Acousticsver7 1 Show

Discussion on Discussion on Acoustics TechnologyAcoustics Technology

By: Romeo Lepiten, ECEBy: Romeo Lepiten, ECE

supplementary encoding by: Anthony Ross G. Naňas, ECTsupplementary encoding by: Anthony Ross G. Naňas, ECT

How do How do we Hearwe Hear

Ear AnatomyEar Anatomy

How do we hearHow do we hear As shown in figure (a), the anatomy of the As shown in figure (a), the anatomy of the

ear; the ear is composed of the outer, ear; the ear is composed of the outer, middle, and inner ears.middle, and inner ears.

The pinna, auditory canal, and eardrum or The pinna, auditory canal, and eardrum or tympanic membrane are parts of the tympanic membrane are parts of the outer outer earear. Sound transmitted through the air or . Sound transmitted through the air or other medium reaches to the pinna and other medium reaches to the pinna and continues through the continues through the auditory canalauditory canal and and strikes the strikes the tympanic membranetympanic membrane. The . The membrane is thus set into vibration.membrane is thus set into vibration.

How do we hearHow do we hear The The middle earmiddle ear, an air-filled chamber, is , an air-filled chamber, is

composed of three tiny bones call composed of three tiny bones call ossiclesossicles.. The ossicles are:The ossicles are:

- the malleous (hammer)- the malleous (hammer)- incus (anvil)- incus (anvil)- stapes (stirrup)- stapes (stirrup)

The vibration of the tympanic membrane The vibration of the tympanic membrane is transmitted to the malleous, incus, and is transmitted to the malleous, incus, and stapes in the middle ear.stapes in the middle ear.

How do we hearHow do we hear The semicircular canals and the cochlea The semicircular canals and the cochlea

are parts of the inner ear, which contains are parts of the inner ear, which contains fluid called the fluid called the perilymphperilymph. The . The semicircular canalsemicircular canal provides the sense of provides the sense of balance, and the balance, and the cochleacochlea contains the contains the organ of organ of CortiCorti, which is the sense organ of , which is the sense organ of hearing.hearing.

How do we hearHow do we hear The figure below shows the cochlea being The figure below shows the cochlea being

straightened out,straightened out,

How do we hearHow do we hear The figure below shows the cochlea’s cross The figure below shows the cochlea’s cross

section,section,

How do we hearHow do we hear The cross section shows a membrane The cross section shows a membrane

called called Reissner’s membraneReissner’s membrane, the , the basilar basilar membranemembrane, and the , and the tectoral membranetectoral membrane..

The cochlea is divided by the basilar The cochlea is divided by the basilar membrane into the membrane into the upperupper and and lower lower galleriesgalleries. .

Between the tectoral and basilar Between the tectoral and basilar membranes are some 23,500 minute hair membranes are some 23,500 minute hair cells.cells.

How do we hearHow do we hear The vibrations from the stapes are The vibrations from the stapes are

transmitted to the inner ear through the transmitted to the inner ear through the oval window, generating waves in the oval window, generating waves in the perilymph fluid.perilymph fluid.

The waves proceed from the oval window, The waves proceed from the oval window, along the upper gallery, around the along the upper gallery, around the cochleal apex, and through the lower cochleal apex, and through the lower gallery, terminating at the gallery, terminating at the round windowround window..

The waves cause displacement of the The waves cause displacement of the basilar membrane, which, in turn, sways basilar membrane, which, in turn, sways the tiny hair cells to and fro, producing a the tiny hair cells to and fro, producing a straining effect on the cells. straining effect on the cells.

How do we hearHow do we hear The straining produces a piezoelectric The straining produces a piezoelectric

effect, and the electricity being produced effect, and the electricity being produced is transmitted to the brain through the is transmitted to the brain through the auditory nerve, producing the perception auditory nerve, producing the perception of hearing.of hearing.

The corresponding electric potential The corresponding electric potential produced is called the produced is called the cochleal potentialcochleal potential..

As long as sound is not transmitted as a As long as sound is not transmitted as a sudden impact, the ear has the capability sudden impact, the ear has the capability to protect itself from intense noise by a to protect itself from intense noise by a reflex action called reflex action called acoustic reflexacoustic reflex..

How do we hearHow do we hear This reflex is the stiffening of the tympanic This reflex is the stiffening of the tympanic

membrane by the action of the two small membrane by the action of the two small muscles in the inner ear call the muscles in the inner ear call the tensor tensor tympanitympani and the and the stapediusstapedius..

The ear can detect sound pressures as low The ear can detect sound pressures as low as 2(10as 2(10-5-5) Pa and, before experiencing ) Pa and, before experiencing pain, to as high as 200 Pa.pain, to as high as 200 Pa.

How Sound Waves travel thru How Sound Waves travel thru Air, How We Hear thru AirAir, How We Hear thru Air simple animated demonstrationsimple animated demonstration

AcousticsAcoustics

AcousticsAcoustics is the physics of sound is the physics of sound

Why do we study acoustics?Why do we study acoustics?

An Acoustical Engineer studies acoustics because he is interested in the fidelity of sound reproduction, the conversion of mechanical and electrical energies into acoustical energy and design of electro-acoustical transducers ( i. e. speakers, microphones)

SoundSound

SoundSound

It is the mental interpretation of the It is the mental interpretation of the auditory sensation produced by the auditory sensation produced by the disturbance of air.disturbance of air.

The normal, everyday sound being The normal, everyday sound being heard is conglomeration of several heard is conglomeration of several frequencies. The audible range is from frequencies. The audible range is from 16 to 20,000 Hz.16 to 20,000 Hz.

Sound below 16 Hz is called infrasonic while sound above 20,000 Hz is called ultrasonic

SoundSound

A sound is propagated through the air, A sound is propagated through the air, compression bands are formed. Compression compression bands are formed. Compression and rarefaction are disturbances over existing and rarefaction are disturbances over existing conditions.conditions.

Pressures are high in the compression band and Pressures are high in the compression band and are lower in the rarefaction bands.are lower in the rarefaction bands.

What accounts for the propagation of sound is What accounts for the propagation of sound is the differential of those pressures above and the differential of those pressures above and below existing atmospheric pressure. Without below existing atmospheric pressure. Without these differentials, no sound can be these differentials, no sound can be transmitted, as ear will perceive no change.transmitted, as ear will perceive no change.

There are three important parameters that There are three important parameters that must be learned to characterize or study must be learned to characterize or study soundsound::

PressurePressure

PowerPower

IntensityIntensity

Plane Plane Acoustic Acoustic WavesWaves

Plane Acoustic WavesPlane Acoustic Waves

Plane acoustic waves are one Plane acoustic waves are one dimensional free progressive waves dimensional free progressive waves traveling in the x-direction, assuming no traveling in the x-direction, assuming no variation of pressure in the y or z variation of pressure in the y or z direction.direction.

The wavefronts are infinite planes The wavefronts are infinite planes perpendicular to the x-axis and they are perpendicular to the x-axis and they are parallel to one another at all times.parallel to one another at all times.

Wave EquationWave Equation

where:where:c = speed of wave propagationc = speed of wave propagationu = instantaneous displacementu = instantaneous displacement

∂ 2u C2 ∂ 2u ∂ t2 ∂ x2

=

Wave EquationWave Equation

General Solution:General Solution:

u(x,t) = A e u(x,t) = A e i (wt – kx)i (wt – kx) + B e + B e i (wt + kx)i (wt + kx)

oror

u(x,t) = A cos (wt – kx) + B cos (wtu(x,t) = A cos (wt – kx) + B cos (wt + + kx) kx)

Elements of Plane acoustic wavesElements of Plane acoustic waves

a.a. Particle Displacement; uParticle Displacement; u

u(x,t) = A e u(x,t) = A e i (wt – kx)i (wt – kx) + B e + B e i (wt + kx)i (wt + kx)

oror

u(x,t) = A cos (wt – kx) + B cos (wt + kx)u(x,t) = A cos (wt – kx) + B cos (wt + kx)

Elements of Plane acoustic wavesElements of Plane acoustic waves

b. Acoustic Pressure; Pb. Acoustic Pressure; P

ρ ∂ u∂ xP= - c2

Acoustic PressureAcoustic Pressure

Considering wave motion along +x axisConsidering wave motion along +x axis

P = P = - - ρρ wcA sin ( wt – kx) wcA sin ( wt – kx) where: where: ρρ = = densitydensity w = 2w = 2ππff

A = displacement amplitude of the A = displacement amplitude of the wavewave

k = wave number ; w/ck = wave number ; w/c

Acoustic PressureAcoustic Pressure

Maximum Differential Pressure; PmMaximum Differential Pressure; Pm

PPmm = = ρρ wcAwcAEffective Acoustic PressureEffective Acoustic Pressure;;

Prms =Pm

2

Sound Sound PowerPower

Sound PowerSound Power

The energy of an oscillating sound wave is composed of The energy of an oscillating sound wave is composed of kinetic and potential energies. When it is compressed, it kinetic and potential energies. When it is compressed, it possesses potential energy possesses potential energy

When it is released against an object place on it, the When it is released against an object place on it, the energy will be transformed to the object as kinetic energy will be transformed to the object as kinetic energy. energy.

Neglecting friction, the resulting kinetic energy will be Neglecting friction, the resulting kinetic energy will be equal to the original potential energyequal to the original potential energy


Sound Power; Pw:Sound Power; Pw:

Pw = ½ S Pw = ½ S ρρ ww22 cA cA22 wattswatts

Sound Sound IntensityIntensity

Sound Intensity; ISound Intensity; I

The intensity of sound waves is defined as the The intensity of sound waves is defined as the average rate at which power is transmitted per average rate at which power is transmitted per unit cross-sectional area in the direction of the unit cross-sectional area in the direction of the traveltravel

1 T∫ Pv ∂t

T

0

I =

or simply:

Pw S

I =

Sound IntensitySound Intensity

Pw S

I =

I = ½ S w2 cA2

Sρ

I =

½ w2 cA2

ρ and:

Prms2

I = ρc


Total Intensity: ITotal Intensity: ITT

IITT = I = I11 + I + I22 + I + I33 + . . . + I + . . . + Inn

Total Pressure: PTotal Pressure: PTT

PPTT22 = P = P11

22 + P + P2222 + P + P33

22 + …+ P + …+ Pnn22

Speed of Speed of Sound Sound

Speed of SoundSpeed of Sound

The speed of a wave is the distance The speed of a wave is the distance that a disturbance travels per unit that a disturbance travels per unit timetime

The speed of a particular type of The speed of a particular type of wave depends on the properties of wave depends on the properties of the medium in which the wave travelsthe medium in which the wave travels

Speed of sound in airSpeed of sound in air

where: ∂ = ratio of the specific heat of air at constant pressure to that of constant volume

P = pressure

p = density

∂Pρ c =

At room temperature and standard At room temperature and standard atmospheric pressure, the speed of atmospheric pressure, the speed of sound in air is sound in air is 341 m/s341 m/s and increases and increases approximately approximately 0.6 m/s0.6 m/s for each degree for each degree centigrade rise, and expressed ascentigrade rise, and expressed as

T = temp in 0K

T C = 20.1

Speed of Sound in Fluids:Speed of Sound in Fluids:

B = bulk modulusB = bulk modulusρρ == densitydensity

c = B ρ

Speed of Sound in Liquids:Speed of Sound in Liquids:

K = compressibility in m secK = compressibility in m sec22/ kg/ kg

c = 1 ρK

Speed of Sound in SolidSpeed of Sound in Solid

Y = Young’s modulusY = Young’s modulusρρ = = densitydensity

c = Y ρ

Velocity of sounds in liquidsVelocity of sounds in liquids

LiquidLiquid Temperature Temperature ((00C)C)

Velocity (m/s) x Velocity (m/s) x 10 10 33

alcohol, ethylalcohol, ethyl 12.512.52020

1.241.241.171.17

MercuryMercury 2020 1.451.45PetroleumPetroleum 1515 1.331.33

Water, sea(36 Water, sea(36 parts/thousanparts/thousan

d salinity)d salinity)1515 1.5051.505

Sound in SolidsSound in SolidsVelocity of sound in some common solid Velocity of sound in some common solid materialsmaterials

MaterialsMaterials Velocity (m/s)Velocity (m/s) Velocity (ft/s)Velocity (ft/s)WoodWood 33003300 1082510825BrickBrick 36003600 1180011800ConcreteConcrete 37003700 1210012100SteelSteel 49004900 16001600GlassGlass 50005000 1640016400AluminumAluminum 58005800 19001900

Note: These figures are approximate, since the listed materials vary in density. Average frequency is used.

Sound Sound Pressure and Pressure and

Sound Sound Pressure LevelPressure Level

Sound Pressure:Sound Pressure:

Sound Pressure (P)Sound Pressure (P) at a point in sound at a point in sound field; is the alternating component of field; is the alternating component of the pressure at the pointthe pressure at the point

Unit: N/mUnit: N/m22 or Pascal (Pa) or Pascal (Pa)

Sound Pressure Level:• Sound Pressure Level (SPL) is equal to 20

times the logarithm to the base 10 of the ratio of the RMS sound pressure to the reference sound pressure (2 x 10-5 N/m2)

( )P Po

P Po

2 SPL = 20 log = 10 log

where: P = RMS sound pressure, N/m2

P0 = reference sound pressure

= 2x10-5 N/m2 or Pascal (Pa)

= 0.0002 µbar

= 2.089 lb/ft2

Sound Intensity Sound Intensity and Sound and Sound

Intensity LevelIntensity Level


Sound Intensity (I)Sound Intensity (I) is the sound energy is the sound energy flux through a unit area, normal to that flux through a unit area, normal to that directiondirection

W 4πr2

I = ; W/m2

for sound coming from a point source (isotropic) in free space


for sound produced at ground levelfor sound produced at ground level

W 2πr2

I = ; W/m2

IL = 10 log IIo

= 10 log ( )P Po

2 = SPL since I ∞P2

Where Io = threshold intensity, W/m2

= 10 -12 W/m2

Sound Power Sound Power and Sound and Sound

Power LevelPower Level


Sound power Sound power of a source; is the total of a source; is the total sound energy radiated per unit timesound energy radiated per unit time

Sound Power Level• Sound Power Level (PWL) is equal to 10

times the logarithm to the base 10 of the ratio of the sound power of the source to the reference sound power (10-12 W)

PWL = 10 log WWo

PWL = 10 log W + 120

where W = sound power in watts

Wo= reference sound power

= 10 -12 W

Relation of SPL and PWLRelation of SPL and PWL

For sound coming from a point For sound coming from a point source (isotropic) in free spacesource (isotropic) in free space

SPL = PWL – 20 log r - 8

• For sound produced at ground level

SPL = PWL – 20 log r - 11

Sample Sample ProblemsProblems

The maximum differential pressure that the ear can tolerate in loud The maximum differential pressure that the ear can tolerate in loud sounds is 28 N/msounds is 28 N/m22. What is the amplitude for sound in air at . What is the amplitude for sound in air at frequency of 1000 Hz? Assume bulk modulus and density equal to frequency of 1000 Hz? Assume bulk modulus and density equal to 1.42 x101.42 x1055 N/m N/m22 and 1.23 kg/m and 1.23 kg/m33 for air respectively. What are the rms for air respectively. What are the rms pressure and intensity?pressure and intensity?

Soln:

c = βρ

c = 1.42x105

1.23

c = 340 m/sec

Pm= ρ wcA

A = Pm

ρ wc

A =340 (1.23)(2π)(1000)

28

A = 10.7 µmPm

Prms=2

Prms=28

2

Prms= 19.8N/m2

I =Prms

2

ρ c

I =19.82

1.23 (340)

I = 0.94 W/m2

Determine the value of the bulk modulus for water from the Determine the value of the bulk modulus for water from the experimental evidence that sound travels 60.4 m in 4x10experimental evidence that sound travels 60.4 m in 4x10-2-2 sec in sec in water of density 1x10water of density 1x1033 kg/m kg/m33..

Soln:

c = βρ

c = 60.4

4x10-2

c = 1.51 x 103 m/sec

β = c2 ρ

β = (1.51x103) (1x103)

β = 2.28 x 109 N/m2

Young’s modulus of copper is 12.2x10Young’s modulus of copper is 12.2x101010 Pa. and the density of Pa. and the density of copper is 8900 kg/mcopper is 8900 kg/m33. Calculate the speed of sound in copper. Calculate the speed of sound in copper

Soln:

c = Yρ

c = 12.2 x1010

8900

c = 3700 m/sec

The intensity of sound at construction site is 0.10 W/mThe intensity of sound at construction site is 0.10 W/m2. 2. Assuming that the are of the eardrum is 0.20 cmAssuming that the are of the eardrum is 0.20 cm22, how much , how much sound energy is absorbed by one ear in an 8 hour work daysound energy is absorbed by one ear in an 8 hour work day

Soln:

Intensity =PowerArea

Power =Energy

time

Intensity =Et A

1hrE = 3600 sec(0.1) (0.20x10-4 m2) (8 hr )

E = 0.06 J

The density of sound in air is given at 1.21 kg/mThe density of sound in air is given at 1.21 kg/m33 and a velocity and a velocity of 344 m/sec for a pressure of 30 N/mof 344 m/sec for a pressure of 30 N/m22. Calculate its intensity.. Calculate its intensity.

Soln:

Intensity =P2

2 ρ c

Intensity =302

2 (1.21) (344)

Intensity = 1.08 W/m2

Sound source radiates uniformly in all directions in air at 20 °C. At a Sound source radiates uniformly in all directions in air at 20 °C. At a distance of 80 m from the source, the sound level is 80 dB. The distance of 80 m from the source, the sound level is 80 dB. The frequency is 440 Hz. What is the displacement amplitude?frequency is 440 Hz. What is the displacement amplitude?

IL = 10 logIIO

Soln:

I = [ antilog ( IL/10 ) ] IO = [ antilog ( 80/10 ) ] ( 1 X 10-12)

c = 20.1 20 + 273 = 344.14

r = 80 m

IL = 80 dB

I = (1/2) ρ c w2 A2

I = 1 X 10-4 W/m2

but,

A = ( 2 I ) / ( ρ c w2 ) so,

A = 0.2486 µm = 0.25 µm

By what factor must the pressure amplitude of the sound wave By what factor must the pressure amplitude of the sound wave be increased in order to increase the intensity by a factor of 9?be increased in order to increase the intensity by a factor of 9?

Soln:

2I =

Prms2

if Prms is tripled 2

Prms =3 Pm

2=

Pm

3

Prms

2I =

(1/3 Prms)2

9 2 I =

Prms2

Ii =Prms

2

2; let

9 I = Ii

Pm should be tripled to increase the intensity by a factor of 9.

If the pressure amplitude of a sound wave is doubled, by what If the pressure amplitude of a sound wave is doubled, by what factor does the intensity of the wave increase?factor does the intensity of the wave increase?

Soln:

2Prms =

2 Pm

if Prms is doubled, I = ?

2 I

i =

Prms2

; let

Prms = 2 Prmsi

2I =

Prmsi2

Prmsi = ( ½ ) Prms

4 2 =

Prms2

I = ( ¼ ) Ii

2Prmsi

=Pm

; let

2I =

( ½ Prms )2

Ii = 4 I

if Pm is doubled, intensity increases by a factor of 4.

A sound wave in air has a frequency of 400 Hz, a wave speed of 344 m/s A sound wave in air has a frequency of 400 Hz, a wave speed of 344 m/s and a displacement amplitude of 0.0075 mm. Calculate the sound and a displacement amplitude of 0.0075 mm. Calculate the sound intensity level.intensity level.

Soln:

f = 400 Hzc = 344 m/sA = 0.0075 mm

IL = 10 logI

1 x 10-12

I = ½ w2 c A2

I = ½ (1.23) (2 x 400)2 (344) (0.0075)2

I = 0.075 W/m2

IL = 10 log0.0075

1 x 10-12

IL = 108.76 dB

The threshold of pain has a sound intensity level of 120 dB. What The threshold of pain has a sound intensity level of 120 dB. What is the equivalent sound intensity?is the equivalent sound intensity?

Soln:

IL = 10 logI

1 x 10-12

IL = 120 dB

I = [ antilog ( 120/10 ) ] ( 1 x 10-12 )

I = 1 W/m2

For a steel rod, Young’s modulus is 2 x 10For a steel rod, Young’s modulus is 2 x 101111 Pa and a density of Pa and a density of 7.8 x 107.8 x 1033 kg/m kg/m33. What is the speed of sound in the steel rod?. What is the speed of sound in the steel rod?

Soln:

c = Yρ

c = 2.0 x1011

7.8 x 103

c = 5063.697 m/sec

A sound wave in air has a displacement amplitude of 0.0140 mm. A sound wave in air has a displacement amplitude of 0.0140 mm. Calculate the pressure amplitude for frequency at 500 Hz.Calculate the pressure amplitude for frequency at 500 Hz.

Soln:

Pm = w c A

Pm = ( 1.23 ) ( 2 x 500 ) ( 330 ) ( 0.0140 x 10-3 )

Pm = 17.85 Pa = 18 Pa

A jet aircraft produces a sound of 140 dB intensity at a distance A jet aircraft produces a sound of 140 dB intensity at a distance of 100 ft. How far away is the intensity at 90 dB?of 100 ft. How far away is the intensity at 90 dB?

Soln:

IL = 140 dB ; r = 100 ft 30.48 m

IL = SPL = PWL – 20 log r - 11

140 = PWL – 20 log ( 30.48 ) - 11

PWL = 180.68 dB

for 90 dB,

90 = PWL – 20 log r - 11

r = antilog [ ( 90 – 180.68 + 11 ) / -20 ]

r = 9638.63 m r = 5.99 mi = 6 mi

A sound intensity level of 55 dB is produced by 10 flutes. What is the A sound intensity level of 55 dB is produced by 10 flutes. What is the number of flutes needed to produce a level of 65 dB under the same number of flutes needed to produce a level of 65 dB under the same circumstance?circumstance?

IL = 10 logI

1 x 10-12

I# = [ antilog ( 65/10 ) ] (1 x 10-12)

65 = 10 logI#

1 x 10-12

I = I# = 3.16 x 10-6 W/m2 per ? flutes

@ IL = 65 dB,

Soln:IL = 55 dB for 10 flutes ; the no. of flutes if IL = 65 dB ?

I10 = [ antilog ( 55/10 ) ] (1 x 10-12)

55 = 10 logI10

1 x 10-12

I = I10 = 31.62 x 10-9 W/m2 per 10 flutes

by R/P;I10 31.62 x 10-9 W/m2

I# 3.16 x 10-6 W/m2=

no. of flutes = 100

A note of frequency 300 vibrations per second has an intensity of A note of frequency 300 vibrations per second has an intensity of 1 1 µµW/mW/m22. What is the amplitude of the air vibrations caused by . What is the amplitude of the air vibrations caused by the sound?the sound?

Soln:f = 300 vib/sI = 1 µW/m2

I = ½ w2 c A2

A = [ ( 2 ) ( 1 x 10-6 ) ] / [ ( 1.23 ) (2 x 300)2 (330)

A = 37.2 x 10-9 m

A = 37 nm

Attributes of Attributes of Sound WavesSound Waves

Attributes of Sound WavesAttributes of Sound Waves

1.1. Pitch Pitch Pitch is that attribute of auditory Pitch is that attribute of auditory

sensation in terms of which sound may sensation in terms of which sound may be ordered on a scale primarily related to be ordered on a scale primarily related to frequency.frequency.

The pitch of a sound depends upon the The pitch of a sound depends upon the frequency of vibration: frequency of vibration: the higher the the higher the frequency, the higher the pitchfrequency, the higher the pitch. . The The subjective unit of pitch is subjective unit of pitch is MelMel. 1000. 1000 Mels Mels is the pitch of 1000 Hz tone at a is the pitch of 1000 Hz tone at a sensation level of 40 dbsensation level of 40 db

2. Timbre2. Timbre Timbre is the quality of a sound related Timbre is the quality of a sound related

to its harmonic structure.to its harmonic structure.Pure tone is a sound composed of only Pure tone is a sound composed of only one frequency in which the sound one frequency in which the sound pressure varies sinusoidally with time.pressure varies sinusoidally with time.

Musical Sounds Musical Sounds (tones) are composed of (tones) are composed of the fundamental frequency and integral the fundamental frequency and integral multiple of fundamental frequency multiple of fundamental frequency (harmonics)(harmonics)

Octave Octave ( a musical terminology) refers to ( a musical terminology) refers to a pitch interval at 2:1. The tone whose a pitch interval at 2:1. The tone whose frequency is twice that of the given tone frequency is twice that of the given tone

3. Loudness3. Loudness

Loudness is the brain’s perception of the Loudness is the brain’s perception of the magnitude of the sound level.magnitude of the sound level.

Loudness level is measured by the sound Loudness level is measured by the sound pressure level of a standard pure tone of pressure level of a standard pure tone of specified frequency which is assessed by specified frequency which is assessed by normal observes of being equally loud.normal observes of being equally loud.

Units used for LoudnessUnits used for Loudness

1.1. PhonPhon - loudness of a tone which when heard at - loudness of a tone which when heard at 1000Hz and the sound pressure level in the free 1000Hz and the sound pressure level in the free progressive wave above 20 progressive wave above 20 µPa.µPa.

2.2. SoneSone – the loudness of 40 phon’s made arbitrarily – the loudness of 40 phon’s made arbitrarily equal to unity and is expected asequal to unity and is expected as

S = 2 (P - 40)/10

Note: a. The phon unit is not additive

b. The sone unit is additive

ExampleExampleWhat is the sum of two 40 phons sound pressure What is the sum of two 40 phons sound pressure

levellevel

40 phons = 1 sone40 phons = 1 sone2 sones = 1 sone + 1 sone2 sones = 1 sone + 1 sone2 sones = 40 phons + 40 phons2 sones = 40 phons + 40 phons

s = 2s = 2(P-40)/10(P-40)/10

2 = 22 = 2(P-40)/10(P-40)/10

P = 50 phonsP = 50 phons

40 phons + 40 phons = 50 phons40 phons + 40 phons = 50 phons

Transmission Transmission of of

SoundSound

Transmission of SoundTransmission of Sound1.1. Transmission of Sound Through Two Media:Transmission of Sound Through Two Media:

Reflected Wave

Incident Wave Transmitted Wave

Medium 1 Medium 2

at x = 0P1 = P2

and V1 = V2

Transmission of SoundTransmission of Sound

Acoustic Pressure: PAcoustic Pressure: P

on medium 1:on medium 1:

P - ρ C2 ∂ u ∂ x

=

P1 - ρ1 C12 ∂ u1

∂ x=


u1 = AI e i (wt – k1

x) + AR e i (wt + k1

x)

but k1 =

at x=0

ww CC11

P1 - ρ1 C12 ∂_

∂ x

= [AI e i (wt – k1

x) + AR e i (wt + k1x)]

P1 i ρ1 C1 e iwt (AI – AR)=


on medium 2:on medium 2:

ww CC22

P2 - ρ2 C22 ∂ u2

∂ x=

u2 = AT e i (wt – k2x) ;but k2 =

P2 - ρ2 C22 ∂_

∂ x

= [AT e i (wt – k2

x)]


at x=0

equate P1 = P2

velocity: vv

P2 i ρ2 C2 w AT e iwt=

i ρ1 C1 w e iwt (AI – AR) i ρ2 C2 w AT e iwt=

ρ1 C1 (AI – AR) ρ2 C2 AT= 1

∂ u ∂ t =

Transmission of SoundTransmission of Soundon medium 1:on medium 1:v1

v1at x = 0

v1

on medium 2:on medium 2: v2

at x = 0 v2

∂ u1 ∂ t =∂ ∂ t = [AI e i (wt – k

1x) + AR e i (wt + k

1x)]

= i w AI e i (wt – k1

x) + i w AR e i (wt + k1

x)]

= i w e i w t (AI + AR)

∂ u2 ∂ t =

∂ ∂ t = [AT e i (wt – k

2x)] = i w AT e i (wt – k

2x)

= i w AT e i w t

Transmission of Sound

equate v1 = v2

Acoustic Impedance: Z- the ratio (real or complex) of pressure and

its velocities.

i w e i w t (AI + AR) = i w AT e i w t

(AI + AR) = AT

P V

=Z=Z ρ C ; rayls

2


substitute Z and eqn 2 in eqn 1:

ρ1 C1 (AI – AR) ρ2 C2 w AT=

Z1 (AI – AR) Z2 (AI + AR)= Z1AI – Z1AR Z2AI + Z2AR= AI (Z1– Z2) AR (Z1+ Z2)=

(Z1– Z2) (Z1+ Z2)

= AI

AR


from eqn 2:

substituting eqn 2’ in eqn 1:

AT - AI = AR 2’

2 Z1

(Z1+ Z2)=

AI

AT 2”


I. Sound Reflection & Transmission Coefficients:

1. Sound Reflection Coefficient; £r

=£r reflected sound energy

incident sound energy


for reflected wave:

for incident wave:

I =

½ w2 cA2

ρ

II = ½ w2 c1AI2

ρ

IR =

½ w2 c1AR2

ρ

=£r½ w2 c1AR

2

ρ ½ w2 c1AI

2

ρ = AR AI

( )2

=£r Z1– Z2

Z1+ Z2[ ]2


1. Sound Transmission Coefficient;

for transmitted wave:

£T

=£T transmitted sound energy

incident sound energy

IT =

½ w2 c2AT2

ρ 2

=£T½ w2 c2AT

2

½ w2 c1AI2

ρ ρ

2

1

=c2AT

2

c1AI

2

ρ ρ

2

1

Z=ρCbut,


Transmission Loss; TL

=£TZ2AT

2

Z1AI

2

from 2”, 2 Z1

(Z1+ Z2)= AI

AT

=£TZ2 Z1

2 Z1

Z1 + Z2

( ) 2

=£T4 Z1 Z2 ( Z1 +Z2 )2

= =TL 10 logIi IT

10 logIT Ii

1 £T = 10 log

1

£T


Transmission of Sound; TL

TL = 10 log1

£T

Transmission of SoundII.For Normal Incidence at Surface of Solids:

Incident Wave

Medium 1 Medium 2

Solid Surface


where:rn = resistive component of ZnXn = reactive component of ZnZn = rn + jXn

=£r (rn - c1)2 + Xn

2ρ 1

(rn + c1)2 + Xn2ρ

1

=£T

4 c1 rnρ (rn + c1)2 + Xn

2ρ 1

1


III. For transmission of Sound Waves from one fluid to another at oblique

incidence:

Incident Wave

Medium 1 Medium 2

Interface


where:θi = angle of incidenceθT = angle of refraction

=£T

4 R1R2 cos θi cos θT

(R2 cos θi + R1 cos θT)2

=£r R2 cos θi – R1 cos θT

R2 cos θi + R1 cos θT( ) 2


IV. For sound waves in air striking at oblique incidence on the surface of a normally reacting solid:

Incident Wave

Medium 1 Medium 2

Solid


where:rn = resistive component of ZnXn= reactive component of Znθi = angle of incidence

=£T

4 R1rn cos θi (rn cos θi + R1)2 + Xn

2 cos2 θi

=£r (rn cos θi – R1)2 + Xn

2 cos2 θi

(rn cos θi + R1)2 + Xn2 cos2 θi


V. Transmission of sound through three mediums

L

Medium 1 Medium 2 Medium 3


L = thickness of medium 2k2 = wave no. of medium 2 =

=£T

4 R1 R3

(R1R3)2 cos2 k2L + (R2 + ) sin2 k2L

Wc2

R1R3

R2

TL = 10 log1

£T

Sound waves in air with an intensity of 0.02 Sound waves in air with an intensity of 0.02 W/mW/m22 is incident normally on a boundary. What is is incident normally on a boundary. What is the absorption of the boundary if the reflected the absorption of the boundary if the reflected sound intensity is 0.005 W/msound intensity is 0.005 W/m22??

α =

Ia = Ii - Ir

Soln:

interface0.005 W/m2

0.02 W/m2

IaIi

= 0.02 – 0.005 Ia = 0.015 W/m2

α = 0.0150.02

α = 0.75

( )

A beam of sound waves is incident normally on a A beam of sound waves is incident normally on a plane interface of air & an unknown body. If half plane interface of air & an unknown body. If half of the sound energy is reflected, what is the of the sound energy is reflected, what is the impedance of the unknown body?impedance of the unknown body?

= (1.21) (341)

interface

unknown bodyair

Zair = ρairCair

Soln:

For air:

Zair = 412.61 rayls=£r 0.5 =£r

Z1– Z2

Z1+ Z2

2 ; but,

0.5 =( )412.61 – Z2

412.61 + Z2

2

Z2 = 70.79 rayls

An acoustic tile panel has a normal specific acoustic An acoustic tile panel has a normal specific acoustic impedance of 1000 – j1300 rayls. Calculate the sound impedance of 1000 – j1300 rayls. Calculate the sound transmission and reflection coefficients for plane acoustic transmission and reflection coefficients for plane acoustic waves in air incident normally of the surface of the tile waves in air incident normally of the surface of the tile panel.panel.

solid surface(tile panel)

air

= (rn - c1)2 + Xn

2ρ 1

(rn + c1)2 + Xn2ρ

1

; 1c1 = Zair = 412.61 raylsρ

Soln:

£r = (rn + c1)2 + Xn

2

(1000 – 412.61)2 + (-1300)2= (1000 – 412.61)2 + (-1300)2

(1000 + 412.61)2 + (-1300)2

£r = 0.55£T + £r = 1

£T = 1 - £r

= 1 – 0.55£T = 0.45

Plane acoustic waves in air strike the surface of Plane acoustic waves in air strike the surface of an acoustic tile panel having an impedance of an acoustic tile panel having an impedance of 1000 – j1300 rayls. Find the reflection coefficient 1000 – j1300 rayls. Find the reflection coefficient for an angle incidence at 80°.for an angle incidence at 80°.

=£r

Soln: (rn cos θi – 1c1)2 + Xn

2 cos2 θi

(rn cos θi + 1c1)2 + Xn2 cos2 θi

ρ ρ

= (1000 cos 80° – 412.61)2 + (-1300)2 cos2 80° (1000 cos 80° + 412.61)2 + (-1300)2 cos2 80°

=£r 0.27

Plane acoustic wave in water is incident normally on the Plane acoustic wave in water is incident normally on the surface of a large steel plate of thickness 0.02m. If the surface of a large steel plate of thickness 0.02m. If the frequency of the wave is 3khZ, find the transmission loss frequency of the wave is 3khZ, find the transmission loss through the steel into water on the opposite sides.through the steel into water on the opposite sides.

steel

H2O H2O

L = 0.02m

=£T

Soln: For water:Z = 1.48 x 106 rayls

For air:Z = 39 x 106 raylsC = 5050 m/sec

R1 = RH2O = R3

R2 = steel

4 R1 R3

(R1R3)2 cos2 k2L + (R2 + ) sin2 k2LR1R3

R2

=£T4 R1

2

4R12 cos2 k2L + (R2 + ) sin2 k2L R1

2

R2

Soln:=£T

4 R12

4R12 cos2 k2L + (R2 + ) sin2 k2L R1

2

R2

=k2Wc2

=2π (3000) 5050

=k2 3.73 rad/m

=£T4 (1.48 x 106)2

4(1.48 x 106)2 cos2 (3.73)(0.02) + (3.9 x 106 + ) sin2 (3.73)(0.02) (1.48 x 106)2

(3.90 x 106)

=£T 0.5

TL = 10 log 1 £T

TL = 10 log 1 0.5

TL = 3.01 dB

Room Room AcousticsAcoustics

Room AcousticsRoom AcousticsRoom Acoustics is concerned with the behavior Room Acoustics is concerned with the behavior of sound within an enclosure space with a view of sound within an enclosure space with a view obtain the optimum acoustic effect on the obtain the optimum acoustic effect on the occupantsoccupants

Requirements:

1. An adequate amount of sound must reach all parts of the room. Most attenuation in this respect needs to be given to those seats further from the source.

2. An even distribution of sound should be achieved throughout the room, irrespective of distance from the source.

3. Other noise which might tend to mask the required sound must be reduced to an acceptable level in all parts of the room.

4. The rate of decoy of sound within the room should be optimum for the required use of the room. This is to ensure clarity for speech and fullness of music.

Optimum Reverberation Time (at 500 to 1000Hz)Optimum Reverberation Time (at 500 to 1000Hz)

Room FunctionRoom Function Reverberation Reverberation Time Time

Recording and Broadcast studiosRecording and Broadcast studios 0.45 – 0.550.45 – 0.55Elementary ClassroomsElementary Classrooms 0.60 – 0.800.60 – 0.80Playhouses, Intimate Drama ProductionPlayhouses, Intimate Drama Production 0.90 – 1.100.90 – 1.10Lecture and Conference RoomsLecture and Conference Rooms 0.90 – 1.100.90 – 1.10CinemaCinema 0.80 – 1.200.80 – 1.20Small TheatersSmall Theaters 1.20 – 1.401.20 – 1.40High School AuditoriumsHigh School Auditoriums 1.50 – 1.601.50 – 1.60General Purpose AuditoriumsGeneral Purpose Auditoriums 1.60 – 1.801.60 – 1.80Churches (Cathedrals)Churches (Cathedrals) 1.40 – 3.401.40 – 3.40

Reverberation TimeReverberation Time

When a sound is emitted in a closed When a sound is emitted in a closed environment such as a room or auditorium it environment such as a room or auditorium it takes a certain amount of time for the intensity takes a certain amount of time for the intensity of sound to dissipate. This is because the sound of sound to dissipate. This is because the sound reflects off the walls and the people and objects reflects off the walls and the people and objects in the room, and dies down only as a in the room, and dies down only as a consequence of the absorption of some of the consequence of the absorption of some of the energy by each object at each reflection.energy by each object at each reflection.

Reverberation TimeReverberation Time is defined as the time it is defined as the time it takes for the intensity of a given steady sound to takes for the intensity of a given steady sound to drop 60 dB from the time the sound source is drop 60 dB from the time the sound source is shut off.shut off.

Reverberation time depends on the total Reverberation time depends on the total acoustic energy pervading the room, the acoustic energy pervading the room, the surface areas of the absorbing material and surface areas of the absorbing material and their absorption coefficient.their absorption coefficient.

A formula that gives good estimates of the A formula that gives good estimates of the Reverberation time is given by:Reverberation time is given by:

t = 0.16 V

-S ln (1-α)

; sec, for room dimensions in meters

t = 0.05 V

-S ln (1-α)

; sec, for room dimensions in feet

where:

V = volume of the room

S = surface area at the boundary

α = absorption coefficient of the boundary

Reverberation time formula derivation:Reverberation time formula derivation:Snowball’s Law

∂ Q∂ t

∞ Q

∂ Q∂ t

= + k Q

growth

decay

Let E = sound energy unabsorbed∂ E∂ t

∞ E

∂ E∂ t

= - k E

Reverberation time formula derivation:Reverberation time formula derivation:∂ E∂ t

= - k ∂t Integrate:

∫ ∂ E E

Eo

E

= - k ∂t ∫0

t

ln E | = - k t |Eo

E

0

t

ln E – ln Eo = - k t

Reverberation time formula derivation:Reverberation time formula derivation:

ln = - k tEEo

eln = e-ktEEo

lne = e-k tEEo

1

= e-k tEEo

1

Reverberation time formula derivation:Reverberation time formula derivation:by definition:

= 10-6EEo

When sound wave is transmitted, the sound energy left after the first reflection is,

E = Eo – EoaE = Eo(1 – a)For n reflections: the sound energy left is,

E = Eo(1 – a)n

= (1 – a)nEEo

2

Reverberation time formula derivation:Reverberation time formula derivation:velocity; V

V = dt

d = V tV = Cair = 341 m/sec

d = (341) tFor n reflections: the distance covered is,

n d = (341) tn = (341) t

dequate: eqn 1 = eqn 2

Reverberation time formula derivation:Reverberation time formula derivation:e-kt = (1-a)n

ln e-kt = ln (1-a)n

-kt ln e = n ln (1-a)1-kt

ln (1-a)n =

(341) td

= -k tln (1-a)

k = -341 ln (1-a)d

-13.82 = t

10-6 = e –[ ]t

Reverberation time formula derivation:Reverberation time formula derivation:Also, in mathematics:

4 Vs

d =V = volume of the rooms = surface area of the room

substitute k in eqn 1:-341 s ln (1-a)

4V

ln (10-6) = ln e –[ ]t-341 s ln (1-a)4V

1

-341 s ln (1-a)4V


t = 0.16 V-s ln (1-a)

; sec metric dimension

translate:Cair = 341 m/sec ft/sec

= 341 m/sec (3.28 ft/m)Cair = 1118.5 ft/sec

-13.82 = t-1118.5 s ln (1-a)4V

t = 0.049 V-s ln (1-a)

; sec english dimension


let: Σ sα = s ln (1-a)

t = 0.16 VΣ sα

t = 0.049 VΣ s α

&

Σ sα = s1a1 + s2a2 + . . . + snan

Other Equations Quantifying Reverberation Other Equations Quantifying Reverberation TimeTime

A. Stephen and Bate Equation A. Stephen and Bate Equation (for ideal (for ideal reverberation time computation)reverberation time computation)

t = r (0.012 V + 0.10703

where:

V = room volume in m3

r = 4 for speech = 5 for orchestra = 6 for choir

Optimum volume/person (in mOptimum volume/person (in m33) for various ) for various types of hall:types of hall:

Types of Hall Optimum Volume/person (in m3)

Concert Halls 7.1Italian Type Opera Houses 4.2 – 5.1Churches 7.1 – 9.9Cinemas 3.1Rooms for speech 2.8

B. Sabines Formula B. Sabines Formula (for actual reverberation time with average (for actual reverberation time with average absorption less than or equal to 0.2)absorption less than or equal to 0.2)

t = 0.16 V

A

t = 0.16 V

sα

and:

For room dimensions in metric:

t = 0.05 V

A

t = 0.05 V

sα

and:

For room dimensions in english:

A lecture room has a volume at 15,000 mA lecture room has a volume at 15,000 m33 if it has 1000 m if it has 1000 m22 of acoustic (of acoustic (αα=0.6), 2400 m=0.6), 2400 m22 at plaster ( at plaster (αα=0.03), 1000 m=0.03), 1000 m22 at concrete (at concrete (αα=0.02), and 4000 m=0.02), and 4000 m22 of wood ( of wood (αα=0.05). Find =0.05). Find the reverberation time of the room.the reverberation time of the room.

t = 0.16 V

Σ sα

Σ sα = s1a1 + s2a2 + s3a3 + s4a4

Σ sα = 1000(0.6) + 2400(0.03) + 1000(0.02) + 4000(0.05)

Σ sα = 892 sabines

t = 0.16 (15000)

892

t = 2.69 sec

Soln:

V = 15,000 m3

A lecture room has a volume at 15,000 mA lecture room has a volume at 15,000 m33 if it has 1000 m if it has 1000 m22 of acoustic (of acoustic (αα=0.6), 2400 m=0.6), 2400 m22 at plaster ( at plaster (αα=0.03), 1000 m=0.03), 1000 m22 at concrete (at concrete (αα=0.02), and 4000 m=0.02), and 4000 m22 of wood ( of wood (αα=0.05). Find =0.05). Find the reverberation time when it holds 200 people.the reverberation time when it holds 200 people.

t = 0.16 V

Σ sα

Σ sα = s1a1 + s2a2 + s3a3 + s4a4 + s5a5

Σ sα = 1000(0.6) + 2400(0.03) + 1000(0.02) + 4000(0.05) + 200(4.7)

Σ sα = 1,832

t = 0.16 (15000)

1833

t = 1.3 sec

Soln:

Find the optimum reverberation time of 500Hz of a living room Find the optimum reverberation time of 500Hz of a living room 20ft long, 13ft wide & 8 ft high with a plaster ceiling 20ft long, 13ft wide & 8 ft high with a plaster ceiling ((αα=0.02)=0.02), a , a carpeted floor carpeted floor ((αα=0.3)=0.3), a wood-paneled sidewall , a wood-paneled sidewall ((αα=0.12)=0.12) & & opposite glass wall opposite glass wall ((αα=0.03)=0.03), an end wall of medium drapery , an end wall of medium drapery ((αα=0.4)=0.4) & a brick fireplace & a brick fireplace ((αα=0.02)=0.02) for the other wall. for the other wall.

t = 0.049 V

Σ sα

Soln:

Σ sα = [(13)(20)(0.02) + (13)(20)(0.3) + (8)(20)(0.12) + (8)(20)(0.03) +(8)(13)(0.4) + (8)(13)(0.02)

Σ sα = 150.88 sabines

t = 0.049 (20)(13) (18)

150.88

t = 0.676 sec

A shower room has a dimension of 5x4x3 m. All the walls A shower room has a dimension of 5x4x3 m. All the walls are at tile, and the door has the same absorption are at tile, and the door has the same absorption coefficient as tile (coefficient as tile (αα=0.03). Find the reverberation time if =0.03). Find the reverberation time if one man is showering and singing “my way”.one man is showering and singing “my way”.

t = 0.16 V

Σ sα

Soln:

Σ sα =(20 + 20 + 12 +12 +15 +15) (0.03) + 4.7

Σ sα =7.52

t = 0.16 (60)

7.52

t = 1.28 sec

A further correction need to be added for higher A further correction need to be added for higher frequency to allow for air absorption, reverberation frequency to allow for air absorption, reverberation time may be expressed as,time may be expressed as,

t = 0.16 V

-s ln (1-α) + X V

X = sound absorption per unit volume of air

X per mX per m33 at a temperature of 20° at a temperature of 20°C

Frequency

(Hz)

30% RHX 10-3

40% RHX 10-3

50% RHX 10-3

60% RHX 10-3

70% RHX 10-3

80% RHX 10-3

1000 3.28 3.28 3.28 3.28 3.28 3.28

2000 1.48 8.2 8.2 6.56 6.56 6.56

4000 39.36 29.52 22.96 19.68 16.4 16.4

p.H. = relative humidity

Calculate the amount of absorption contributed at 2000Hz by the Calculate the amount of absorption contributed at 2000Hz by the 30,000m30,000m33 of air at a cathedral when the relative humidity is 60%. of air at a cathedral when the relative humidity is 60%. If its reverberation time empty at 2000Hz is 4seconds, find the If its reverberation time empty at 2000Hz is 4seconds, find the number of square meters of absorbent in the structure.number of square meters of absorbent in the structure.

Soln:

let A = -s ln (1-a)

t = 0.16 V

A + XV

A = 0.16 V

t- XV

A = 0.16 (30,000)

4- (6.56 x 10-3) (30,000)

A = 1003.2

Important Equations on Enclosed RoomsImportant Equations on Enclosed Rooms

1. Direct Field Sound Intensity : I1. Direct Field Sound Intensity : IDD

ID = W Q

4 π d2 : w/m2

where:

W = actual radiated power rating of the source in watts

Q = directivity

d = distance from source in meters

2. Reverberant Field Sound Intensity : Ir

Ir = 4 Wa

R

R =

S α

1- α

where:

R = room constant = absorbing power of the roomα = average room absorption coefficient

S = total surface area of the boundery

How much acoustic power must a public address system be able How much acoustic power must a public address system be able to put out in order to create SPL = 100 dB for a musical show in a to put out in order to create SPL = 100 dB for a musical show in a 101055 m m33 auditorium with reverberant time of 1.2 sec. auditorium with reverberant time of 1.2 sec.Soln:

t = 0.16 V

A

A = 0.16 V

t

A= 0.16 (105)

1.2

A= 13,333.3

IL = SPL = 100 dB

IL = 10 log I

10-12

I = (antilog ) 10-12 1W

10 I = 0.01 w/m2

Ir = 4 W

R R = A

W = Ir A

4

W = (0.01) (13,333.3) 4

W = 33.33 watts

What is the reverberant sound level at the 100 mWhat is the reverberant sound level at the 100 m33 room, room, reverberation time of 1.2 sec and moderately vigorous reverberation time of 1.2 sec and moderately vigorous speaking voice can produce acoustic power on the order of speaking voice can produce acoustic power on the order of 100 100 µW?µW?

Soln:

Ir = 4 Wa

R

t = 0.16 V

A

A = 0.16 V

t

A = 0.16 (100)

1.2

A = 13.33

4 (100x10-6)

A = R

Ir = 4 Wa

A

Ir = 13.33

Ir = 30 µw/m2

IL = 10 log I

10-12

IL = 10 log 30x10-6

10-12

IL = 75 dB

PUBLIC ADDRESS-SYSTEM DESIGN PUBLIC ADDRESS-SYSTEM DESIGN AND LAYOUTAND LAYOUT

Choose the number, type, and location of the speakers Choose the number, type, and location of the speakers as well as the amplifier rating for a factory sound as well as the amplifier rating for a factory sound system. The factory floor served by the system is 80ft system. The factory floor served by the system is 80ft (24.4m) long and 100ft (30.5m) wide. It is desired to use (24.4m) long and 100ft (30.5m) wide. It is desired to use the public-address system for both voice and music the public-address system for both voice and music amplification.amplification.

Calculation Procedure:1. Compute the number of speakers required.

The factory floor area is L X W = 80 X 100 = 8000ft2 (743.2m2). Using this area, enter Table 17 at the line marked Factories. This line shows that an 8000ft2 (743.2m2) factory requires four speakers.

Calculation Procedure:2. Select the type of speaker to use.

Table 17 shows that the reentrant type horn speaker is recommended for factories having floor areas of 8000 or more ft2 (743.2 m2). With only four speakers, each can operate at high output in a high-level speaker system. With a large number of speakers, eight or more, they can be the low-output type. This is termed a low-level speaker system. For this factory having four speakers, a high-level speaker system probably would be best.

Calculation Procedure:3. Select the public-address-system amplifier rating.

Table 17 shows that a 50-W amplifier would be suitable for this factory. The output required for a public-address-system amplifier depends on the size and type of area served by the sound system. Where music is to be played over the sound system, a record player can be built into the amplifier housing. Where clarity of the amplified sound is critical, anti-feedback controls are available on most standard amplifiers.

Calculation Procedure:3. Select the public-address-system amplifier rating.

Table 17 shows that a 50-W amplifier would be suitable for this factory. The output required for a public-address-system amplifier depends on the size and type of area served by the sound system. Where music is to be played over the sound system, a record player can be built into the amplifier housing. Where clarity of the amplified sound is critical, anti-feedback controls are available on most standard amplifiers.

4. Select the amplifier power source.

Standard sound amplifiers can be operated from 110- to 125-V 60-Hz or 115-V 25-Hz alternating current, or 115-, 4.6-, or 12-V direct current.

Calculation Procedure:5. Choose the amplifier Input devices.

There are five major types of input devices: (a) Microphones – crystal, dynamic, or velocity, omni-directional, bidirectional, or unidirectional (cardioid); (b) record player: automatic or manual; (c) tape/CD player: single- or multiple-track; (d) radio tuner: AM or FM; (e) tone generator: produces a tone signal for factory work shifts, lunch periods, etc.; also used as an electronic siren for alarm applications; bell sounds for church belfry.

Calculation Procedure:6. Choose the output-tap impedance values.

The output-tap impedance value depends on the type of output devices selected in step 5 because there must be an impedance match. In general, three types of output taps are used: (a) Direction connection – 4-, 8-, and 16-Ω taps; (b) constant-voltage line-transformer connection with 70-, 100-, or 140-V taps; (c) constant-impedance line-transformer connection with 250- and/or 500-Ω taps. Select one or more taps to give the desired impedance match with the output device or devices.

Calculation Procedure:7. Choose the type of speaker connection.

There are three main types of speaker connections: (a) Direct connection to the amplifier output taps corresponding in impedance value (ohms) to the impedance value (ohms) of a single speaker or a number of speakers in series, parallel, or series-parallel; (b) connection to the amplifier constant-voltage output taps at 70, 100, 140 V, etc., through constant-voltage line-matching transformers; (c) connection to amplifier high-impedance output taps of 250- or 500-Ω impedance through constant-impedance line-matching transformers.

Calculation Procedure:8. Select the speaker locations.

Locate the speakers on a plan of the area served. In churches, theaters, and auditoriums, place the speakers well forward of the microphones to prevent feedback (squealing).

Related Calculation: Use this general method to choose sound systems for auditoriums, ballrooms, churches, classrooms, offices, stores, factories, funeral parlors, restaurants, nightclubs, stadiums, gymnasiums, etc. For impedance values of the equipment selected, consult the manufacturer’s engineering data The method presented here is one recommended by Electrical Construction and Maintenance magazine.

Calculation Procedure:

How Speakers Work

image above shows a basic speaker and its parts as well as its cross-section

How Speakers Work

A speaker is essentially the final translation machine -- the reverse of the microphone. It takes the electrical signal and translates it back into physical vibrations to create sound waves. The voice coil is a basic electromagnet. Running electrical current through the wire creates a magnetic field around the coil, magnetizing the metal it is wrapped around. The field acts just like the magnetic field around a permanent magnet: It has a polar orientation -- a "north" end and a "south" end -- and it is attracted to iron objects. But unlike a permanent magnet, in an electromagnet you can alter the orientation of the poles. If you reverse the flow of the current, the north and south ends of the electromagnet switch. This is exactly what a stereo signal does -- it constantly reverses the flow of electricity.

ExercisesExercises

1. The lower the frequency of the wave,

a ) the higher its velocity c) the smaller its amplitude b) the longer its wavelength d) the shorter its period

2. An entirely longitudinal wave is

a) a water wave b) a sound wave c) an electromagnetic wave d) a wave in a sketched string

3. Sound cannot travel through a a) vacuum b) gas c ) liquid d) solid

4. An 80 dB sound relative to a 30 dB sound is more intense by a factor of

a) 5 b) 50 c) 500 d) 105

5. A sound intensity level of 55 dB is produced by 10 flutes. The number of flutes needed to produce a level of 65 dB under the same circumstances is

a) 20 b) 60 c) 100 d) 200

6. What quantity is carried by all types of waves from their source to the place when they are eventually absorbed?

a) energy b) power c) wave motion d) displacement

7. How many times more intense that the 60 dB sound of a person talking loudly is the 100 dB sound of a power lawn mower?

a) 1000 b) 100 c) 10,000 d) 100,000

8. Sound wave whose intensities exceed about 1w/m2 cause damage to the ear. How many decibels is this equivalent to?

a) 110 dB b) 120 dB c) 10 dB d) 100 dB

9. A certain jet aircraft produces sound of 140 dB intensity at a distance of 100 ft. How many miles away is the intensity at 90 dB?

a) 6 mi b) 100 m c) 100 ft d) 4 km

10. A certain person speaking normally produces a sound intensity of 40 dB of a distance of 3ft. If the threshold intensity for reasonable audibility is 20 dB, how far away can the person be heard clearly?

a) 10 ft b) 5 ft c) 30 ft d) 100 ft

11. A sound has an intensity of 9.2 µw/m2. What is the sound intensity level of this sound?

a) 92.8 dB b) 90 dB c) 69.6 dB d) 75.8 dB

12. A sound has a sound intensity level of 65.3 dB. What is the intensity of this sound?

a) 3.4 µw/m2 b) 10 µw/m2 c) 5.8 µw/m2

d) 10 µw/m2

13. A note of frequency 300 vib/sec has an intensity of 1.0 µw/m2. What is the amplitude of the air vibrations caused by this sound?

a) 36 nm b) 72µm c) 20 nmd) 80 µm

14. For a steed rod, Young’s modulus is 2x1011 Pa and a density of 7.8kg/m3. What is the speed of sound in the steel rod?

a) 5050 m/sec b) 5064 m/sec

c) 6000 m/secd) 4850 m/sec

15. The threshold of pain has a sound intensity level of 120 dB. What is the equivalent sound intensity?

a) 10 w/m2 b) 1.0 w/m2

c) 0.5 w/m2

d) 0.1 w/m2

16. If the pressure amplitude is a sound wave is doubled, by what factor does the intensity of the wave increase?

a) 2 b) 3

c) 4 d) 5

17. By what factor must the pressure amplitude of a sound wave be increased in order to increase the intensity by a factor of 9?

a) 2 b) 3 c) 4d) 5

18. Sound wave in air has a displacement amplitude of 0.0140 mm. Calculate the pressure amplitude for frequency at 500 Hz.

a) 18.2 Pa b) 20 µN/m2

c) 28.4 Pa d) 50 µN/m2

19. A sound wave in air has a frequency of 400 Hz, a wave speed at 344 m/sec and a displacement amplitude of 0.00750 mm. Calculate the sound intensity level for this sound wave.

a) 80.2 dB b) 83.3 dB

c) 93.2 dB d) 109 dB

20. Sound source radiates uniformly in all directions in air at 20 0C. At a distance 80 m from the source, the sound level is 80 dB. The frequency is 440 Hz. What is the displacement amplitude?

a) 0.25 µm b) 25 µm

c) 250 µm d) 2.5 µm

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