Puzzler

1

An airplane flies in a straight line from airport A to airport B, then back again in a straight line from B to A. It travels with a constant engine speed and there is no wind. Will its travel time for the same round trip be greater, less, or the same if throughout both flights wind blows from A to B. Assume that the engine speed is always the same.

Solution: It takes longer with the wind

2

Example: A à B = 150 miles, vplane= 50 mph, vwind = 25 mph

No windA à B : 150/50 = 3 hrsB à A : 150/50 = 3 hrsTotal : 6 hrs

With wind @ 25 mphA à B : 150/75 = 2 hrsB à A : 150/25 = 6 hrsTotal : 8 hrs

Introduction to �Audio and Music Engineering

Lecture 9

•  Acoustic resonances in tubes•  Waveform and timbre•  Acoustic resonances in 3-d•  Room modes•  Reverberation

3

Acoustic ModesAcoustic waves obey the same wave equation as a string – just change the variables.

4

d 2p(x ,t )dt 2

= c 2 d 2p(x ,t )dx 2

1-d

x

p = pressure

Boundary conditions: open end à p = 0closed end à p = maximum

Solutions of 1-d Acoustic wave equation

5

Both sine and cosine satisfy the wave equation.

How do we know which solution to choose?

p(x ,t )= cos(nω0t ) sin(nπ xL)*or*cos(nπ x

L)

⎡

⎣⎢

⎤

⎦⎥

Oscillation in time

Choose the one that satisfies the boundary conditions.

One-half of a sine wave

Open Open

L

n = 1

sin(πx/L)

f1 =c2L

Same as fixed-fixed string!

flute

Pressure

Pressure à 0 at open end

Higher modes

6

n = 2

f2 = 2c2L

= cL= 2 f1

Mode frequencies of open-open tube are the same as those for a fixed-fixed string.

L

n = 3

f3 = 3c2L

= 32cL= 3 f1

sin(2πx/L)

1 cycle of sine wave

sin(3πx/L)

3/2 cycles of sine wave

Modes of open-open tube are multiples of one-half of a sine wave.

Closed-Open �Boundary Condition

7

Closed OpenL

Clarinet

Closedp = max

Openp = 0

n = 1 1/4 cycle of cosine

L = λ4

λ1 = 4L

f λ = c !!!so!!f1 =cλ1

= c4L

n = 3 3/4 cycle of cosine

L = 34λ

λ3 =4L3

f3 =cλ3

= 3 c4L

= 3 f1

n = 5 5/4 cycle of cosine

L = 54λ

λ5 =4L5

f5 =cλ5

= 5 c4L

= 5 f1

Summary

8

open-open

L

closed-open

f1 =c2L fn = n

c2L

f1 =c4L fn = (2n −1)

c4L

n = 1,2,3 …C4L ≈ 66 cm

L ≈ 60 cm

D3“concert”

n = 1,2,3 …

Only odd harmonics

All harmonics

closed-closed

261.6 Hz

146.8 Hz

Closed Closed

Waveform and timbre

9

ClarinetA few cycles

Note that the odd numbered harmonics have the greatest amplitudes. This is because the clarinet bore supports the odd numbered harmonics of the fundamental mode.

Spectrum

1 3 5 7

2 4 6

D3 “concert”

Flute Timbre

10

A few cyclesFlute C4

1 2 3 4 5 6

The Open-Open boundary condition of a flute supports all harmonics of the fundamental mode.

C4

Acoustic resonances in higher dimensions

11

x

y

z

Lx

Ly

Lz ∂2p∂t 2

= c 2 ∂2p∂x 2

+ ∂2p∂y 2

+ ∂2p∂z 2

⎛

⎝⎜⎞

⎠⎟

p(x ,y ,z ,t )= X (x )Y (y )Z (z )⋅cos(ωt )

Acoustic wave equation in 3-d:

Solutions:

Separable into functions of x, y, z

p(x ,y ,z ,t )= Almn cos(lkx x )⋅cos(mky y )⋅cos(nkz z )⋅cos(ωt )

p = max at each wall

kx = π/Lx , l = 0,1,2 …

ky = π/Ly , m = 0,1,2 … kz = π/Lz , n = 0,1,2 … k 2 = kx

2 + ky2 + kz

2

ω lmn = clπLx

⎛

⎝⎜

⎞

⎠⎟

2

+ mπLy

⎛

⎝⎜

⎞

⎠⎟

2

+ nπLz

⎛

⎝⎜

⎞

⎠⎟

2

ω010 = cπLy

ω111 = cπLx

⎛

⎝⎜

⎞

⎠⎟

2

+ πLy

⎛

⎝⎜

⎞

⎠⎟

2

+ πLz

⎛

⎝⎜

⎞

⎠⎟

2

for example:

12

Keeping''f ,λ 'and'ω ,k ''all'straight

ω = 2π f angular frequency(radians per second)

f = 2πω

frequency (cycles/sec)

k = 2πλ

wavenumber (radians/meter)

λ = 2πk

wavelength (meters/wave)

λ f = c2πk

ω2π

= ωk= c

n = 1

sin(πx/L)

à

λ = 2πk

= 2L→ L = πk→ k = π

L

ω = ck = c πL

λ = 2L λ f = c → f = c2L

remember!

Resonances in rooms

13

flmn =ω lmn

2π= c2π

lπLx

⎛

⎝⎜

⎞

⎠⎟

2

+ mπLy

⎛

⎝⎜

⎞

⎠⎟

2

+ nπLz

⎛

⎝⎜

⎞

⎠⎟

2

= c2

lLx

⎛

⎝⎜

⎞

⎠⎟

2

+ mLy

⎛

⎝⎜

⎞

⎠⎟

2

+ nLz

⎛

⎝⎜

⎞

⎠⎟

2

Room Dimensions: 3 meters high 4 meters wide 5 meters long

C = 340 m/sec

Lx 4mLy 5mLz 3m

l m n f(l,m,n)1 0 0 42.50Hertz0 1 0 34.000 0 1 56.671 1 0 54.430 1 1 66.081 0 1 70.831 1 1 78.572 0 0 85.000 2 0 68.000 0 2 113.332 1 0 91.552 0 1 102.162 1 1 107.672 2 2 157.14

Direct versus reverberant sound

14

Radius of reverberation

Reverberation time

15

1:0.2s,smallcontrolroom2:0.5s,classroom3:2s,mechanicalworkshop4:5s,church5:10s,largesportsvenue

T60

16

T60 measured at 1 kHz for room with and without carpeting

T60 ≈ 3.5 sec

T60 ≈ 2.0 sec

Sabine-Franklin-Jaeger �Theory of Room Acoustics

•  T60 = time required for sound to decay 60 dB•  V = Volume of room•  As = equivalent area of an open window resulting from

all sound absorption in room

17

T60 = 6log104VcAs

⎛

⎝⎜

⎞

⎠⎟

Volumeofairlostpersecond=cAs

AsV

•  T60 is essentially the time it takes to empty all of the air in the room out of the window 4x

Practical Sound Treatment

18

AbsorpNonCoefficientsFrequency(Hz) 2”StudiofoamPyramid 4”StudiofoamPyramid

125 0.13 0.27250 0.18 0.50500 0.57 1.011000 0.96 1.132000 1.03 1.114000 0.98 1.12

T60 =0.049VA

A = S1α1 + S2α2 + ...

α

Sabine Equation

V=Volumeofroom(V3)A=totalabsorpNoninroom(sabins)

secs