ME459 – Notes 10 Sound Propagation & Wave Equation

Home , Particle velocity, Sound power, Wave equation

Wikipedia: Sound is "(a) oscillation in pressure, stress, particle displacement, particle velocity, etc., propagated in a medium with internal forces: elastic or viscous, or the superposition of such propagated oscillation. (b) Auditory sensation evoked by the oscillation in (a).

Sound can be viewed as a wave motion in air or other elastic media. In this case, sound is a stimulus. Sound can also be viewed as an excitation of the hearing mechanism that results in the perception of sound. In this case, sound is a sensation.

Physics of sound Sound can propagate through a medium such as air, water and solids as longitudinal (compression) waves, and also as a transverse wave in solids (alternate shear stress).

The sound waves are generated by a sound source, such as the vibrating diaphragm of a stereo speaker. The source creates oscillations in the surrounding medium. As the source continues to vibrate the medium, the vibrations propagate away from the source at the speed of sound, thus forming the sound wave: At a fixed distance from the source, the pressure, velocity, and displacement of the medium vary in time. At an instant in time, the pressure, velocity, and displacement vary in space. Note that the particles of the medium do not travel with the sound wave. The vibrations of particles in the gas or liquid transport the vibrations, while the average spatial position of the particles over time does not change.

Sound waves are characterized by these generic properties:  Frequency, or its inverse, wavelength  Amplitude, sound pressure or intensity  Speed of sound (c)  Direction

Sound that is perceptible by humans has frequencies from about 20 Hz to 20,000 Hz. In air at standard temperature and pressure, the corresponding wavelengths of sound waves range from 17 m to 17 mm. Sometimes speed and direction are combined as a velocity vector; wave number and direction are combined as a wave vector.

Sound measurement is performed with microphones. Do note that sound (as seen next) is a perturbation of pressure (in time and space) in a stationary (calm) medium.

To measure sound level or magnitude one typically uses a decibel (db) = 1/10 bel.

A human’s perceived sound level is proportional to the log scale of sound pressure (p’).

since 2 p as the L 10log RMS p 10 p ref 2 power in a sound wave is ~ pressure .

The threshold sound pressure of human’s hearing at 1 kHz is -6 pref = 20 10 Pa

If the sound power drops to ½ of the reference, then 2 pRMS Lp 10log102  10log 10  0.5  10*(  .3)   3 dB pref While if the sound power doubles, then 2 pRMS Lp 10log102  10log 10  2  10*(  .3)   3 dB pref

Also 80 dB means  10,000 larger than reference 22 pp    8 logRMS    RMS   1084 p  10 p  0.01 Pa 10 pp    RMS ref ref   ref 

While 120 dB means  1 million larger than reference 1 pp    6 logRMS  RMS  1066 p  10 p  1 Pa 10     RMS ref ppref   ref  LOUD!

Typical sound pressure levels of common noise sources (copied from Acoustic Noise Measurements – Bruel & Kjaer)

(Based on handwritten Lecture Notes 18-19 by Dr. Joe Kim) Mathematical analysis of sound propagation and acoustic pressure fluctuations considers

a) Inviscid media ( gas has no viscosity)

b) Adiabatic process – no thermal energy exchange between gas (air) particles  a constant entropy process.

The equation of state for an ideal gas is:

PRT G (1) where P and T are the gas absolute pressure and absolute temperature, RG is the gas constant and  is the ratio of specific heats (=Cp/Cv).

The media (gas) is considered stagnant (with zero mean velocity) and the pressure is equal to

P Paa  p''      (2)

Where the sub index a stands for ambient condition and the ‘ stands for a fluctuation (or perturbation) about the mean value.

Conservation of mass The equation for conservation of mass in a compressible fluid is D    v 0 (3) Dt t where v=(vx, vx, vz) is the acoustic velocity of a particle in a static medium. Above  is the divergence operator.

Here, a small term like (’v) is neglected.

Conservation of momentum For an inviscid fluid (no viscosity), the momentum equations of a fluid reduce to Dvv    vv    P (5) Dt t where  is the gradient operator. Ignoring advection of fluid motion and linearization of the temporal change in momentum leads to Euler’s equation:

 v    p' (6a) a t

These are three independent equations written in the Cartesian coordinate system as

vx p'''vy  p  vz  p ;;     (6b) at  x a  t  y a  t  z which shows the particle acceleration (change in momentum) is proportional to the gradient of the perturbed (dynamic) pressure along that direction.

Now, derive the conservation of mass Eq. (4) with respect to time to obtain

And substitute Euler’s Equation into Eq. (7) to obtain 2  '   p'0  (8a) t 2 And recall    2 is the Laplacian operator. Hence, Eq. (8a) becomes 2  ' 2 p' (8b) t 2

Changes in pressure and density define the sound speed co as p'  R T c2 (9)  ' Go

For air (molecular weight=29) RG=286.7 J/(kg K) and  =1.4, and at a temperature of 300 K (~27 C), the speed of sound co =347 m/s

Substitute Eq. (9) into Eq. (8b) to obtain

2 2 2 2 1p '2  p '  p '  p ' 2 2 p'  2  2  2 (10a) co  t  x  y  z

This is the WAVE Equation or the equation of propagation of acoustic pressure in Cartesian coordinates (x,y,z).

In polar coordinates (r,, z):

The wave Eq. (10) is solved in the domain with specified particular boundary conditions in the closure of the domain as well as with initial conditions.

Once the pressure field p’ is obtained, the velocity field (v) follows  v 1 from Euler’s Eqn.   p' (6) t a

Harmonic wave propagation The solution of the wave equation is of the general form

ixx x i x iyy y i y p'x,,, y z t  Ax e  A x  e A y  e  A y  e  (11) izz z i z i t i t  Az e A z  e A t  e A t  e  where i is the imaginary unit. Above is a characteristic 1/length=wave number andis a 1/time=frequency scale.

(w/o reflective boundaries) Let iy ixx y izz it (12) p'x,,, y z t  A e e e e 

Substitute into Eq. 10(a) to obtain 2 2 1' p 2  2222 (*) 22 p'  2 x  y  z =  cto  co

ˆˆˆ where the wave number () is a vector:  xi   y j   z k .

MEEN 459 – Notes 10 Sound propagation and the wave Eqn. L. San Andrés © 2019 8 it Similarly for the velocity vector, let vv 'x,, y z e and sub into  v Euler’s equation    p' to obtain a t 11 v''' pp    (13) i a ic o   a

Since  co .

Note above the relationship between frequency () and wave number (). For example, given co =347 m/s, then

Freq  co  Hz rad/s 1/m m 20 125.6 0.36 17.34 500 3,142 9.05 0.68 5,000 31,420 90.5 0.068 20,000 125,700 362 0.0174

Recall sound that is perceptible by humans has frequencies from about 20 Hz to 20,000 Hz. The higher the frequency is, the smaller is the wave length () [full period]

MEEN 459 – Notes 10 Sound propagation and the wave Eqn. L. San Andrés © 2019 9 WAVE PROPAGATION IN A DUCT For sound propagation along the x-direction1 (only), the equations are 22pp'' v  p'  c2 (15a) and  x  (15b) tx22o a tx

Above coG  R T [m/s] is the sound speed.

The solution of PDE is of the form pv'(,)()()x t x t (16) Sub into Eq. (15a) to get

v()()tx22 co   (17) v()()tx 22 or vv()()()t 0 &  x    x  0 (18) where  /co (19)  is a frequency [1/s] &  is the inverse of the wave length [m].

The solution of the ODEs (18) is (for example) it v()t C tcos t  S t sin t  D t e (20a)

i x i x ()xC xcos  x  S x sin  x  Ae  Be (20b)

i x i  x i  t Hence, p'(,)()()x t x v t  Ae  Be e (21a) And the propagation speed is 1 p ' 1 ix i x i t (21b) vx   Ae  Be e ico  a x c o  a

1 Note the wave equation is identical to that derived for vibrations of a taut string (see Lecture Notes 3).

Example 1. Duct with one open end

At x=0, specified velocity Aei x Be i x it x0 AB  vx(0, t ) v o e  (0) v ( t )    v o (22a) cco a o a

At x=L, open end  no pressure perturbation (a pressure release condition) i L i L (22b) p'(,)()()()L t 0  L v t  L  0  Ae  Be

Hence, the two equations for finding A and B are

11  Ac    ao (23) i L i L    vo e e   B  0 

with  ei L  e i L  2cos( L ) (24) recall: ei acos( a )  i sin(a); e i a  cos( a )  i sin(a) 11 then A c v ei L; B c v e i L (25) a o o a o o

 cv i x i  x i  ta o o i x L  i x  L i  t p'(,)xt  Ae  Be e  e  e e  

eia cos( a ) i sin(a) Since  2ia sin( ) ia ecos( a ) i sin(a)

sin  xL  it Then  p'(,)x t i a c o v o e  (26) cos( L)

Similarly, the propagation speed is

1' p cos xL  it vxo   v e  (27) icoa  x cos(  L) where the wave number co .

Natural frequencies and mode shapes From the system of Eqns. (23), note the characteristic equation is  2cos( L )  0  cos L 0 (28) having an infinite number of solutions. The wave numbers and natural frequencies are

2nn 1  2 1 c      c   o ; (29) n22LL n n o n1,2.....

nsin nxL n1,2,... (30) or x L  35  x  L     x  L  1sin  ;  2  sin   ;  3  sin   2LLL   2   2  shown below.

Fig 3. Natural modes shapes (x) for duct with one end open

Acoustic impedance Recall the definition of an impedance Z = effort/flow (31) In an electrical system  Z = voltage/current = V/I; while in a mechanical system  Z=Force/velocity or Torque/angular speed.

p' sin  xL  Z  ia c o  i  a c o tan    x  L (32) vx cos  x L

At the inlet of the duct, x=0, the acoustic impedance is

Z(x 0)  i a c o tan L  (33)

The relationship is imaginary, hence reactive (not adding energy)

Recall in an electric system, with a resistor element Z= V/I  R (real #) > 0 dissipates energy as the voltage and current are in phase (power = I2 R).

Not so for a capacitance (C) or an inductance (L) where V=1/C q (charge) and V = L dI/dt, with I=dq/dt (the temporal change in chargeq)

These two elements are conservative, i.e. storing energy, and with the power in a full cycle = 0.

At x=0, specified velocity Aei x Be i x it x0 AB  vx(0, t ) v o e  (0) v ( t )    v o (34a) cco a o a

At x=L, closed end  no velocity i L i L Ae Be vvx( L , t )00  (L) ( t )   (34b) coa

Hence, find A and B from

11  Ac    ao (35) i L i L    vo e e   B  0 

with  ei L  e i L  2 i sin( L ) (36)

11 then A c v ei L; B c v e i L (37) a o o a o o

i x i  x i  t Sub Eq. (37) into p'(,)xt  Ae Be e to get

cos xL  Then  p'  i c v eit (38) (,)x t a osin( L) o And the propagation speed is

1' p sin  xL it vxo   v e  (39) icoa  x cos(  L) where the wave number co .

Natural frequencies and mode shapes The characteristic equation is  2iL sin( )  0 (39) having an infinite number of solutions. The wave numbers and natural frequencies are

n c     cn   o ; (40) nLL n n o n0,1,2.....

Associated to each natural frequency, the mode shapes are (see Eqn. (38):

ncos nxL n0,1,2,... (41)

Fig 3. Natural modes shapes (x) for duct with one end closed

Acoustic impedance The acoustic impedance is

p' cos xL  1 Z  ia c o  i  a c o tan    x  L  (42) vx sin   x L

At the inlet of the duct, x=0,

Z(x 0)  i a c o cot L  (33)

The relationship is imaginary, hence reactive (not dissipating or adding energy).

Example 3. More on acoustics of organ pipe (duct with one end closed)

In equal temperament, the octave is divided into equal parts on the logarithmic scale.

12 tone scale A4 A3 =1/2 A4A5 = 2A4

440   "A4"   "La4"        466.16 1 1 "A#4" "La#4"       1 220 1 880      493.88 "B4" "Si4"   2 233.08 2 932.32     523.25 "C5" "Do5"   3 246.94 3 987.76      554.37  "C#5"   "Do#5"  4 261.625 4 1.046·103  587.33  "D5"   "Re5"  f  Hz 5 277.185 5 1.109·103 s   fs     622.25  Hz f 2  Hz  "D#5"   "Re#5"    6 293.665 s 6 1.175·103  659.25 2  "E5"   "Mi5"  7 311.125 7 1.244·103  698.46  "F5"   "Fa5"    8 329.625 8 1.319·103     739.99 3  "F#5"   "Fa#5"    9 349.23 9 1.397·10  783.99 10 369.995 10 1.48·103  "G5"   "Sol5"         830.61 3  "G#5"   "Sol#5"  11 391.995 11 1.568·10 12 415.305 12 1.661·103

wave length for A4 1 1 1 7.967  "A4"  1 78.864   2 8.441 "A#4" 2 74.438   3 8.943  "B4"  4 9.475 3 70.26  "C5"  5 10.038 4 66.316   2πfs 1  "C#5"  κ   6 10.635 c 5 62.594 c m o  "D5"  o λ   6 59.081 cm 7 11.267 f   s "D#5" 8 11.937 7 55.765    "E5"  9 12.647 8 52.636  "F5"  10 13.399 9 49.681   "F#5" 11 14.196 10 46.893   11 44.261  "G5"  12 15.04   12 41.777  "G#5"  n12 j1n1  

f sj1 rat  ratio between notes = 1.059 j f sj An equal temperament is a musical temperament, or a system of tuning, in which the frequency interval between every pair of adjacent notes has the same ratio. In other words, the ratios of the frequencies of any adjacent pair of notes is the same 12 max( rat ) 1.059 min( rat ) 1.059 rat  2 j1n  1

 "A4"  900    "A#4"  G5  "B4"  800  "C5"  F5   "C#5" 700 E5    "D5"  D5   600  "D#5"  C5  "E5" 

Frqequency (Hz) B4  "F5"  500 A4    "F#5"  400  "G5"  0 5 10 15    "G#5"  Musical scale (A4) Example Acoustic Vibrations - ORGAN PIPE L San Andres (c) SP 19 MEEN 459

PHYSICAL Parameters molecular weight for gas MW  29 2 8314.34 J m R    286.701 G 2 gas constant MW kg K Ks

pipe length γ  1.4 ratio of specific heats T  300 K temperature Pa  1 bar L  78.41 cm 0.5 m exact length for A4 co  γRGT  347.008 s sound speed CLOSED DUCT Pa kg ρ   1.163 a 3 RGT m

( a) natural frequencies & mode shapes .5 using separation of variables, pxt() = ϕ()vtx  () (1) with κω= co Substitute into the field Eq. (0) to obtain the following two ODEs: 2 d 2 ϕ  k ϕ = 0 (2a) 2 dx The solution to the ODEs is ϕ()x = Axcos()κx  Bxsin()κx (3a) 2 d 2 (2b) v  ω v = 0 vt()= Acos()ωt  B sin()ωt (3b) 2 t t dt

Satisfy the boundary conditions.

At left end x=0, dp/dx=0 (zero velocity). dϕ = A κsin()κ0  B κcos()κ0 = 0 then BX = 0 dx x x and ϕ(x )= cos()κx (4) is the equation for the shape funcion.

At the right end, x=L, the duct is closed, hence p'=0 (zero velocity) dϕ Set T  4000 N at x=L = sin()κL = 0 (5a) j0n  dx example jπ = characteristic equation. The roots are κj  T 1 L κ  ()0 4.007 8.013 12.02 m And thus, the natural frequencies are ωn  coκj rad/s The longer the pipe is, the lower the j natural frequency T ωn  ()Hz0 221.278 442.556 663.834  rigid body mode + hamonics of first A3 2π A3 A4 A5 j1n  2π 2L λ  wave lengths  The mode shape j CLOSED DUCT κj j functions are ϕ1x( ) cos() 0 ϕ2x( ) cos[]π()x1 ϕ3x( ) cos[] 2π()x1 natural mode shapes 1  0    x=0: rigid 1.568 boundary λ    m 0.5 0.784    0.523  0 0

 0   0.5   ω x=L n 221.278 f     Hz closed duct (v=0) n 2π 442.556   1   0 0.2 0.4 0.6 0.8 663.834 x   Mode 1 L λ2 Mode 2  1 (mode 3) Mode 3 L 0.784m L kg Solution for a specific velocity and frequency ρ  1.163 m a 3 c  347.008 m o s L 0.784m recall natural frequencies Set a velocity perturbation at x=0 T f ()Hz0 221.278 442.556 663.834 m f 430 Hz n   vo  1 s pitch lower than A4 3 1 build pressure and velocity waves ω  f2 π  2.702 10 ω 1 s κ   7.786 c m period 1 o T   2.326 ms f wave length 2π λ   0.807m κ

p_  ρacovo  403.449Pa constant for excited pressure

from lecture notes 10

cos[]κ()xL  iωt take real part & divide by constant value p_ and vo pxt()  p_ ie sin()κ L  cos[]κ()xL pxt()  1 sin()ωt sin[]κ()xL  iωt sin()κL vxt()   v e  o sin()κL   sin[]κ()Lx  vxt()  1 cos()ωt   sin()κL  κL  6.105 λ  1.029 Note pressure is 90 away out of phase with velocity < 2π L power() x t  pxt()vxt  () sin(κL )  0.177 near resonance

1 S   5.64 max sin()κL

For time-varying solution N_periods 2 for analysis

tmax  T N_periods

t N  30 total number of frames or steps max  4 Δt   1.55 10 s time step N

k10 FRAME S  6 t_k  kΔt max BUILD: pressure &velocity waves vs time f 430 Hz L 0.784m   T 2.326ms  λ  0.807m velocity pressure

t_k 5 5  0.667 T vx0() px0() 1 0 0 vx t_k px t_k

 5  5 0 0.2 0.4 0.6 0 0.2 0.4 0.6 λ  1.029 x L x L power p_ 403.449Pa 20 1 power() x 0  5.64 0 sin()κL power x t_k  20 T f  ()Hz0 221.278 442.556 663.834  0 0.2 0.4 0.6 n x

Waves of velocity, pressure and power vs. 0.4 xx L  31.364cm time at 1

 3 T 2.326 10 s

5 T

0 0

 5  3  3  3  3 0 110 210 310 410

time (s) 1 f 430 v s p power PRESSURE AMPLITUDE FREQUENCY RESPONSE at varios spatial locations x

ω graph f  1000H cos ()xL  max p c   o  T 1 p_ p ()xω  1 f  ()0 221.278 442.556 663.834 F  ω  n s sin L natural frequencies:  co  p_ 403.449Pa 10

1 1 Amplitude of pressure/p_ Amplitude 0.1 0.1 3 0 200 400 600 800 110

frequency (Hz) NOTE resonances x=0 at each natural frequency. x=0.25L x=0.5L x=0.75L VELOCITY FREQUENCY RESPONSE at varios spatial locations x natural frequencies: ω sin ()Lx  T 1 v   f  ()0 221.278 442.556 663.834 co  n s v v ()xω  1 o F  ω  sin L NOTE resonances m  co  at each natural frequency v  1 o s

1 1 Amplitude of velocity/vo 0.1 0.1 3 0 200 400 600 800 110

frequency (Hz) x=0 x=0.25L x=0.5L x=0.75L More notes from Dr. Joe Kim

Sound intensity level

SoundIntensitypwbey

To measure sound power perunit area

Spacer

J k

to Two microphones

Probe Briiel Kjar Sound Intensity

H I I Re p't I the CPI't Definition Sound intensity acoustic power perunit area

Recall in HW 2 I etzke.CAtvJ tzReCFV Ep.werEIoreehrelocityp jI pressure Of area For the measurement of the sound intensity p can be measured with a microphone and I with various sensors it Here is measured with two microphones based on Euler's equation Post 3 FinitedifferenceApproximation FDA r f ie.ir D t mic I mic 2

p Pitt 2 and A

Int I Rel ie l

qq.tc.ggRe fi P pz Pip P Pz p Pil

Define Sir P Pz i cross spectrum Su Pip complex Sit numbers g pip I RT Auto spectrum real Su pipe numbers IE.e.i.hdtmcs.LT Rolling Tire at 21 mph

Tire driven by a roller Aperture 8 reference microphones 8 by 8 scanning microphone array (10 cm spacing) Hologram height: 6 cm

16 Point Measurement with 5 cm Sampling Space y

x Smooth Roller Surface

32 Point Measurement with 5 cm Sampling Space

Acoustics and Signal Processing Laboratory, Department of Mechanical Engineering Rolling Tire: NAH Results at 21 mph (128 Hz, n = 3)

Projected sound fields on source plane (z = 0) at 128 Hz: (a) acoustic pressure, (b) particle velocity in z-direction, (c) active sound intensity in z-direction, and (d) reactive sound intensity in z-direction.

Acoustics and Signal Processing Laboratory, Department of Mechanical Engineering