424  

I. Energy conservation and Bernoulli's EquationII. Bulk modulus and the speed of soundIII. Viscosity

425  

I. Energy conservation and Bernoulli's EquationConsider a tube of moving fluid with constant density ρ

By the Continuity Equation, we know that mass is conserved.Consider the time interval dt.At End 1, the fluid travels distance v1dt.At End 2, the fluid travels distance v2dt.

'End 1':area A1

pressure P1

height h1

velocity v1

'End 2':area A2

pressure P2

height h2

velocity v2ρ

v1dt v2dt

426  

The total amount of mass dm tht enters and exits during interval dt is

dm = massvolume

× area × lengthtime

× time

= ρ ⋅A1 ⋅v1 ⋅dt (enters) = ρ ⋅A2 ⋅v2 ⋅dt (exits)

During the flow from End 1 to End 2, the fluid may:- change kinetic energy (KE1 ⇒ KE2 )- change potential energy (U1 ⇒U2 )- have work W1 done on it by pressure P1

- have work W2 done on it by pressure P2

(Note: work done on the fluid is positive; work done by the fluid is negative.)

427  

By conservation of energy, we expectKE1 +U1 +W1 = KE2 +U2 +W2 'Eq 1'

KE1 = dm( ) v12

2

KE2 = dm( ) v22

2U1 = dm( )gh1

U2 = dm( )gh2

W1 = Force ⋅distance

Note pressure = forcearea

and velocity = distancetime

So W1 = P1A ⋅v1dt

428  

W1 = P1A ⋅v1dt

= P1A ⋅v1dt ⋅ρρ

=P1 dm( )

ρSimilarly,

W2 =P2 dm( )

ρSubstitute all of these into Eq 1:

dm( ) v12

2+ dm( )gh1 +

P1 dm( )ρ

= dm( ) v22

2+ dm( )gh2 +

P2 dm( )ρ

v12

2+ gh1 +

P1

ρ= v2

2

2+ gh2 +

P2

ρThat is, at any point along the path:v2

2+ gh + P

ρ= constant

429  

Rewriting: v2

2+ gh + P

ρ= constant

Symon separates this into 2 terms: " pρ+ u"

Symon calls this "−G"

Bernoulli's Equation: as velocity increases, pressure decreases.

430  

II. Bulk modulus and the speed of soundRecall Euler's equation of motion for a moving fluid [Slide 418]

∂v∂t

+ v ⋅∇v +

∇pρ

=fρ

Fext /Vol pressure densityLet the fluid be at rest and not accelerating:∇p0

ρ0

=f0ρ0

'Eq 0', Euler's Equation for a fluid at rest

Now induce some sound waves: perturb the fluid slightly so thatp0 ⇒ p0 + p ' ≡ pρ0 ⇒ ρ0 + ρ ' ≡ ρ Note these prime symbols do NOT indicate a derivative. These are just Symon's notation for "small perturbation" p ' << p0

ρ ' << ρ0

431  

For small velocity and small perturbations, we will assume

p '( )2 = 0

ρ '( )2 = 0p 'v = 0v ⋅∇v = 0Substitute these into Euler's Equation for moving fluids∂v∂t

+ v ⋅∇v + 1

ρ0 + ρ '( ) ⋅∇ p0 + p '( ) = 1

ρ0 + ρ '( ) ⋅ f

≈ 0, neglect{ }

Expand these as

1ρ0 + ρ '( ) =

1

ρ0 1+ ρ 'ρ0

⎛⎝⎜

⎞⎠⎟

= 1ρ0

1− ρ 'ρ0

+ ...⎛⎝⎜

⎞⎠⎟

⎧

⎨⎪⎪

⎩⎪⎪

⎫

⎬⎪⎪

⎭⎪⎪

432  

∂v∂t

+ 0 + 1ρ0

∇p0

− ρ 'ρ0

2 ∇p0 + 1ρ0

∇p '− ρ 'p0

2 ∇p ' ≈ f0ρ0

− ρ 'ρ0

2 f0

f0ρ0

f0

All that remains is:∂v∂t

≈ − 1ρ0

∇p ' 'Eq 1'

Now make the same substitutions and approximations in the Continuity Equation [Slide 416],∂ρ∂t

+∇⋅ ρv( ) = 0

∂ ρ0 + ρ '( )∂t

+∇⋅ ρ0 + ρ '( ) v = 0

433  

∂ρ0

∂t+ ∂ρ '

∂t+∇⋅ ρ0

v( ) +∇⋅ ρ ' v( ) = 0

at rest,ρ0 ≠ ρ0 t( )

⎧⎨⎩

⎫⎬⎭

≈ 0, neglect{ }

We are left with:∂ρ '∂t

+∇⋅ ρ0

v( ) = 0

∂ρ '∂t

+ ρ0

∇⋅ v + v ⋅

∇ρ0 = 0

at rest,ρ0 ≠ ρ0 xi( )

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪We are finally left with:∂ρ '∂t

= −ρ0

∇⋅ v 'Eq 2'

434  

Recall [Slide 398]:

strain = − 1Bdp = dV

V= − dρ

ρIn this section, dp is p ' (means "small pressure") andand dρ is ρ ' (means "small density") so

removing the unnecessary dVV

and recalling that B is bulk modulus,

strain = − 1Bp ' = − ρ '

ρ0

That is:

ρ ' = ρ0p 'B

Substitute this into Eq 2:

∂ ρ0p 'B

⎛⎝⎜

⎞⎠⎟

∂t= −ρ0

∇⋅ v

435  

∂p '∂t

= −B∇⋅ v 'Eq 3'

Rewrite Eq 1:∂v∂t

≈ − 1ρ0

∇p '

Equations 1 and 3 are the fundamental equations for sound waves,coupling pressure and velocity variations in space and time.

Next goal: Find an expression for the speed of sound. CombineEq 1 and Eq 3 to get an equation of the form

∇2 something( )− 1something else( )2

∂2 something( )∂t 2 = 0

Since this is the wave equation, the "something else" must be thespeed of sound.

436  

To uncouple the equations, take ∇⋅ Eq 1( ) :

∇⋅ dv

dt⎛⎝⎜

⎞⎠⎟ =∇⋅ − 1

ρ0

∇p '

⎛⎝⎜

⎞⎠⎟

ddt∇⋅ v( ) = − 1

ρ0

∇2p ' 'Eq 4'

Now take ddt

Eq 3( ) :

ddt

dp 'dt

⎛⎝⎜

⎞⎠⎟ =

ddt

−B∇⋅ v( )

d 2p 'dt 2 = −B d

dt∇⋅ v( )

Plug in Eq 4 here

d 2p 'dt 2 = −B − 1

ρ0

∇2p '⎛⎝⎜

⎞⎠⎟

d 2p 'dt 2 = B

ρ0

∇2p '

437  

d 2p 'dt 2 = B

ρ0

∇2p '

∇2p ' = 1Bρ0

d 2p 'dt 2

We see that

speed of sound( )2 = Bρ0

speed of sound = "c"= Bρ0

Vocabulary :

Mach number = vc

, where v is whatever velocity is characteristic of

the problem (for example, average fluid velocity). Typically forv < c, the fluid can be treated as incompressible.

438  

What to use for bulk modulus B?

Recall B ≡ − dpdV

V

Recall from Thermodynamics that for physical many-particle systems,most transformations are isothermal (constant temperature) oradiabatic (no heat transfer). These two have different relationships

between p and V , so their dpdV

's are different.

439  

Isothermal AdiabaticConstant temperature, so the No heat transferideal gas law pV = RT becomes: pV γ = constantpV = constant. where γ = cp / cV These are specific heats at constant volume or constant pressure. --To find B, differentiate both sides-- d(pV ) = d(constant) d(pV γ ) = d constant( )dpV + pdV = 0 dpV γ + pγV γ −1 = 0

− dpdV

V = p − dpdV

V = pγ

-- Conclude: --Bisothermal = p Badiabatic = pγ

440  

Empirical data on the speed of sound for different materialshas shown that in most cases Bsound

transmission= Badiabatic so if in doubt,

use B = pγ .

441  

III. Viscosity...the internal friction that inhibits the flow of a fluid.To characterize it, consider 2 parallel glass plates of area A separated by adistance d. A fluid is between them.

Bottom plate has velocity = 0.Top plate has velocity v = vx.To maintain this there must be aforce Fv applied across the top plate.That is, it must be shearing.

Fluid directly below top plate has vel ≈ v.Fluid directly above bottom plate has vel ≈ 0.We expect a smooth increase of fluid velocity magnitude from bottom to top.

442  

We expect the force Fx needed to produce this velocity pattern to beFx ∝ AFx ∝ v

Fx ∝1d

Put these together:

Fx =ηAvd

η, called the coefficient of viscosity, is a constant of proportionalitythat depends upon the fluid.

Note that "poise," the units of η are not MKS: kgm − s

= 10 poise.

Most materials have coefficients in the range of centipoise = 10−2 poise = 10−3 kg/m-s

443  

Approach the limit where d ⇒δ y and v⇒δvx

Then Fx =ηAδvxδ y

η = Fx / Aδvx /δ y

Typical values: water = 1 centipoise blood = 4 centipoise ketchup = 5 ×104 centipoise peanut butter = 1.5 ×105 centipoise road tar = 1010 centipoise

444  

I. Poiseuille's LawII. Sound waves

445  

I. Poiseuille's LawGoal: show that for viscous fluid, the throughputthrough a pipe is proportional to the (pipe radius)4 .Consider a pipe of radius a, length .Assume flow is laminar: in sheets, layers

Define Q = throughput of the pipe = discharge rate = volume expelledtime

Q = vA cross-sectional area of pipe fluid velocity, which depends on radiusSo to find throughput Q, we need to find v.The issue is: in a viscous fluid, v depends on location in the pipe.

2a

What is throughput?

446  

Consider the fluid in a layer from r = 0 to r = r '.The frictional force on its boundary is

Fvisc =ηAdvdr '

=η 2πr '( ) dvdr '

If the flow is at constant speed, acceleration = 0, soFtot = ma = 0.In that case Fvisc must be balanced by another force: this is associatedwith the pressure differential between the 2 ends of the pipe.

2r

dr

447  

The pressure difference along the pipe is Δp = FA=Fpressureπa2 .

Fvisc + Fpressure = 0 :

η2πr ' dvdr '

+ Δpπa2 = 0

dv = − 12η

Δprdr

To find the velocity of the fluid at radius r ', integrate from r ' to pipeboundary a, then plug in v a( ) = 0 :

dvr '

a

∫ = − 12η

Δp

r drr '

a

∫

v(a)− v(r ') = − 12η

Δpa2 − r '2

2

v(r ') = Δp4η

a2 − r '2( )

448  

Recall Q = vALet dQ be throughput from a ring at radius r of thickness dr.dQ = vdA

= Δp4η

a2 − r2( ) 2πrdr( )

Total throughput:

Q = dQ0

a

∫ = Δp4η

⋅2π a2 − r2( )r dr0

a

∫

= πa4Δp8η

Poiseuille's Law, valid only for laminar flow.

449  

Define: Reynolds number R, for fluid velocity v, density ρ, viscosity η,flowing in a pipe of diameter 2a:

R ≡ ρvη

2a( )

R is dimensionless.R ≤ 2000: laminar flowR > 2000: turbulent flow

450  

II. Sound wavesRecall that the 2 fundamental equations for sound are ∂v∂t

= − 1ρ0

∇p ' 'Eq 1'

small pressure fluctuations from ambient fluid density in equilibrium fluid velocity (not wave velocity)and∂p '∂t

= −B∇⋅ v 'Eq 2'

We eliminated v from them to get

∇2p '− 1Bρ0

⎛⎝⎜

⎞⎠⎟

∂2 p '∂t 2 = 0 'Eq 3'

451  

We could instead have eliminated p ' from them to get

∇2v − 1Bρ0

⎛⎝⎜

⎞⎠⎟

∂2v∂t 2 = 0 'Eq 4'

Goal now:1) show that the solutions are plane waves in both p ' and v2) show that v is parallel to the direction of wave propagation,that is: sound waves are longitudinal, not transverse

452  

Let us understand plane waves in 2 steps:first describe the plane, then describe the wave.Consider a plane moving in the n direction:

Identify 2 points on it,r = x, y, z( ) andr0 = x0, y0, z0( )They are joined by vector r − r0( )

x

z

n

y

r

r0

r − r0

453  

By definition of a plane (i.e., flat surface), for any r and r0 on the plane,r − r0( ) must be ⊥ to n.

So r − r0( ) ⋅ n = 0

x − x0( )nx + y − y0( )ny + z − z0( )nz = 0xnx + yny + znz = x0nx + y0ny + z0nz

This is a constant that defines the orientiation of the plane.So a plane satisfies the equation:n ⋅ r = constant, in space, at any time

That was the plane. Now the wave:Recall that any function that maintains a constant phase as it travels is,by definition, a wave. Let the phase be ξ ≡ n ⋅ r - ct

454  

One way to describe sound is as a pressure wave, which satisfiesEq 3 on Slide 450. The solution to that should look likep ' = f n ⋅ r − ct( ) f is an arbitrary shape n ⋅ r builds in the nature of a plane As with 1-dimensional waves on strings, the spatial (r ) and temporal (t) parts of the phase have to be coupled in order for it to propagate. We name the coupling parameter "c." We saw on Slide 372 that this coupling is the phase velocity, which is the velocity that a viewer traveling alongside the wave must have in order to see the phase remain constant.

455  

We next demonstrate directly that this p ' = f n ⋅ r − ct( ) = f ξ( ) satisfies

the wave equation (Eq 3 on Slide 450): ∇2p '− 1Bρ0

⎛⎝⎜

⎞⎠⎟

∂2 p '∂t 2 = 0.

Compute the LHS:If p ' = f ξ( ),∇p ' = df

dξ∇ξ = df

dξn

Then ∇2p ' = ∇⋅ dfdξ

n⎛⎝⎜

⎞⎠⎟= ddξ

dfdξ

n⎛⎝⎜

⎞⎠⎟⋅∇ξ

= d 2 fdξ 2 n +

dfdξ

dndξ

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥⋅∇ξ

0 n

= d2 fdξ 2 n ⋅ n =

d 2 fdξ 2

456  

Similarly compute the RHS,∂2 p '∂t 2 = c2 ∂2 f

∂ξ 2

We conclude that the function p ' = f n ⋅ r − ct( ) satisfies the

wave equation ∇2p '− 1Bρ0

⎛⎝⎜

⎞⎠⎟

∂2 p '∂t 2 = 0 for any f , if c = B

ρ0

.

457  

By a similar treatment we can solve Eq 4,

∇2v − 1Bρ0

⎛⎝⎜

⎞⎠⎟

∂2 v∂t 2 = 0

with any wave that has the form v =h n '⋅ r − ct( )

This is phase ξ'.Note that the direction of this plane's motion is n', not a priori the same as n.

Note this wave h is a vector function, conceptually like f but with 3 equations

implied: one each for vx , vy , and vz .

We have met Goal 1 on Slide 451, which was to show that the solutionsto the sound wave equation in 3-d are plane waves in p ' and v.Now we go after Goal 2, to show that these waves are longitudinal.

458  

Recall Eq 1 [Slide 450]∂v∂t

= − 1ρ0

∇p ' 'Eq 1'

Plug in p ' = f ξ( ), v =h(ξ '), and c = B

ρ0

:

∂v∂t

= ∂h

∂ξ '∂ξ '∂t

= ∂h

∂ξ '−c( ) = − ∂

h

∂ξ 'Bρ0

∇p ' = ∂ f∂ξ

∇ξ = ∂ f∂ξ

n

Eq 1 becomes:

− ∂h

∂ξ 'Bρ0

= − 1ρ0

∂ f∂ξ

n

∂h

∂ξ '= n

Bρ0

∂ f∂ξ

This must be true for all r , at all t

459  

∂h

∂ξ '= n

Bρ0

∂ f∂ξ

'Eq 5'

The RHS is a function of ξ and is constant, for constant ξ.So the LHS must be constant when ξ is constant.So ξ' must be a function of ξ, and the range of allowed functionsis limited by maintaining the plane wave.For simplicity let us choose ξ ' = ξ.Then n '⋅ r − ct = n ⋅ r − ctso n ' = n

Now rewrite Eq 5, setting ξ ' = ξ.

∂h

∂ξ= n

Bρ0

∂ f∂ξ

Integrate, set integration constant = 0 because p ' = 0 when v = 0(that is, when there is no wave).

460  

h = n

Bρ0

f

But h = v and f = p ':

v = nBρ0

p '

Conclusions:By construction, n is the direction of the wave's velocity, which is

cn = Bρ0

n.

Here we see that n is also the direction of the fluid particles' velocity v.This demonstrates Goal 2: sound waves are longitudinal;and the pressure variations p ' arise from the velocity variations.

461  

I. Wave vectorII. Sound wave powerIII. Other wave formsIV. Normal modes of a fluid in a boxV. Features of sound waves in cavities

462  

I. Wave vector

Define the wave vector k for 3-dimensional waves (like sound)

analogously to the wave number k (for 1-dimensional waves on string)k ≡ ω

cn

n is the direction of propagation of the wave front ω is the wave's angular frequency c here is the speed of the wave, not the speed of lightThen p ' = f n ⋅ r − ct( ) becomes

= f cωk ⋅ r − ct⎡

⎣⎢⎤⎦⎥

= f cωk ⋅ r −ωt( )⎡

⎣⎢⎤⎦⎥

But c and ω are fixed.

= fk ⋅ r −ωt( )

463  

II. Sound wave powerThe force that causes the pressure variations moves the particles back and forth.Power =

F ⋅ v

F v, so power is transmitted.Symon uses the confusing notation P to mean "Power per area"

So P = PowerArea

=F

Area⋅ v

= FArea

v (scalars)

= p ⋅v for pressure p

Average over one cycle of the wave:

P = p0 + p '( )v But p0 is equilibrium pressure, a constant, and v = 0

464  

P = p 'v

But v = p 'Bρ0

n

P =p '( )2

Bρ0

Bρ0 is a constant that can come outside the averaging operation

P = 1Bρ0

p '( )2

The average power per area of a sound wave is the energy per unit area persecond, travelling in direction n.

465  

III. Other wave formsWave shapes that satisfy the wave equation

∇2p '− 1c2

∂2 p '∂t 2 = 0

take their form from the shape of the initial disturbance. Thus they could be:

−spherical: p ' = 1rf r − ct( )

−cylindrical: p ' = 1rf r − ct( )

What motivates the radial factor in front of function f?

In a classical wave, energy = amplitude 2 .As spherical or cylindrical waves move outward, they cover anincreasingly larger area, so their amplitude per area must decrease in order toguarantee energy conservation.The area of a wave front in a spherical system grows as r2.The area of a wave front in a cylindrical system grows as r.

466  

IV. Normal modes of a fluid in a boxRelevance: acoustics and solids, but also electromagnetic waves in waveguidesand quantum mechanical particles' probability density in a potential well.Consider a fluid in a box of volume Lx ⋅Ly ⋅Lz .The fluid can oscillate but must maintain the boundary conditions.Oscillations described by the "sound wave"equation apply to any 3-dimensional wave:

∇2p '− 1c2

∂2 p '∂t 2 = 0

Our goal: solve this by separation of variables(in 2 steps), for various boundary conditions.

∇2p ' = 1c2

∂2 p '∂t 2 'Eq 1'

Step 1: separate spatial dependence from time dependence.Guess p ' =ψ x, y, z( )T t( )

LxLy

Lz

467  

p ' =ψ x, y, z( )T t( )∇2p ' = ∇2ψ ⋅T

∂2 p '∂t 2 =ψ ∂2T

∂t 2

⎫⎬⎪

⎭⎪ Substitute these into Eq 1.

∇2ψ ⋅T = 1c2ψ

∂2T∂t 2

÷ by ψ T:∇2ψψ

= 1c2T

∂2T∂t 2

Both sides must equal the same constant. Call it "− ω 2

c2 "

LHS RHS∇2ψψ

= −ω2

c2 1c2T

∂2T∂t 2 = −ω

2

c2

∇2ψ + ω 2

c2 ψ = 0 'Eq 2' ∂2T∂t 2 +ω 2T = 0

Solving Eq 2 requires Step 2. T = Acosωt + Bsinωt

468  

Step 2: separate all the spatial dimensionsGuess ψ =X x( ) ⋅Y y( ) ⋅Z z( )

So ∇2ψ = ∂2

∂x2 +∂2

∂y2 +∂2

∂z2

⎛⎝⎜

⎞⎠⎟XYZ

= ∂2 X∂x2

⎛⎝⎜

⎞⎠⎟YZ + X ∂2Y

∂y2

⎛⎝⎜

⎞⎠⎟Z + XY ∂2Z

∂z2

⎛⎝⎜

⎞⎠⎟

Eq 2 becomes:

∂2 X∂x2

⎛⎝⎜

⎞⎠⎟YZ + X ∂2Y

∂y2

⎛⎝⎜

⎞⎠⎟Z + XY ∂2Z

∂z2

⎛⎝⎜

⎞⎠⎟+ ω 2

c2 XYZ = 0

÷ by XYZ

1X

∂2 X∂x2

⎛⎝⎜

⎞⎠⎟ + 1Y

∂2Y∂y2

⎛⎝⎜

⎞⎠⎟ + 1Z

∂2Z∂z2

⎛⎝⎜

⎞⎠⎟ = −ω

2

c2

"− kx2 "

Term 1⎧⎨⎩

⎫⎬⎭

"− ky

2 "Term 2

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

"− kz2 "

Term 3⎧⎨⎩⎪

⎫⎬⎭⎪

To ensure that this is true for all x, y, z( ),each term on the LHS must be a constant

⎧⎨⎩

469  

So − kx2 − ky

2 − kz2 = −ω

2

c2

kx2 + ky

2 + kz2 = ω 2

c2 'Eq 6'

Now the partial derivatives become totals:

Term 1: 1X

∂2 X∂x2

⎛⎝⎜

⎞⎠⎟= −kx

2

∂2 X∂x2 + kx

2X = 0

X = Cx coskxx + Dx sin kxxSimilarly,Term 2 produces Y = Cy coskyy + Dy sin kyyTerm 3 produces Z = Cz coskzz + Dz sin kzz

470  

Put this together to getp ' =ψT = XYZT

We now consider the boundary conditions for a closed rectangular box:Particles' velocity cannot transport them beyond the walls:1) vx x = 0( ) = 0

2) vx x = Lx( ) = 0

3) vy y = 0( ) = 0

4) vy y = Ly( ) = 0

5) vz z = 0( ) = 0

6) vz z = Lz( ) = 0

471  

To prepare to apply boundary condition (1), recall one of the2 fundamental equations for sound waves [Eq 1 on Slide 432]:∂v∂t

= − 1ρ0

∇p '

In 1-dimension:∂vx∂t

= − 1ρ0

∂p '∂x

Substitute p ' = XYZT .

Note: ∂p '∂x

= kx ⋅ −Cx sin kxx + Dx coskxx[ ]⋅YZT∂vx∂t

= − kxρ0

⋅ −Cx sin kxx + Dx coskxx[ ]⋅YZ ⋅ Acosωt + Bsinωt[ ]Integrate, choose constant of integration = 0:

vx = − kxρ0

⋅ −Cx sin kxx + Dx coskxx[ ]⋅YZ ⋅ dt∫ Acosωt + Bsinωt[ ]

1ωAsinωt − 1

ωBcosωt

472  

vx = − kxYZρ0ω

−Cx sin kxx + Dx coskxx[ ] Asinωt − Bcosωt[ ]Now apply boundary condition (1). To achieve it, Dx = 0. Then X = Cx coskxx.Notice this means that function X is maximum at x=0: an "anti-node"To emphasize: pressure p' ∝ X is a maximum at x=0.

Thus, the statement " p x = 0( ) = max" is an alternative form ofBoundary Condition (1).

Similarly, we can replace Boundary Condition (2) [vx x = Lx( ) = 0]

with " p x = Lx( ) = max"

coskxLx = ±1 kxLx = π ( = 0,1,2,...)

kx =πLx

473  

Similarly for functions Y and Z we find:Dy = 0Dz = 0

ky =mπLy

(m = 0,1,2,...)

kz =nπLz

(n = 0,1,2,...)

Now recall the constraint, Eq 6 [Slide 469]:

kx2 + ky

2 + kz2 = ω 2

c2

Solve for allowed frequencies ω :

ω = c kx2 + ky

2 + kz2⎡⎣ ⎤⎦

12

= cπ 2

Lx2 + m

2

Ly2 + n

2

Lz2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

12

Call these the "ω mn "

Every combination of integers (,m,n) except 0,0,0( ) gives anallowed normal mode of this resonant closed rectangular cavity.

474  

Build the pressure wavefunction from the information we have up to this point:p ' = TXYZ

= Acosω mnt + Bsincosω mnt( )Cx cos π xLx

Cy cosmπ yLy

Cz cos nπ zLz

Rename ACxCyCz ⇒ ARename BCxCyCz ⇒ B

p ' = Acosω mnt + Bsincosω mnt( )cos π xLx

cosmπ yLy

cos nπ zLz

Similarly,

vx =π

Lxρ0ω mnAsinω mnt − Bcosω mnt[ ]sin π x

Lxcosmπ y

Lycos nπ z

Lz

vy =mπ

Lyρ0ω mnAsinω mnt − Bcosω mnt[ ]cos π x

Lxsin mπ y

Lycos nπ z

Lz

vz =nπ

Lzρ0ω mnAsinω mnt − Bcosω mnt[ ]cos π x

Lxcosmπ y

Lysin nπ z

LzThese are the four equations of motion for a fluid in a closed rectangular cavity.

475  

II. Features of sound waves in cavities1) If a cavity is long and narrow ("organ pipe"), then Lx >> Ly or Lz ,and the dominant tones will occur for m = 0, n = 0, and all .2) If the cavity has one end open, that boundary condition is harder todefine. Typically we choose p' at that end to be a node. (Lord Rayleigh

found that the node actually occurs 8d3π

beyond the end, where d

characterizes the diameter of a circular pipe.

Consider a rectangular pipe with one end open:

x = 0This faceis open.

x = LxThis faceis closed.

476  

Recall p ' = XYZTWe focus here on function X = Cx coskxx + Dx sin kxxBoundary condition (1): X(x = 0) = 0 : The open face is a pressure node. Requires Cx = 0Boundary condition (2): X(x = Lx ) = max : The closed face is a velocity antinode, as previously sin kxLx = ±1

kxLx =2+1( )π

2

kx =2+1( )π

2LxThe boundary conditions on functions Y and Z are as previously, so

ω = c kx2 + ky

2 + kz2⎡⎣ ⎤⎦

12 ⇒ω mn = cπ

2+1( )2

4Lx2 + m

2

Ly2 + n

2

Lz2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

12

These are the normal frequencies of an open-ended pipe.

477  

3) For sound propagating through a pipe that is open at both ends:Choose the coordinate system such that the open ends are along z.p ' = XYZT

= Acos π xLx

cosmπ yLy

cos kzz −ωt( )

These are standing waves This is the propagating term

It is still true that ω = c kx2 + ky

2 + kz2⎡⎣ ⎤⎦

12 , so

kz = ± ω 2

c2 − πLx

⎛⎝⎜

⎞⎠⎟

2

− mπLy

⎛

⎝⎜⎞

⎠⎟

2⎡

⎣⎢⎢

⎤

⎦⎥⎥

12

,m = 0,1,2,...

= ± ωc

1− πcωLx

⎛⎝⎜

⎞⎠⎟

2

− mπcωLy

⎛

⎝⎜⎞

⎠⎟

2⎡

⎣⎢⎢

⎤

⎦⎥⎥

12

478  

kz = ±ωc

1− πcωLx

⎛⎝⎜

⎞⎠⎟

2

− mπcωLy

⎛

⎝⎜⎞

⎠⎟

2⎡

⎣⎢⎢

⎤

⎦⎥⎥

12

So for this wave, the phase velocity cm = ωkz

= c 1− πcωLx

⎛⎝⎜

⎞⎠⎟

2

− mπcωLy

⎛

⎝⎜⎞

⎠⎟

2⎡

⎣⎢⎢

⎤

⎦⎥⎥

− 12

Conclude:i) cm > c, so guided waves travel faster than unguided ones

ii) cm is undefined when 1− πcωLx

⎛⎝⎜

⎞⎠⎟

2

− mπcωLy

⎛

⎝⎜⎞

⎠⎟

2

= 0

so there is a cutoff frequency:

ω cutoff =πcLx

⎛⎝⎜

⎞⎠⎟

2

+ mπcLy

⎛

⎝⎜⎞

⎠⎟

2⎡

⎣⎢⎢

⎤

⎦⎥⎥

12

Below this frequency, propagation is attentuated.