTk and the expression for propagation speed (adiabatic assumption) c = yields ro

g Po gr( okrT ) c== =g rTk rroo where r is the gas constant for a particular gas æök ç÷ R m r ==o èøamu Eq A9.17 M æöm ç÷av èømamu

Ro the universal gas constant = 8314 J/(kg-˚K) = 8.315 J/(mol-˚K) M = average molecular weight (weighting corresponding to the fraction by volume of the total number of molecules) k the Boltzmann’s constant = 1.381 x 10-23 J/˚K -27 mamu is the mass of 1 atomic mass unit = 1.661 x 10 kg

mav is the average mass per molecule

For Air: Mixture for dry air by volume is 78% N2 - Molecular weight = 28 21% O2 - Molecular weight = 32 1% Argon - Molecular weight = 40

M = (0.78)(28) + (0.21)(32) + (0.01)(40) = 28.96 R 8314 J /(kg -o K) r = o = = 287.1 J /(kg -o K) M 28.96 d ++2527 g ====1.4 d 55 d = # of excited degrees of freedom and d = 5 for both N2 and O2 (3 translational and 2 rotational)

oo c= γrTk = (1.4)( 287.1JkgK /(- ))( 273.16 K) = 331.4 m/s 2 J N× m (kg× m / s )× m m2 m Note unit: = = = = kg kg kg s2 s

Oelze ECE/TAM 373 Notes - Chapter 5 pg 11 LIQUIDS g P g BB In general, for a gas c = o and, for a liquid c ==TA where ro rroo

BT = isothermal bulk modulus BA = adiabatic bulk modulus, that is BBAT= g

*************************** Example 5.3 *************************** Determine sound speed in fresh water assuming an adiabatic process. Sol: From Appendix A10 for fresh water at 20˚C 9 BT = 2.18´ 10Pa= 2.18 Ga P 3 ro = 998 kg/m g =1.004 Thus, the propagation speed assuming an adiabatic process is 1.004 2.18x 109 Pa g BT ( )( ) c = = 3 =1481 m/s ro 998 kg/m ******************************************************************************

SOLIDS In solids, two types of waves can propagate, that is, longitudinal and shear, each having a different propagation speed:

Y (1-s ) cL = rso (112+)( - s) Y cS = where ccLS> 21rso ( + ) where (see Appendix A10) Y = Young’s modulus ç What is Young’s modulus? Relates to strain in material. s = Poisson’s ratio ç What is Poisson’s ratio? Compress in one direction how much does it strain in transverse direction. Push down on something, how does it come out the sides?

Density and propagation speed ranges are Gases: 3 ro »1 kg / m co » 100 -1,000 m / s

Liquids: 3 ro »1,000 kg / m co » 1, 000 - 2,000 m / s Solids Oelze ECE/TAM 373 Notes - Chapter 5 pg 12 3 ro » 2,000 -10,000 kg / m cL » 2,000 -10,000 m / s (Longitudinal or Bulk) cS » 1,000 - 5,000 m / s (Shear or Bar)

Let’s calculate the acoustic pressure in a gas for a typical acoustic wave. Assume the solution for the 1D wave equation in terms of the particle displacement. ¶¶22xx = c2 ¶¶tx22 If the acoustic wave is traveling in one direction then we have a solution of the form:

xxw( xt,) =o cos( t- kx) wp2 where k == is the wave number (radians per meter, 1/m) c l w = 2πf is the (radians per second, 1/s) c is the propagation speed (m/s).

You may recall from our consideration of the continuity equation that we came up with a relation between the condensation, s, and the particle displacement. ¶x s =- (in one direction) ¶x By combining the Equation of State with the equation above, we can obtain the acoustic pressure in terms of the particle displacement: ¶x pcs= r 2 ® pc=-r 2 0 0 ¶x

The instantaneous condensation (from the solution to the 1D wave equation) is ¶x ( xt, ) ¶ sxt( ,) =-=--(xwo cos( t kx)) = -xwoktkxsin ( - ) ¶¶xx and the instantaneous acoustic pressure is

22 2 2 æöw pxt( ,) ==rrxwooo csxt( ,) c(-- k sin ( t kx)) =--rxoocksin ( w tkx) =--rxooctkxç÷sin ( w ) èøc

=--rxwooctkxsin ( w ) =ptkxo sin (w -)

In pxt( ,) =--rxwoo c sin ( w t kx) =ptkxo sin (w -) , po is the amplitude acoustic pressure, and the magnitude of the amplitude acoustic pressure is:

pcooo= r wx

However, most of the time, the absolute brackets are omitted when representing the magnitude of the amplitude acoustic pressure (phase is incorporated into the argument). So, what does this mean in real terms: Oelze ECE/TAM 373 Notes - Chapter 5 pg 13

*************************** Example 5.2 ***************************

If an individual speaks (in air at 20˚C) at a Level (SPL) of 70 dB, at a frequency of 1 kHz, then what are the magnitudes of the amplitude acoustic pressure and amplitude particle displacement? æöp ANSWER: Sound Pressure Level (see Eq 5.12.2) is defined as SPL= 20log o where p is 10 ç÷ o èøpref the peak (or rms) amplitude pressure and pref is the peak (or rms) reference amplitude pressure (“rms” is also referred to as “effective”). Note that the numerator and denominator must be either peak or rms, not mixed. The airborne pressure reference (see Table 5.12.1) is 28.9 µPa (peak) or 20 æöp p µPa (rms); this is a SPL of 0 dB. Therefore, 70= 20log o , o ==1070/ 20 3162.3, 10 ç÷ èøpref pref

po = (3162.3)( 28.9 µPa) = 0.0914 Pa(peak) or po = (3162.3)( 20 µPa) = 0.0632 Pa(rms).

That means that the particle displacement is: p 0.0914 Pa o (peak). xo = = 3 = 35 nm rwoc (1.21 kg / m )(343 m / s)(2p x 1000 r / s) Note that the reference amplitude acoustic pressure (peak) at 1 kHz is 0.11 Å (1 Å = 0.1 nm). ******************************************************************** P po Po

Po

time

Back to xxw( xt,) =o cos( t- kx) .

¶x ( xt, ) The instantaneous particle velocity is uxt( ,) ==--wx sin ( w t kx) ¶t o ¶uxt( , ) The instantaneous particle acceleration is axt( ,) ==--wx2 cos( w t kx) ¶t o ¶x ( xt, ) The instantaneous condensation is sxt( ,) =-=-- kxw sin ( t kx). ¶x o The respective magnitude expressions are for plane waves (these are useful to know!):

po uoo==wx roc

Oelze ECE/TAM 373 Notes - Chapter 5 pg 14 2 w po auoo==wwx o = roc

wxoou po skoo==x ==2 ccro c

*************************** Example 5.2 ***************************

If an individual speaks (in air at 20˚C) at a Sound Pressure Level (SPL) of 70 dB, at a frequency of 1 kHz, then what are the magnitudes of the amplitude particle velocity, amplitude particle acceleration and amplitude condensation?

ANSWER: Using information from the previous example, magnitudes (peak values) of the amplitude particle velocity is 2.20x10-4 m/s, amplitude particle acceleration is 1.38 m/s2 and amplitude condensation is 6.72x10-7. Small…

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A significant parameter in nonlinear acoustics and what is called shock wave theory is the Mach number (see Problem 5.7.1 in Kinsler et al). Based on the parameters introduced already, the Mach u number is defined as M = o . Note how this compares to the magnitude of condensation: c

wxoou po skoo==x ==2 ccro c

Oelze ECE/TAM 373 Notes - Chapter 5 pg 15