Physics 1251 The Science and Technology of Musical Sound

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Physics 1251 The Science and Technology of Musical Sound. Unit 3 Session 35 Pipes, Voice and Percussion Review . Physics 1251 Unit 3 Session 35 Pipes, Voice and Percussion. Foolscap Quiz:

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### Physics 1251The Science and Technology of Musical Sound

Unit 3

Session 35

Pipes, Voice and Percussion

Review

Physics 1251 Unit 3 Session 35 Pipes, Voice and Percussion

Foolscap Quiz:

What are the first three harmonics that “speak” in the harmonic series of a Tympani tuned to A3 (110 Hz)?

Answer: approximately 220 Hz, 330 Hz, 440 Hz. The fundamental 110 Hz is missing.

Physics 1251 Unit 3 Session 35 Pipes, Voice and Percussion

1′ Lecture:

• The pitch of pipes and the voice is determined by a harmonic series.
• Percussion most often have no pitch because they lack a harmonic series.
• Vibration recipes of pipes, the voice and percussion arise from the modes of vibration the air column, of the vocal folds and of the membrane, plate, or block, respectively.
Physics 1251 Unit 3 Session 35 Pipes, Voice and Percussion

The key to understanding the

Sound of Music!

6

2

4

1

3

7

5

Harmonic Series produce a sense of musical intonation.

Physics 1251 Unit 3 Session 35 Pipes, Voice and Percussion

80/20The timbre of an instrument’s sounds depends on its vibration recipe.

fn = n f1

Pitched

Amplitude

f1

2f1

3f1

4f1

fn m = xn m f1

Unpitched

Amplitude

f01

Frequency

Physics 1251 Unit 3 Session 26 Sound in Pipes

1′ Lecture:

• Sound in pipes can produce standing waves in the air column.
• Standing waves in air columns produce pressure nodes and displacement nodes (and antinodes) at different places.
• A change in the acoustic impedance of the air column produces a reflection.
• Organ pipes and the flute are examples of open or unstopped pipes.
Physics 1251 Unit 3 Session 26 Sound in Pipes

Standing Waves in a Cylindrical Pipe:

• A Closed or Stopped Pipe – the pressure wave reflects without inversion, but the displacement wave inverts upon reflection.
• Thus, a pressureanti-node will occur at the wall; but, on the other hand, a displacementnode will occur at the same place.
Physics 1251 Unit 3 Session 26 Sound in Pipes

Comparison of Pressure and Displacement Standing Wave in a Double Stopped Pipe

λ/4

λ/4

Pressure Wave

Displacement Wave

Physics 1251 Unit 3 Session 26 Sound in Pipes

80/20Acoustic Impedance:

Z = p/U

Acoustic Impedance is the ratio of the pressure p of a sound wave to the flow U (= u S) that results.

For a plane wave in a tube of cross section S (m2) in air the acoustic impedance is:

Z = ρv/S = 415/ S rayl

Physics 1251 Unit 3 Session 26 Sound in Pipes

80/20For Stopped Pipe:

Nna = odd number = 2n-1, n=1,2,3,4 …

λn = 4L/ Nna = 4L / (2n-1)

fstopped = f2n-1 = v/ λ2n-1 = (2n-1) v/ 4L

80/20Only odd harmonics of fstopped 1 = v/4L.

Physics 1251 Unit 3 Session 26 Sound in Pipes

80/20For Open Pipe:

Nna = even number = 2n, n=1,2,3,4…

λn = 4L / Nna = 4L/(2n) = 2L/ n

fopen = fn = v/ λn = n ‧ v/2L

80/20All harmonics of fopen1 = v/2L [= 2 fstopped 1 ]

Physics 1251 Unit 3 Session 26 Sound in Pipes

End Correction for Open Pipe without Flange

δ ≈ 0.6 a for a ≪λ ; δ ≈ 0 a for a > λ / 4

L + δ

δ

Physics 1251 Unit 3 Session 26 Sound in Pipes

Transverse Flute

80/20The transverse flute is a cylindrical open pipe.

Mouthpiece is open

Physics 1251 Unit 3 Session 26 Sound in Pipes

Summary:

• fopen = fn = n ‧ v/2L
• fstopped = f2n-1 = (2n-1) v/ 4L
• Stopped and open cylindrical pipes have different timbres.
• Impedance: Z = p/U
• An abrupt change in Z is responsible for the reflections that lead to standing waves in pipes.
Physics 1251 Unit 3 Session 27 Flutes et cetera

1′ Lecture:

• Flutes and flue pipes are driven by fluid flow instabilities at their mouth.
• Standing waves in open air columns of flutes determine the pitch.
• Open holes in the flute tube change the effective length of the air column.

f1 f2 f3 f4

♩ ♪ ♫

fn

~

~

Physics 1251 Unit 3 Session 27 Flutes et cetera

The Flute

• The transverse flute is acoustically driven by the fluid flow instabilities whose frequency is controlled by the feedback of the resonances of the pipe.

Standing wave frequencies

Flow Instability

Feedback

Physics 1251 Unit 3 Session 27 Flutes et cetera

Transverse Flute

80/20The flute is driven by air flow against the edge of the embrochure.

Air flow

Embrochure

Physics 1251 Unit 3 Session 27 Flutes et cetera

Edge Tone

80/20An air stream striking against an edge produces a fluctuating instability in flow.

Air Stream

Edge

The flow alternates sides.

Physics 1251 Unit 3 Session 27 Flutes et cetera

Why does the stream oscillate?

• When the stream bends to the left, the stream moves faster on the right side.
• Bernoulli’s Principle tells us that the faster the flow, the lower the pressure.
• Therefore, the left-flowing stream will bend back to the right
Physics 1251 Unit 3 Session 27 Flutes et cetera

Bernoulli Effect

• 80/20The pressure in a fluid decreases as the velocity increases.
• Hold the foolscap by the edge and blow across the top. What do you observe?
Physics 1251 Unit 3 Session 27 Flutes et cetera

fedge = 0.4 vjet / 2 b = 0.2 vjet /b

Edge Tone

u = 0.4 vjet

b

b

vjet

u

Physics 1251 Unit 3 Session 27 Flutes et cetera

Feedback from the acoustic standing wave locks the frequency of the oscillation if the edge tone is near the fundamental frequency.

fedge = 0.2 vjet /b

fn = n v/ 2L; fedge≈ fn

Displacement wave

Physics 1251 Unit 3 Session 27 Flutes et cetera

The Problem with Flutes:

• Only about 1% of the energy of the air stream produces sound.
• Playing louder means more air flow.
• More air flow means higher jet velocity
• Edge tone goes sharp
• Worse in Recorder than in Transverse Flute
• Player must “lip” tone into tune

Leff

Physics 1251 Unit 3 Session 27 Flutes et cetera

How does one “play” the notes?

• By effectively changing the length of the air column.
• Opening holes introduces reflections that change the standing wave length.

Displacement wave

f n′ = n ‧ v/2Leff

♯♩

♯♩

Physics 1251 Unit 3 Session 27 Flutes et cetera

Cross Fingering

80/20The position and size of the open holes modify the effective length of the air column and consequently the pitch.

Physics 1251 Unit 3 Session 27 Flutes et cetera

Why does the size of the hole matter?

Z =p/U

Impedance = pressure/flow

Displacement →Flow U:

Z

Z ′

Physics 1251 Unit 3 Session 27 Flutes et cetera

Summary:

• Flutes and flue pipes are open columns of air, with fn = n v/2L, n = 1,2,3,4….
• Flue pipes are excited by flow instabilities of the air steam in the embrochure or fipple.
• The frequency range is selected by the edge tone.
• The pitch is determined by the effective length of the pipe.
• Open holes determine the effective length of the pipe.
Physics 1251 Unit 3 Session 28 Clarinets et cetera

1′ Lecture:

• Reed instruments are stopped pipes.
• The clarinet has a cylindrical bore and is a stopped pipe; consequently, only odd harmonics are significant.
• Conical pipes exhibit all harmonics, even in stopped pipes.
• The saxophone, oboe and bassoon‒all have conical bores.
Physics 1251 Unit 3 Session 28 Clarinets et cetera

Comparison of Flute and Clarinet Registers

• Overblown flutes jump from a fundamental f1= v/2L to an octave f2 = 2f1 in the second register; an octave (2x) and a perfect fifth (3/2) f3 = 3 f1 =3 (v/2L) in the third register.
• Overblown clarinets jump from a fundamental f1 = v/4L to an octave (2x) and a fifth (3/2)‒“the twelfth‒” in the second register, because only odd harmonics produce standing waves in a stopped cylindrical pipe.

f1 f2 f3 f4

♩ ♪ ♫

fn

~

~

Physics 1251 Unit 3 Session 28 Clarinets et cetera

Reed Instruments

• The reed produces a pulsation in the pressure admitted to the pipe; the pressure standing wave feeds back to control the oscillations of the reed.

Standing wave frequencies

Reed pulsations

Feedback

Air flow

Tonguing

Physics 1251 Unit 3 Session 28 Clarinets et cetera

The Single Reed

80/20The reed opens and closes like a valve, pressurizing the pipe when open, closing due to the Bernoulli effect when the air flows.

Reed

Physics 1251 Unit 3 Session 28 Clarinets et cetera

Hard and Soft Reeds

80/20A hard reed is one for which the frequency is determined by its stiffness and dimensions.

A soft reed flexes easily and vibrates at the frequency of external pressure fluctuations.

Soft Reeds

Hard Reed: Harmonica

Clarinet

Oboe

Air flow

Physics 1251 Unit 3 Session 28 Clarinets et cetera

The Double Reed

80/20The reed opens and closes like a valve, pressurizing the pipe when open, closing due to the Bernoulli effect when the air flows.

Pressure Pulses

Reed Tip

Physics 1251 Unit 3 Session 28 Clarinets et cetera

Bernoulli Effect

• 80/20The pressure in a fluid decreases as the velocity increases.
• Thus, as the air flows past the reed, it is forced closed.

Bernoulli Effect

Pressure inverts

Physics 1251 Unit 3 Session 28 Clarinets et cetera

80/20Feedback from the pressure standing wave locks the frequency of the oscillation of the reed.

f2n-1 = (2n-1) v/ 4L′

Pressure wave

L′ = L + 0.3 d

0.3 d

Physics 1251 Unit 3 Session 28 Clarinets et cetera

80/20For a stopped conical pipe:

fn≈ nv / 2(L′ + c)

if c << λ

L′ = L + 0.3 d

L′

d

c

0.3 d

Physics 1251 Unit 3 Session 28 Clarinets et cetera

Summary:

• Reed Instruments are stopped pipes.
• L′ = L + 0.3 d
• f2n-1 = (2n-1) v/4L′ for stopped cylindrical pipes such as the clarinet.
• fn = n v/ 2(L′+c) for stopped conical pipes such as the saxophone, oboe, bassoon, etc.
• Soft reeds act as pressure valves that respond to the frequency fed back from the standing waves of the pipe.

f1 f2 f3 f4

♩ ♪ ♫

fn

~

~

Physics 1251 Unit 3 Session 29 Brass Instruments

Brass Instruments

• The lips produce a pulsation in the pressure admitted to the pipe; the pressure standing wave feeds back to control the oscillations of the plays lips.

Lip-valve pulsations

Standing wave frequencies

Feedback

Physics 1251 Unit 3 Session 29 Brass Instruments

The Lip Valve

80/20Brass instruments are played by the player’s lips.

Breath pressure, muscle tension and pressure feedback from the pipe determine the frequency of the opening and closing of the lips.

Louis Armstrong – trumpet (1901-1971)

Physics 1251 Unit 3 Session 29 Brass Instruments

Lip Valve

• The lips of the player act as a valve that admits pressure pulses into the pipe.
• The frequency is determined by the breath air pressure, the lip tension and the resonances of the pipe.
Physics 1251 Unit 3 Session 29 Brass Instruments

80/20Brass Instruments

are stopped pipes.

• The player’s lips produce a displacement node (pressure antinode) at the mouthpiece.
• A displacement anti-node (pressure node) exists at the bell.

Winton Marsalis Trumpet

Physics 1251 Unit 3 Session 29 Brass Instruments

The Mouthpiece

Cup Volume

80/20The Cup Volume and the diameter of the constriction leading to the back bore are more important than the shape of the cavity.

Diameter

Physics 1251 Unit 3 Session 29 Brass Instruments

80/20The Brass mouthpiece lowers the high frequency resonances.

Resonance for Combination Pipes

f

Cone with mouthpiece

Cone

Physics 1251 Unit 3 Session 29 Brass Instruments

The pitch is changed by pipe length and excitation of resonances.

By means of slides and valves the length is changed.

Pedal Tone

Physics 1251 Unit 3 Session 29 Brass Instruments

Resonance for Combination Pipes

f

Cone/ Cylinder %

0/100%

25/75%

40/60%

50/50%

20/80%

100/0%

Physics 1251 Unit 3 Session 29 Brass Instruments

Resonances for Combination Bores in Brass Instruments

80/20A 50% cylindrical ‒ 50% conical bore has a nearly harmonic series.

x

Physics 1251 Unit 3 Session 29 Brass Instruments

Exponential Horn

The Bell

a = ao exp(m x)+ b

80/20m is called the “flare constant.”

Larger m means more rapid flare.

a

x

Physics 1251 Unit 3 Session 29 Brass Instruments

Bessel Horns

The Bell

a = ao e-(εx) +b

80/20Called “Bessel Horns” because the standing wave follows a Bessel Function.

Physics 1251 Unit 3 Session 29 Brass Instruments

Summary:

• Brass Instruments are stopped pipes.
• The pipe bore is designed to give resonances that are harmonic.
• The pedal tone (the lowest note) is not harmonic.
• The player’s lips are a soft reed.
• The pitch is changed by changing the length and exciting resonances.

1′ Lecture:

• The pitch of a wind instrument is determined by the length and shape of its air column.
• The effective length of the air column is controlled with holes, valves and slides.
• Feedback from the resonances of the pipe select the frequency of oscillation of the jet, reed or lip-valve.
• The excitation, transmission and emittance of the sound in the horn determine the timbre of the instrument.

Transverse Flute

80/20The flute is driven by air flow against the edge of the embrochure hole.

80/20A pressure node exists at the open hole.

Air flow

Embrochure

Air flow

Tonguing

The Single Reed

80/20The reed opens and closes like a valve, pressurizing the pipe when open, closing due to the Bernoulli effect when the air flows.

80/20A pressure anti-node exists at the reed.

Reed

Air flow

The Double Reed

80/20The reed opens and closes like a valve, pressurizing the pipe when open, closing due to the Bernoulli effect when the air flows.

80/20A pressure anti-node exists at the reed.

Pressure Pulses

Reed Tip

The Lip Valve

80/20Brass instruments are played by the player’s lips that form a lip valve.

80/20A pressure anti-node exists at the player’s lips.

Louis Armstrong – trumpet (1901-1971)

6fO

5f1

6f1

5f1

5fO

5f1

4f1

4fO

4f1

3f1

3fO

3f1

3f1

2f1

2fO

2f1

fO

f1

f1

f1

f1

Comparison of Wind Instruments

f

Pedal Tone

fo =

(1+ξ)v/4(L+c)

L

f1 = v/2L

f1 = v/4L

f1 = v/2(L+c)

Other Woodwinds

Flute

Clarinet

Brass

c

Open Cylinder Np – Np

fn = nf1

f1 = v/2L

Stopped Cylinder Ap – Np f2n-1 = (2n-1)f1 f1= v/4L

Stopped Cone Ap – Np fn = nf1 f1= v/2(L+c)

Stopped Combination Ap – Np fn = nf0f0= (1+ξ)v/4(L+c)

Comparison of Wind Instruments (cont’d.)

fo =

(1+ξ)v/4(L+c)

L

f1 = v/2L

f1 = v/4L

f1 = v/2(L+c)

Other Woodwinds

Flute

Clarinet

Brass

c

80/20In the flute, feedback from the acoustic standing wave locks the frequency of the oscillation if the edge tone is near the fundamental frequency.

Displacement wave

fedge = 0.2 vjet /b

fn = n v/ 2L; fedge≈ fn

Pressure inverts

80/20IIn reed instruments, feedback from the pressure standing wave locks the frequency of the oscillation of the reed.

f2n-1 = (2n-1) v/ 4L′

Pressure wave

L′ = L + 0.3 d

0.3 d

Feedback from Resonaces

• 80/20The pitch of a wind instrument is determined by the influence on the jet/reed/lip-valve of feedback from the pressure/displacement standing waves in the pipe.

f1 f2 f3 f4

♩ ♪ ♫

fn

~

~

Wind Instruments

• A jet produces a fluctuating air flow, while a reed or the lips produce pressure pulsations, the frequencies of which are controlled by feedback from standing waves in the horn.

Standing waves in horn

Flow fluctuations or Pressure pulsations

Feedback

Effect of Excitation

• The mode of excitation of the horn significantly influences the harmonic recipe of the air column.
• The harmonics will only be as strong as the excitation of the jet/reed/lip-valve.

The Mouthpiece

Cup Volume

80/20The Cup Volume and the diameter of the constriction leading to the back bore are the most important factors in determining the frequency spectrum of the mouthpiece.

Diameter

Driven Pipe Vibration Recipe

Pipe Spectrum

A

Mouthpiece Spectrum

A

Driven Pipe Spectrum

A

Frequency

Effect of the Pipe

• A pipe is three dimensional; therefore, 3-D modes of oscillation are possible in the pipe.
• 80/20Only those modes with frequency above a Cut-off Frequency fc will exist in the pipe.

f > fc for propagation.

Modes of Vibration of a Column of Air

(0,0)

D

(1,0)

(2,0)

Cut Off Frequency fc = qn m v/D; for f < fc no propagation q00 = 0; q10 = 0.59; q20 = 0.97

Effect of Modes on Spectrum

• More modes implies more intensity.
• Most influential in high f harmonics.
• Shape and relative diameter of pipe influence modes.
• Thus, a square organ pipe has a different timbre than does a round organ pipe because of the modes.

Reflections from the array of holes in a woodwind affect the relative strength of the high frequency harmonics in the pipe.

Displacement wave

Reflections from holes (closed and open)

Effect of Holes on Transmission

• Larger holes have greater effect.
• A “high pass filter:” Low frequencies tend to be reflected more and high frequencies transmitted more.
• The holes make a “brighter” sounding instrument.

Reflections from joints and imperfections affect the relative strength of the high frequency harmonics in the pipe.

Reflections

f1 f2 f3 f4

♩ ♪ ♫

fn

~

~

Filtering of Wind Instrument Sound

• The vagaries of transmission of the various frequency components in the pipe produce a filtering effect on the frequency spectrum of the sound.

Transmission through horn

f1 f2 f3 f4

♩ ♪ ♫

fn

~

~

Radiation of Sound from Wind Instruments

• The radiation characteristics of the bell “shape” the harmonic recipe and strongly influence the timbre of the instrument.

80/20The diameter of the mouth and the flare rate of the bell determine the radiation characteristics of brass instruments.

• The larger the bore diameter, the more intense the low frequency harmonics.
• The more rapid the flare, the more the low frequencies are reflected, and thus, the more high frequency harmonics are radiated.

Trumpet

Cornet

Flugel Horn

x

Exponential Horn

The Bell

a = ao exp(m x)+ b

80/20m is called the “flare constant.”

Larger m means more rapid flare.

a

x

Bessel Horns

The Bell

a = ao e-(εx) +b

80/20Called “Bessel Horns” because the standing wave follows a Bessel Function.

Mutes

• The French Horn player’s hand modifies the radiation characteristics of the horn, as well as the effective flare.
• Mutes reduce the effective area of the horn and, therefore, reduce the intensity.
• Mutes tend to reduce more the first and second harmonic of the pipe than higher frequency harmonics due to their internal modes of oscillation.
• Mutes make brass sound “thin and reedy.”

Summary:

• The pitch of a wind instrument is determined by the length and shape of its air column.
• Feedback from the resonances of the pipe select the frequency of oscillation of the jet, reed or lip-valve.
• The excitation, transmission and emittance of the sound in the horn determine the timbre of the instrument.

1′ Lecture:

• The vocal folds, located in the larynx, produce vibrations in the vocal tract.
• The vocal tract is a stopped air column approximately 17 cm long. It resonates at a fundamental frequency of about 500 Hz.
• The shape of the vocal tract provides an acoustic filter of the harmonics produced by the vocal folds.

Anatomy of the Human Voice

80/20The vocal tract is the instrument of the human voice.

Vocal Tract:

• Lungs—source of air

Pharynx

• Trachea—wind pipe
• Larynx—voice box

Larynx

Trachea

• Pharynx—mouth and nose

Lungs

Anatomy of the Human Voice

80/20The sound of the human voice originates in the larynx.

Larynx

Larynx

Closed

Open

Anatomy of the Human Voice

80/20The larynx (or voice box) contains the vocal folds.

Vocal Folds

Larynx

Anatomy of the Human Voice

80/20The vocal folds rapidly open and close, introducing pulsations of air into the vocal tract.

Vocal Folds

Lower Vocal Tract

Anatomy of the Human Voice

80/20The vocal folds rapidly open and close, introducing pulsations of air into the vocal tract.

Vocal Folds

Lower Vocal Tract

Phonation

The Vocal Folds--Function

80/20The vocal folds are controlled by muscle and actuated by air moving between them, closing due to the Bernoulli Effect, opening by tension.

When flow is interrupted folds open.

Air Flow

Vocal Folds

The Vocal Folds

80/20The pressure waveform produced by the action of the vocal folds is an asymmetrical sawtooth, rich in harmonics.

80/20The fundamental frequency of the voice is determined by the properties of the vocal folds, not the vocal tract.

Vocal Folds snap open

are pulled shut by air flow

are pulled shut by air flow

are pulled shut by air flow

are pulled shut by air flow

are pulled shut by air flow

Resonances of Vocal Tract

L≈ 17 cm

f1 = v/4L = 354/(4 ‧ 0.17) = 521 Hz f3 = 3f1 , f5 = 5f1 …

80/20The Vocal Tract is a “lossy” stopped pipe ~17 cm long with a fundamental frequency of ~500 Hz.

Formants of Vocal Tract

Amplitude

Formant

L≈ 17 cm

Frequency

f1 = v/4L = 354/(4 ‧ 0.17) = 521 Hz f3 = 3f1 , f5 = 5f1 …

80/20The Vocal Tract filters the spectrum generated by the vocal folds; the frequency filter is called the Formant.

Speech

• 80/20The individual units of speech are called phonemes.
• The classes of (English) phonemes are:
• Unvoiced Plosives‒ p, t, k (c, q, x)
• Voiced Plosives‒ b, d, g
• Fricatives‒ unvoiced/voiced: f/v, th/th,
• Sibilants‒ unvoiced/voiced: s(c)/z, sh/zh (j), h/kh
• Liquids‒ l, r
• Nasals‒ m, n, ng
• Semi-vowels‒ w, y
• Vowels‒ a, e, i, o, u

80/20The shape of the Vocal Tract determines the frequency of the Formants.

“ah”

“eh”

“oh”

“oo”

Summary:

• The vocal folds, located in the larynx, produce vibrations in the vocal tract.
• The vocal tract is a stopped air column approximately 17 cm long, that resonates at ~500, 1500 and 2500 Hz.
• The shape of the vocal tract provides an acoustic filter, called the formant, that modifies the amplitude of the harmonics produced by the vocal folds.
Physics 1251 Unit 3 Session 32 The Singing Voice

1′ Lecture:

• The pitch range of the singing voice is determined by the properties of the vocal folds.
• The intelligibility of words is due to the relationship of the first two formants.
• Modification of the shape of the vocal tract significantly affects the timbre of the singing voice.
Physics 1251 Unit 3 Session 32 The Singing Voice

The Mechanics of the Vocal Folds

80/20The properties of the vocal folds determine their vibration frequency.

Larynx

Larynx

Closed

Open

Physics 1251 Unit 3 Session 32 The Singing Voice

The Mechanics of the Vocal Folds

80/20The properties of the vocal folds determine their vibration frequency.

Vocal Folds

Larynx

fvocal = 1/2π√k/ m

Physics 1251 Unit 3 Session 32 The Singing Voice

The Mechanics of the Vocal Folds

80/20The properties of the vocal folds determine their vibration frequency.

fvocal = 1/2π√k/ m

Vocal Folds

Density ρ

k = fold stiffness m = effective mass

For a cord:

f = 1/2L√T/ μ T = σ (t‧d) μ = ρ(t‧d) f = 1/2L√ σ / ρ

Stress σ

Length L

fvocal = 1/2π√k/ m

Physics 1251 Unit 3 Session 32 The Singing Voice

The Mechanics of the Vocal Folds

80/20The properties of the vocal folds determine their vibration frequency.

fvocal = 1/2π√k/ m

f = √ σ / (4L2 ρ)

k = fold stiffness m = effective mass

k = π2σ m/ L2 ρ = π2 T/ L m = ρ L(t‧d)

For a cord:

f = (1/2L)√T/ μ T = σ (t‧d) μ = ρ(t‧d) f = (1/2L)√ σ / ρ

L ≈ 0.017 m ρ ≈ 1040 kg/m3 σ ≈ 12 kPa f ≈ 100 Hz m ≈ 200 mg; T≈ 0.14 N

Physics 1251 Unit 3 Session 32 The Singing Voice

The Mechanics of the Vocal Folds

80/20The properties of the vocal folds determine their vibration frequency.

f1 = (1/2L)√ σ / ρ

• 80/20Conclusions:
• Resting length, stress and density set voice range
• Stress (tension) can be increased external to the vocal fold or internal to it.
• Overall, increased tension increases stiffness, pitch
Physics 1251 Unit 3 Session 32 The Singing Voice

Anatomy of the Human Voice

80/20During adolescent the vocal folds grow longer and the voice lowers in pitch.

Vocal Folds lengthen at puberty

f1 = √ σ / (4L2 ρ)

f 1 = 1700/L (mm)

Pitch lowers at puberty.

Physics 1251 Unit 3 Session 32 The Singing Voice

Anatomy of the Human Voice

80/20The vocal folds comprise muscle, lamina propria and epithelium.

Cover

Body

Epithelium

Lamina Propria (3 layers)

Thyroarytenoid Muscle

Physics 1251 Unit 3 Session 32 The Singing Voice

80/20Pitch is raised by increasing tension on vocal folds, both external to the vocal fold (Cricothyroid muscle) and internal to it (Thyroarytenoid muscle).

f1 = (1/2L)√ σ / ρ

The nature of the stress in the vocal fold (internal or external tension) permits phonation in different registers.

Physics 1251 Unit 3 Session 32 The Singing Voice

80/20Vocal Registers:

f1 = (1/2L)√ σ / ρ

σ =σexternal + σinternal

Terminology

Speaking: Pulse Modal Falsetto

(alternative) Fry Middle Whistle Stohbass flageolet

Physics 1251 Unit 3 Session 32 The Singing Voice

Vowels and Formants

80/20The relative frequency of the 1 st and 2 nd vowels formants are unique to various vowels.

i

I

ε

æ

e

Second formant frequency

Λ

D

U

u

c

First formant frequency

Physics 1251 Unit 3 Session 32 The Singing Voice

Control of Formants

80/20Tongue and lip placement and the shape of the pharanx are most important in vowel formation.

“Corner Vowels”

D

i

u

A

A

A

f

f

f

Physics 1251 Unit 3 Session 32 The Singing Voice

Harmonics align with Formants

Singers’ Formant

Formants and Singing

• Vowel modification shifts formats. •Alignment of formants with harmonics intensifies pitch. •Dilation of vocal tract causes Singer’s Formant.
Physics 1251 Unit 3 Session 32 The Singing Voice

Summary:

• The pitch range of the singing voice is determined by the size, tension, and density of the vocal folds.
• Vocal registers and breaks in the voice result from changing modes of oscillation of the vocal folds.
• Vowels are distinguished by the frequency relationship of the first two formants.
• Modification of the vocal tract shape sets the timbre of the singing voice.
Physics 1251 Unit 3 Session 33 Percussion

1′ Lecture:

• Percussion instruments are instruments that are struck.
• The timbre of their sound is determined by their vibration recipe.
• Their vibration recipe is determined by the modes of oscillation that are excited.
• Often percussion instruments do not have pitch.
Physics 1251 Unit 3 Session 33 Percussion

80/20The timbre of an instrument’s sounds depends on its vibration recipe.

fn = n f1

Pitched

Amplitude

f1

2f1

3f1

4f1

fn m = xn m f1

Unpitched

Amplitude

f01

Frequency

Surface density σ

Surface Tension S

Physics 1251 Unit 3 Session 33 Percussion

The Oscillation of a Clamped Membrane

Mode: (0,1)

d

f0 1 = v/λ; v = √(S/ σ)

f0 1 = x0 1 /(π d) ‧ √(S/ σ)

x0 1 = 2.405

Surface density σ= mass/area σ= density ‧ thickness

Surface Tension S= force/length

Surface density σ

Surface Tension S

Physics 1251 Unit 3 Session 33 Percussion

The Modes of Oscillation of an (Ideal) Clamped Membrane

Mode: (0,1)

f0 1 = x0 1 /(π d) ‧ √(S/ σ)

x0 1 = 2.405

Mode: (1,1)

Mode: (2,1)

f1 1 = (x1 1 / x0 1) f0 1

x1 1 / x0 1 = 1.594

f2 1 = (x2 1 / x0 1) f0 1

x2 1 / x0 1 = 2.136

Physics 1251 Unit 3 Session 33 Percussion

The Modes of Oscillation of a Clamped Membrane

Mode: (0,1) xn m / x0 1 : 1

(1,1)1.594

(2,1)2.136

(0,2)2.296

(3,1)2.653

(1,2)2.918

(4,1)3.156

(2,2)3.501

(0,3)3.600

(5,1)3.652

Physics 1251 Unit 3 Session 33 Percussion

80/20Membrane Acoustics:

• The overtones of a circular membrane clamped at the edge are not harmonic and, therefore, they have no pitch.

fnm = (xn m /x01)f01

• The frequencies fnm of a membrane are (1) proportional to the square root of the ratio of surface tension of the head to the surface density ∝√(S / σ) and (2) inversely proportional to its diameter ∝1/d.
Physics 1251 Unit 3 Session 33 Percussion

Ideal vs Real Membranes:

80/20Real membranes have a lower frequencies than predicted for ideal membranes because of air loading; the lowest frequencies are lowered the most.

Physics 1251 Unit 3 Session 33 Percussion

Mode Excitation:

80/20Only those frequencies for which the modes were excited will appear in the vibration recipe.

80/20The highest frequency that can be excited by a mallet that is in contact with the surface for a period of Tcontact is

f max= 2/Tcontact

Physics 1251 Unit 3 Session 33 Percussion

Mode Excitation:

80/20The highest frequency that can be excited by a mallet that is in contact with the surface for a period of Tcontact is

f max= 2/Tcontact

Tcontact = ½ Tperiod= 1/(2fmax )

Physics 1251 Unit 3 Session 33 Percussion

vbend

h: thickness

Bending Wave in a Plate

ρ: density

E: Young’s Modulus

• Density ρ= mass/volume

• vL = √E/(.91 ρ)

vbend = √[1.8 f h vL ]

• Young’s ModulusE= stress/elongation =stiffness

fnm = 0.0459 h vL( ynm /d)2

Physics 1251 Unit 3 Session 33 Percussion

The Modes of Oscillation of a Flat Cymbal

Mode: (2,0) fn m / f0 1 : 1

(0,1)1.730

(3,0)2.328

(1,1)3.910

(4,0)4.110

(5,0)6.30

(2,1)6.71

(0,2)3.600

Physics 1251 Unit 3 Session 33 Percussion

80/20 Plate Acoustics:

• The overtones of a circular plate clamped in the center are not harmonic and, therefore, have no pitch.

fn m = (yn m /y20)2 f20

• The frequencies fnm of a circular plate are (1) proportional to the thickness ∝h and (2) to the square root of the ratio of the stiffness and the density ∝√E/ρ and (3) inversely proportional to the square of the diameter ∝1/d2 .
Physics 1251 Unit 3 Session 33 Percussion

Summary:

• Percussion instruments are instruments that are struck.
• Their vibration recipe is often not harmonic and, therefore, they do not have a definite pitch.
• For ideal circular edge-clamped membranes: fnm ∝(xnm /d)√(S/σ).
• For circular plates free at the edge: fnm ∝h ‧ (ynm /d) 2 √(E/ρ).
• The maximum frequency excited by a mallet is f max= 2/Tcontact.
Physics 1251 Unit 3 Session 34 Percussion with Pitch

1′ Lecture:

• Piano strings exhibit inharmonicity because of the stiffness of the wire.
• Some percussion instruments have pitch.
• Pitch results from a harmonic series of overtones.
• Tympani and Tabla are pitched drums.
• Orchestra Chimes, Glockenspiel, Xylophone, Marimba and Vibraphone have intonation.

fn = n f1

Pitched

Amplitude

f1

2f1

3f1

4f1

fn m = xn m f10

Unpitched

Amplitude

f01

Physics 1251 Unit 3 Session 34 Percussion with Pitch

80/20The task of producing pitch in a percussion instrument is an exercise in manipulating the overtones into a harmonic series.

Frequency

Tension T

L

Linear density μ

Physics 1251 Unit 3 Session 34 Percussion with Pitch

The Modes of vibration of an ideal string are harmonic.

• Linear density μ= mass/length

The stiffness of the wire increases the frequency of the higher frequency harmonics.

• Tension T= force

fn = n /(2 L) ‧ √(T/ μ) n= 1, 2, 3, 4, 5, 6, 7….

₧ = 3986¢ Log(nf1 /440) + I(₧)

I(₧) = Inharmonicity

Physics 1251 Unit 3 Session 34 Percussion with Pitch

Inharmonicity of Piano

40¢

20¢

Inharmonicity

-20¢

Pitch (¢)

Because of the inharmonicity of strings the octaves are “stretched” in a piano.

Physics 1251 Unit 3 Session 34 Percussion with Pitch

Tension device

Tension pedal

Surface density σ

Surface Tension S

Physics 1251 Unit 3 Session 34 Percussion with Pitch

The mass of air moved by the membrane adds to the effective surface density, lowering the frequency.

Air mass

Physics 1251 Unit 3 Session 34 Percussion with Pitch

80/20The kettle of Tympani modifies the membrane frequencies by the interaction of the air resonances with the surface modes.

Modes of air vibration

Strike point

Physics 1251 Unit 3 Session 34 Percussion with Pitch

The Modes of Oscillation of Tympani

Mode: (0,1) fn m/f01 : 1

(1,1)1.594

(2,1)2.136

(0,2)2.296

(3,1)2.653

(1,2)2.918

(4,1)3.156

(2,2)3.501

(0,3)3.600

(5,1)3.652

2f0

3f0

4f0

5f0

6f0

(0,1)

f0

(1,1)

(2,1)

(3,1)

(4,1)

(0,3)

(3,2)

(2,2)

(5,1)

(0,2)

(1,2)

(0,1)

(1,1)

(2,1)

(3,1)

(4,1)

(0,3)

(3,2)

(5,1)

(0,2)

(1,2)

(2,2)

Physics 1251 Unit 3 Session 34 Percussion with Pitch

80/20Tympani achieve pitch by (1) suppression of “radial” modes; (2) modification of other mode frequencies by air loading and the effect of the kettle ; (3) attenuation of the lowest mode.

Amplitude

Frequency

Physics 1251 Unit 3 Session 34 Percussion with Pitch

Metalophones:

Glockenspiels, Xylophones and Marimbas

Bar

h thickness

w width

L Length

Density ρ = mass/volume

Young’s Modulus E= Force/elongation

Physics 1251 Unit 3 Session 34 Percussion with Pitch

Metalophones:

Glockenspiels, Xylophones and Marimbas

Longitudinal Waves in a Bar

vL = √E/ ρ

Longitudinal Wave Velocity

node

Anti-node

Anti-node

fn = n/2L√E/ ρ like an open pipe

Density ρ = mass/volumeYoung’s Modulus E= Stress/Elongation

End Clamped

f1= 0.1782 fo

f2= 1.116 fo

f3=3.125 fo

Physics 1251 Unit 3 Session 34 Percussion with Pitch

Bending Modes in Bars:

Physics 1251 Unit 3 Session 34 Percussion with Pitch

Free Ends

Bending Modes in Bars:

f1= 1.133 fo

f2= 3.125 fo

f3=6.125 fo

fo∝ h/L2

.224 L

Physics 1251 Unit 3 Session 34 Percussion with Pitch

Free Ends

Orchestral Chimes

End Plug

f1= 1.133 fo

f2= 3.125 fo

f3=6.125 fo

λ/4

Vibraphone

Physics 1251 Unit 3 Session 34 Percussion with Pitch

Mode Frequencies in Undercut Bar:

Undercut Bar in Xylophone, Marimba and Vibraphone

Xylophone

f1/f1 = 1.00f2 /f1 = 3.00f3 /f1 =6.1

Marimba/Vibes

f1 /f1 = 1.00f2 /f1 = 4.00f3 /f1 =6.5

Physics 1251 Unit 3 Session 34 Percussion with Pitch

80/20What is the different between a Xylophone, a Marimba and a Vibraphone?

• The depth of the undercut: a marimba is undercut more than a xylophone.
• The first harmonic of a xylophone is 3x the fundamental, for a marimba and “vibe” it is 4x.
• The xylophone sounds “brighter” and the marimba more “mellow.”
• Vibes have a tremolo mechanism.
Physics 1251 Unit 3 Session 34 Percussion with Pitch

Summary:

• Piano strings exhibit inharmonicity because of the stiffness of the wire.
• Some percussion instruments have pitch.
• Pitch results from a harmonic series of overtones.
• Tympani and Tabla are pitched drums.
• Orchestra Chimes, Glockenspiel, Xylophone, Marimba and Vibraphone have intonation.
• Marimba are undercut more than xylophones.