PHY2464 - The Physical Basis of Music
PHY
-2464
PHY-2464
Physical Basis of Music
Presentation
Presentation 13
13
Basics
Basics of
of Sound
Sound in
in Pipes
Pipes
Adapted
Adapted from
from Sam
Sam Matteson’s
Matteson’s
Unit
3
Session
Unit 3 Session 26
26
Sam
Sam Trickey
Trickey
Feb.
Feb. 24,
24, 2005
2005
PHYPHY-2464
Pres. 13 Basics of Sound in Pipes
The Issue • A string with fixed ends has harmonic vibrations
given by fn = [n/(2L)] × √(T/ µ), n= 1, 2, 3, …
• These are standing waves
• Organ flue pipes, flutes, pan-pipes, and song
whistles are familiar examples of sound in pipes
• In each of these pipes there is a clear fundamental
and a set of apparently harmonic overtones.
How are standing waves formed and
maintained in pipes?
PHY2464 - The Physical Basis of Music
PHYPHY-2464
Pres. 13 Basics of Sound in Pipes
Standing Waves:
In strings – reflected waves combine to produce
standing waves through
•
Cancellation - destructive interference at nodes
and
•
Addition - constructive interference at
antinodes.
PHYPHY-2464
Pres. 13 Basics of Sound in Pipes
Reminder - The distance between neighboring nodes &
antinodes is ¼ λ. [ dN-A = ¼ λ ]
Second Harmonic
f2 = vstring / λ2 = vstring / L
L = λ2
Node
Node
Antinode
λ2 /4
λ2 /4
Node
Antinode
λ2 /4
λ2 /4
PHY2464 - The Physical Basis of Music
PHYPHY-2464
Pres. 13 Basics of Sound in Pipes
Standing Waves in a Cylindrical Pipe:
•
A Closed or Stopped Pipe –
• the pressure wave reflects without inversion
at the stopped end
but
• the displacement wave inverts upon reflection
PHYPHY-2464
Pres. 13 Basics of Sound in Pipes
Standing Waves in a Cylindrical Pipe:
•Thus, a pressure anti-node will occur at the wall;
but, on the other hand,
• a displacement node will occur at the same place.
Nothing exotic here • Displacement node just means that air molecules
right at the wall MUST be still
• Pressure anti-node means that the gas is most
squeezed right at the wall
PHY2464 - The Physical Basis of Music
PHYPHY-2464
Pres. 13 Basics of Sound in Pipes
Reflection of a Sound Wave in a Stopped Pipe:
A pressure anti-node (max) appears at a wall.
Pressure Wave
+
+
Pressure Node
-
λ/4
PHYPHY-2464
Pressure AntiAnti-node
λ/4
Pres. 13 Basics of Sound in Pipes
Reflection of a Sound Wave in a Stopped Pipe:
A displacement node forms at a wall.
Displacement AntiAnti-node
Displacement Wave
Displacement Node
λ/4
λ/4
PHY2464 - The Physical Basis of Music
PHYPHY-2464
Pres. 13 Basics of Sound in Pipes
Comparison of Pressure and Displacement Standing Wave in
a Double Stopped Pipe
Pressure Wave
λ/4
λ/4
Displacement Wave
PHYPHY-2464
Pres. 13 Basics of Sound in Pipes
Standing waves in a pipe are an example
of the property of
Interference
of sound waves.
PHY2464 - The Physical Basis of Music
PHYPHY-2464
Pres. 13 Basics of Sound in Pipes
Recall:
•
fλ=v
•
Thus, f = v/ λ; L = Nna λ/4
•
Nna = number of node→anti-node loops
•
so f = Nna v/4L
•
Nna = 2 n ;
fn = 2 n v/4L
fn = n v/2L
In double-stopped pipe.
PHYPHY-2464
Pres. 13 Basics of Sound in Pipes
Why does a sound wave reflect?
Because of an abrupt change in a
property of the medium.
PHY2464 - The Physical Basis of Music
PHYPHY-2464
Pres. 13 Basics of Sound in Pipes
Acoustic Impedance:
Z = p/U
Acoustic Impedance is the ratio of the pressure p of
a sound wave to the flow U that results.
U = volume/time crossing the area S
U
Area =S
PHYPres. 13 Basics of Sound in Pipes
PHY-2464
Acoustic Impedance:
For a plane wave in air at atmospheric pressure and room
temperature in a tube of cross section S (m2) the
acoustic impedance is:
Z = ρv/S ≈ 415/ S
The units are Acoustic Ohms, symbol ΩA
MAIN POINT
Z ∝ 1/ S
PHY2464 - The Physical Basis of Music
PHYPHY-2464
Pres. 13 Basics of Sound in Pipes
Acoustic Impedance and Reflection:
The pressure from reflection in medium “1”
by medium “2” is
pout = R x pin
R = (Z2 – Z1 )/ (Z2 + Z1)
R = reflection coefficient
At a rigid wall U = 0 (no displacement) irrespective
of the pressure and, thus,
Z2 → ∞
R≈1
PHYPHY-2464
Pres. 13 Basics of Sound in Pipes
But what about an open pipe?
Does the sound reflect?
Yes! Z = p/U
p drops suddenly near the end of the pipe
to essential pAtmosphere (S → ∞)
Thus, Z ≈ 0, R ≈ -1
PHY2464 - The Physical Basis of Music
PHYPHY-2464
Pres. 13 Basics of Sound in Pipes
Comparison of Pressure Standing Wave in a Stopped (One
End) versus an Open Pipe
λ/4
λ/4
3λ/4
λ/4
4λ/4
L
PHYPHY-2464
Pres. 13 Basics of Sound in Pipes
For Stopped Pipe:
Nna = odd number = 2n-1, n=1,2,3,4 …
λn = 4L/ Nna = 4L / (2n-1)
fstopped = f2n2n-1 = v/ λn = (2n-1) v/ 4L
Only odd harmonics of fstopped 1 = v/4L.
PHY2464 - The Physical Basis of Music
PHYPHY-2464
Pres. 13 Basics of Sound in Pipes
For Open Pipe:
Nna = even number = 2n, n=1,2,3,4…
λn = 4L / Nna = 4L/(2n) = 2L/ n
fopen = fn = v/ λn = n x v/2L
All harmonics of fopen 1 = v/2L [= 2 fstopped 1 ]
PHYPHY-2464
Pres. 13 Basics of Sound in Pipes
End Correction for Open Pipe
δ ≈ 0.6 a for a << λ ; δ ≈ 0 a for a > λ / 4
a Radius
L+ δ
δ
PHY2464 - The Physical Basis of Music
PHYPHY-2464
Pres. 13 Basics of Sound in Pipes
Summary:
•
Stopped Pipes:
fstopped = f2n(2n-1) v/ 4L
2n-1 = v/ λn = (2n-
Only odd harmonics of fstopped 1 = v/4L.
•
Open Pipes: fopen = fn = v/ λn = n x v/2L
All harmonics of fopen 1 = v/2L [= 2 fstopped 1 ]
•
Acoustic impedance goes as 1/S
•
End corrections: add ≈ 0.6 pipe radius for each open
end to get effective length