2UNIT OUTLINETo achieve this outcome students should demonstrate the knowledge and skills to;explain sound as the transmission of energy via longitudinal pressure waves;mathematically model the relationship between wavelength, frequency and speed of propagation of sound waves using v = fλexplain the difference between sound intensity (Wm-2) and sound intensity level (dB)calculate sound intensity at different distances from a source using an inverse square law.explain resonance in terms of superposition of a travelling sound wave and its reflectionanalyse, for strings and open and closed resonant tubes the fundamental as the 1st harmonic and subsequent harmonicsexplain qualitatively, in terms of electrical and electromagnetic effects the operation ofmicrophones, including electret - condenser, crystal, dynamic and velocity microphonesdynamic loudspeakersexplain qualitatively the effects of baffles and enclosures for loudspeakersinterpret frequency response curves of microphones, speakers, simple sound systems and hearing including loudness (phon)evaluate the fidelity of microphones and loudspeakers in terms of purpose, frequency response and qualitatively constructioninterpret qualitatively the directional spread of various frequencies in terms of different gap width or obstacle size including the significance of the magnitude of the λ/w ratiouse safe and responsible practices when working with sound sources and sound equipment
3Chapter 1 Topics covered: Wave nature of Sound. Transverse Waves. Longitudinal Waves.Sound Production, Transmission, Detection and Absorption
41.0 The Wave Nature of Sound Waves are a method of TRANSFERRING ENERGY from one place to another.Some waves (eg. Sound, Water Waves) need a MEDIUM through which to travel.The MEDIUM (eg. air, water), although disturbed by the passage of the waves, does NOT suffer any PERMANENT DISTORTION due to the wave’s movement through it.The “WAVE” nature of SOUNDSound Waves require matter (either solid, liquid or gas) as their mediumThis means, of course, no one can hear you scream in space.There are two basic types of waves:TRANSVERSE WAVES.LONGITUDINAL WAVES.
5Direction of motion of wave 1.1 Transverse WavesTransverse waves are characterised by having the individual particles of the medium through which the wave travels, moving perpendicular to the direction of motion of the wave.Direction ofmotion of themedium’sparticles1.1 Transverse WavesDirection of motion of waveNotice the “medium” does not move along with the wave.Pick a spot and follow its motion.
61.2 Longitudinal WavesLONGITUDINAL WAVES are characterised by having the individual particles which make up the medium through which the wave travels, moving parallel to the direction of motion of the wave.Sound is a LONGITUDINAL WAVE.Individual particlesof MediumDirection of Wave MotionDirection of Motion ofParticles of MediumAgain, notice the “medium” does not move along with the wave.Pick a spot and follow its motion.
7Question 1In the sentences below, options are given within the brackets. Only one of the options will be correct. Circle the best option.A sound wave is a [torsional / transverse / longitudinal] wavein which the air particles move [at right angles to / parallel to /by spiralling around] the direction of propagation of the wave.The wave transmits [energy / air particles / wave maxima]from the source to the receiver.
81.3 Sound ProductionSound is produced by making an object vibrate (move backward and forward).As the object vibrates back and forth, it pushes on the air particles immediately next to it, creating a series of COMPRESSIONS and RAREFACTIONS which move outward from the source.This moving chain of compressions and rarefactions form a Sound Wave.CompressionRarefactionThe faster the object vibrates, the higher the frequency of the sound.
9Consider a dust particle one metre in front of a loudspeaker that is producing a constant tone sound wave.Question 2Which one of the following statements and diagrams (A to D below) best describes the motion of the dust particle?The dust particle oscillates in a vertical direction.The dust particle travels away from the speaker with the wave.The dust particle remains stationary as the wave passes.The dust particle oscillates in a horizontal direction
101.4 Sound TransmissionSound is transmitted from one place to another through a MEDIUM.The medium may be solid, liquid or gas.Generally the DENSER the medium the FASTER the speed of sound.Sound is transmitted through a medium by causing the particles of the medium to be disturbed from their mean or average positions as the wave passes by.The particles making up the medium DO NOT move along with the sound wave.The medium suffers no permanent “effect” from having a sound wave pass through it.
11A particle of dust is floating at rest 10 cm directly in front of a loudspeaker that is not operating.The loudspeaker then emits sound of frequency of 10 Hz and speed of 330 ms–1.Question 3Which one of the following statements best describes the motion of the dust particle?A. It vibrates vertically up and down at 10 Hz remaining on average 10 cm in front of the loudspeaker.B. It vibrates horizontally backwards and forwards at 10 Hz remaining on average 10 cm in front of the loudspeaker.C. It travels away from the loudspeaker at 330 ms–1 while moving horizontally backwards and forwards at 10 Hz.D. It remains at rest.
121.5 Sound Detection Sound is detected by making a receiver vibrate. Sound detection occurs in devices such as microphones where incoming sound waves cause the production of an electrical signal.Sound is detected by humans using our ears, in particular our Cochlea, a circular canal lined with clumps of hairs. Each hair clump is designed to react to a particular frequency.Sound Level Meters are used to measure Sound Intensity Levels, which are displayed in decibels (dB).The output from most sound level meters is adjusted to mirror the ear’s response by using the so called dB(A) scale. (see Slide Frequency Response Graphs)
131.6 Sound AbsorptionWhen a sound strikes a barrier, it is either reflected off, transmitted through or absorbed by, that barrier.The amount of reflection, transmission or absorption depends upon the nature of the barrier.The physical absorption of sound, as measured by the ABSORPTION COEFFICIENT (A.C.), occurs when the energy of the wave is transformed into other forms of energy (eg. Heat) within the absorbing material.The A.C. varies with frequency.Hard, rigid, non-porous materials have low A.C.’sSoft, pliable, porous materials have high A.C.’s
14Chapter 2 Topics covered: Amplitude. Period. Frequency. Wavelength. Wave Speed.Sound Waves in Air.
152.0 AmplitudeAmplitude is a measure of the size of a disturbance above or below a mean or average value.In sound wave representations, the amplitude is measured as a variation in air pressure (P), above or below the normal atmospheric pressure.This method allows sound to be presented as a transverse rather than a longitudinal wave.Point of Max.Pressure aboveAtmospheric, a COMPRESSIONPTimeAmplitudeAtmosphericPressureThe unit for P is the PASCAL (1 Pa = 1 Nm-2).Point of Min.Pressure belowAtmospheric, a RAREFACTIONHuman ears interpret Amplitude as the “Loudness” of a sound:Large amplitude = loud sound,Small amplitude = soft sound
162.1 FrequencyFrequency (symbol f ) is most generally defined as the number of events which occur during a time interval.In terms of Sound Waves it represents the number of complete sound waves passing a given point in a given time.In the SI system, frequency is defined as the number of events or cycles per second.The UNIT for frequency is the HERTZ (Hz), where 1 Hz = 1 cycle per secondPTimeLow FrequencyHigh FrequencyHuman ears interpret frequency as the “pitch” of a sound:High frequency = high pitch,Low frequency = low pitch
172.2 Period Period (T) = 0.02 s f = 1/T = 1/0.02 = 50 Hz Time0.020.04Period (symbol T) is defined as the time it takes for one event to occur.It is the time it takes for one complete sound wave to pass a given point.Period is the measure of a time interval, thus has the unit seconds (s).Period and frequency are the inverse of one another thus:Period (T) = 0.02 sf = 1/T= 1/0.02= 50 HzThus, a wave of period 0.02 s has a frequency of 50 Hz
182.3 WavelengthλWavelength, (symbol , Greek Letter LAMBDA), is a measure of the distance between two adjacent points on a wave undergoing similar motions.Thus the distance between two adjacent compressions or two adjacent rarefactions would be 1 wavelength.CompressionRarefactionλλWavelength is a distance measure, hence the unit for is metres (m).
19Rachel and Bruce have assembled some laboratory equipment and are planning a series of sound-related experiments.An audio-signal generator is used to drive a small loudspeaker, which emits sound uniformally in all directions. The audio power from the loudspeaker is kept constant at all frequencies used in the experiments. A sound levelmeter is used to measure sound intensity. This is shown in Figure 2.Initially, the frequency of the signal generator is set to 476 Hz.The speed of sound at the time of the experiment was 340 ms-1.Question 4Calculate the wavelength of the 476 Hz sound wave. Include a unit in your answer.λ = v/f = 340/476 = 0.71 m
202.4 Wave SpeedThe relation is summarised in the so called “WAVE EQUATION”,v = fWave Speed (symbol v) is a measure of how quickly a “wave train” is moving.where; v = Speed (ms-1), f = Frequency (Hz) = Wavelength (m).The wave speed is dependent on the frequency and wavelength of the wavetrain.The speed of Sound in Air is temperature dependent and is approx 340 ms-1 at 200CSound travels faster through denser mediums
21Roger, an instrument maker, is constructing and testing pipes for a pipe organ. He measures the speed of sound in air at the time of the test to be 333 ms–1.Question 5One pipe is designed to produce the note middle C (256 Hz).Which one of the following best gives the wavelength corresponding to middle C?A mB mC. 1.3 mD. 2.6 m
222.5 Sound Waves in AirIn air, the passage of a sound wave causes a series of COMPRESSIONS and RAREFACTIONSThis means the air particles must vibrate back and forth around their MEAN, AVERAGE or CENTRAL POSITION.In areas of above average air pressure (Compressions), the particles are packed CLOSE TOGETHER.So only small scale vibrations are needed for them to transfer their “information” (sound wave energy), to adjacent particles.In areas of below average air pressure (Rarefactions) the particles are SPREAD APART.So large scale vibrations are needed for information to be transferred to adjacent particles.Energy lost per transfer is low and the sound travels a greater distance than at normal air pressureThe energy lost per transfer is high and so sound travels a lesser distance than at normal air pressure.
23Chapter 3 Topics covered: Sound Intensity. Sound Intensity Level. The Decibel Scale.Frequency Response Graphs.Sound Intensity versus DistanceHuman Response
243.0 Sound Intensity Area (A) Sound Energy (P) The INTENSITY of a sound is DEFINED as THE RATE OF FLOW OF ENERGY through an area perpendicular to the direction of travel of the sound wave.Area (A)Sound Energy (P)The rate of flow of energy is the definition of POWER.In this case the power is ACOUSTICAL POWER.Thus SOUND INTENSITY is defined as POWER/AREA.Mathematically: I = P/A Where,I = Sound Intensity (Wm-2)P = Total Acoustical Power (W)A = Area (m2)
25It is a cold, windless morning and three hot-air balloons hover above a park. Each balloon is stationary and indirect line of sight, with no obstacles near them, as shown in Figure 3. Balloon A is equipped with a 100 Wsiren, which emits a 2000 Hz tone uniformally in all directions. On board balloons B and C are students withsound measuring equipment.Question 6Which of the following is the best estimate of the sound intensity of the siren as measured at balloon B?A. 0.5 Wm-2B. 2.5 × 10-2 Wm-2C. 8.0 × 10-4 Wm-2D. 2.5 × 10-5 Wm-2100 W spread over a sphere of radius 100 m gives a sound intensity of 100/(4π(100)2) = 8.0 x 10-4 Wm-2
263.1 BelsAlexander Graham Bell.The Bel (symbol B) is a unit of measurement of ratios, such as power levels and voltage levels.It is mostly used in telecommunications, electronics and acoustics.It was invented by engineers at the Bell Telephone Laboratory to quantify (give a number to) the reduction in audio level over a 1 mile length of standard telephone cable.It was named in honour of Alexander Graham Bell.The bel was too large for everyday use, so the decibel (dB), equal to 0.1 Bel, became the more commonly used unit.
273.2 DecibelsThe decibel is not a unit in the sense that a metre or a kilogram is.Metres and kilograms are defined quantities of distance and mass. They never change.A decibel is a RELATIONSHIP between two values of POWER.We could use scientific notation, but a comparison between 2.3 x 101 and 4.7 x 1012 is still awkward.For convenience, we find the RATIO between the two numbers and convert that into a logarithm.Power difference = log4.7 x 10122.3 x 101= BDecibels are designed for talking about numbers of vastly different magnitudes, eg., 23 Watts vs. 4,700,000,000,000 Watts.With such vast differences, the most difficult problem is getting the number of zeros right.To make life a little easier, we can get rid of the decimal by multiplying the result by 10, so;So comparing the numbers above on this basis, you find that the larger number is 113dB bigger than the smaller number.
283.3 Sound Intensity Level S.I.L. = 10 Log I Io Sound Intensity measured in Wm-2 and Sound Intensity Level measured in decibels (dB) are NOT the same.Sound Intensity Level is DEFINED as the Logarithm of the ratio of the intensity of a sound to that of a reference sound.The intensity of the reference sound has a value of 1 x Wm-2, and is the minimum audible sound intensity at 3000 Hz. Corresponds to displacement of air particles by 100 billionth of a metre.The decibel scale ranges from 0 dB (the softest audible sound) to approx 140 dB (sound causing pain/ear damage)SIL’s for various objects or eventsMathematically:S.I.L. = 10 Log I Iowhere:S.I.L. = Sound Intensity Level (dB)I = Sound Intensity (Wm-2)Io = 1 x Wm-2
29An isolated siren emits sound of 3000 Hz uniformly in all directions An isolated siren emits sound of 3000 Hz uniformly in all directions. At a point 20 m from the siren, the sound intensity is measured to be 1.0 × 10–3 Wm–2.Question 7Which one of the following best gives the sound intensity level (in dB) at this point?A. 1.0 × 10–3 dBB. 9.0 dBC. 90 dBD. 100 dBThe sound intensity is measured at a distance of 60 m from the siren.Question 8Which one of the following best gives the sound intensity (in Wm–2) at 60 m?A. 3.3 × 10–3 Wm–2B. 1.1 × 10–4 Wm–2C. 3.0 × 10–2 Wm–2D. 9.0 × 10–3 Wm–2
303.4 Comparing Sound Intensity Levels The Sound Intensity Level formula can also be used to determine CHANGES in dB levels between two intensities labelled I1 and I2.Thus the equation becomes:SIL = 10 Log I2/I1When used in this form, the reference term (Io) is not used, and I1 and I2 are the two sound intensities being compared.Let I1 = 1 Wm-2and I2 = 10 Wm-2S.I.L. = 10 Log I2/I1= 10 Log 10/1= 10 dBThis 10 dB increase in S.I.L. is perceived by the Human Ear as a Doubling in the LOUDNESS of the sound.In fact, every 10 dB increase leads to a doubling of the loudness.So an 80 dB sound will be perceived as twice as loud as a 70 dB soundNOTE: Loudness is a subjective, non measurable quantity, used by humans to characterise and compare sounds.
313.5 The Decibel ScaleRemember when comparing S.I.L’s we use SIL = 10 Log I2/I1The decibel scale is used for a number of reasons:1. The human ear responds to a vast range of sound intensities (from Wm-2 to 102 Wm-2 - a range of or one hundred thousand billion units).2. In order to bring this range to a more manageable size, the log of intensities is used, so the range now becomes 0 dB to 140 dB.3. Luckily the ear also responds to sound intensities in a logarithmic rather than a linear fashion, as shown in last section.Let I1 = 100 Wm-2and I2 = 200 Wm-2S.I.L. = 10 Log I2/I1= 10 Log 200/100= 3 dBThus, if the sound intensity doubles, this leads to a 3 dB increase in S.I.L.This is about the smallest change in S.I.L. detectable by the human ear.So, if you replace your 100 W speakers with far more expensive 200 W ones, you will barely notice any difference !!!!!!
32Question 9By how many decibels will the sound intensity level at balloon C be lower than at balloon B?Doubling the distance quartered the intensity.Each time the intensity was halved the sound level reduced by 3 dB, so the total reduction was 6 dB.Balloons B and C move so that they are at equal distances from balloon A.The sound intensity at balloon C is now measured as 1.0 × 10-2 Wm-2.Question 10What is the sound intensity level (dB) at balloon B?SIL = 10 log I/Io = 10 log (1.0 × 10-2 )/(1.0 × 10-12) = 100 dB
333.6 Frequency Response Graphs The Human Ear and various Musical/Electrical devices (eg. Microphones) respond to different Audible Frequencies in different ways.We don’t hear each frequency with equal loudness.In order for the Ear to perceive various frequencies at the SAME LOUDNESS, they must be played at VARYING SOUND INTENSITY LEVELS.This is best shown on a Frequency Response GraphThus a 20 Hz sound needs to be played at 25 dB for the ear to hear itat the same loudness as a 4000 Hz sound played at 1 dBThe ear is most sensitive at the “lowest” point on the graph in this case 4000 Hz.
343.7 Sound Intensity vs Distance If a sound source is small (a so called POINT SOURCE), the sound it produces radiates out equally in all directions.This has the effect of producing an expanding sphere of sound. (NB. The surface area of a Sphere = 4r2)If the source is operating at a fixed power level, the sound intensity/unit area will decrease as the area (of the expanding sphere) increases.The rate at which the intensity drops off is inversely proportional to the square of the distance from the source:Mathematically: I 1/r22rI/4Point SourcerISo doubling the distance from the source leads to the intensity dropping in to ¼ of its original value.
35At a distance of 4. 0 m from a loudspeaker, a sound intensity of 1 At a distance of 4.0 m from a loudspeaker, a sound intensity of 1.25 x 10-4 Wm-2 is detected.Question 11What sound intensity would be detected at 1.0 m from the source?I ∝ 1/r2, therefore, decreasing distance to one quarter increases intensity by a factor of 16.16 x 1.25 x 10-4 = 2.0 x 10-3 Wm-2Question 12What sound intensity level would be detected at 1.0 m from the source?SIL =10 log I/IO = 10 log (2.0 x 10-3)/(1.0 x 10-12) = 93 dB
363.8 Human ResponseThe human ear responds to sound in the range from about 20 Hz to 20 kHz.Music covers a wider range of frequencies from about 50 Hz to 12 kHzHuman speech ranges from about 100 Hz to 8 kHzThe ear is most sensitive at about 4 kHz and has the lowest threshold of pain at the same frequencyThe distance between the red and green lines represents the range of audible sound for each frequency
373.9 Phons The ear is not equally responsive to all frequencies. The curves represent equal loudness as perceived by the average human earThe Phon is defined as a unit of apparent loudness, equal in number to the intensity in dB of a 1 kHz tone judged to be as loud as the sound being measured.The ear is less sensitive to low frequencies and this discrimination against lows becomes steeper for softer soundsSound intensity in dB does not reflect changes in the ear’s sensitivity with frequency and sound levelThus 50 phon means:“as loud as a 50 dB, 1kHz tone”Curve for the threshold of hearingand 100 phon means:“as loud as a 100 dB, 1 kHz tone
38The graph in Figure 1 shows the relationship between sound intensity level (dB), frequency (Hz) and loudness.Sound intensity level (dB) of a note of Hz is measured by a sound meter to be 60 dB.Question 13Which one of the values below best gives the loudness in phon at this point?A. 20 phonB. 40 phonC. 60 phonD 80 phonQuestion 14The loudness scale (phon) specifically takes account of which one of the following factors?A. Intensity of sound, as perceived by human hearing, is inversely proportional to distance from the source.B. The perception of sound by human hearing is logarithmic, rather than linear, compared to sound intensity.C. The perception of the intensity of sound by human hearing varies with frequency.D. Human hearing has a very limited range of frequencies that it can hear.
40Direction of Reflected 4.0 ReflectionSound (like any other wave) undergoes reflection when it strikes a wall or barrier.It will follow the laws of reflection: i =rwhere i is measured between the direction of the incoming wave train and the Normal and r is measured between the Normal and the direction of the reflected sound waves.NormalDirection of incomingSound WavesDirection of ReflectedSound WavesIncomingCompressionAngle of IncidenceAngle of Reflection
41Two physicists are discussing the design of a new theatre for use by a school choir. The design requirementis for good acoustic properties; in particular, for even distribution of sound over the whole frequency rangethroughout the theatre.A plan of the theatre to be used is shown in Figure 2.One of the physicists wants to line the walls of the audience area of the theatre with heavy sound-absorbing curtains.Question 15Which one of the following states why this is a good idea?A. The curtains will reduce the effect of diffraction through the stage opening, hence producing better quality sound.B. Without the curtains, different frequencies will reflect differently from the walls, causing distortion due to diffraction effects.C. Without the curtains, different frequencies will reflect differently from the walls, causing distortion due to interference effects.D. Without the curtains, there would be multiple paths from the speaker to each member of the audience, thus causing distortion and sound loss due to interference effects in some parts of the theatre.
424.1 Refraction SOUND REFRACTION Refraction is the bending of waves when they enter a medium where their speed is differentSOUND REFRACTIONMore Dense MediumSound waves, unlike light waves, travel faster in denser materials, such as solids and liquids, than they travel in air.When sound waves leave a solid, their velocity and wavelength decrease and they are bent towards the normal to the surface of the solid.Less Dense MediumFor sound waves in air, their speed is temperature dependent. (v = T)During a temperature inversion, the sound will refract back toward the ground.For normal conditions sound will refract away from the ground, producing a sound shadow as shownSound ShadowSound Shadow
434.2 DiffractionFor visible light, λ is about 550 nm or 5.5 x 10-7 m, so it needs to pass through a VERY NARROW gap to produce a diffraction effect.Diffraction is a phenomenon demonstrated by all waves and is best described as the bending of waves as they pass around objects or through gaps or openings.The extent of diffraction depends on the ratio between the wavelength of the wave and the size of the object, gap or opening. This is called the λ/w ratioDiffraction is rarely seen or experienced in the visual world but is part of everyday experience in the aural (hearing) world.The easiest large scale diffraction effect observable is that of ocean waves bending as they pass around an island or headland.For sound of frequency 4000 Hz (when the ear is at its most sensitive), λ = 8.75 cm, so sound can (and does) diffract around everyday objects.
444.3 Diffraction Around Corners Since sound waves have wavelengths in the centimetre to metre range, sound can, and does, suffer diffraction in the world in which we live.BuildingApproachingSound WavesSound ShadowDiffractedThis is because the houses we live in, and the objects we surround ourselves with, have similar dimensions to the wavelengths of sound.He can hear what the ladies are talking about without being able to see themWe can hear mum shouting to turn down the stereo in part because her sound waves are diffracted around the house.
454.4 Diffraction through Gaps When a series of straight waves approaches a gap, such as a doorway, those waves will suffer a change in direction in passing through the gap, ie. they will suffer diffraction.The EXTENT OF DIFFRACTION depends upon the ratio of wavelength () to the gap width (w).If the wavelength and gap are about the same size, appreciable diffraction will occur.Barrier withNarrow GapDiffractedSound WavesApproachingSound WavesIf is very much bigger (or smaller) than w, no diffraction effects will occur.Thus:1. Maximum diffraction occurs when /w is between about 0.1 and 502. No diffraction effects occur when /w 1 or /w 1Gap Width w
46Question 16In the paragraph below, options to complete each sentence are given within the brackets. Circle the correct option in each case.Jamie is listening to the sound of an orchestra through a small gap ina partly open sliding door. When the sound wave travels through thegap,[constructive interference / destructive interference /diffraction] occurs and spreading of the wave results. High pitched(frequency) instruments such as flutes experience [more / thesame / less] spreading than lower pitched instruments. As the sizeof the gap decreases, the angle of spreading will [increase / notchange / decrease].
47(High Frequency) Sound 4.5 Diffraction EffectsShort Wavelength(High Frequency) SoundHall with open doorsBand playing insideLong Wavelength(Low Frequency) SoundAn observer at A will hear both highand low frequency soundsBandAAn observer at B will hearlow frequency sound ONLYBSince wavelength and frequency are so closely related, the extent of diffraction for sound can be thought of in terms of the frequency of sound.Low Frequency (Bass) sounds are generally diffracted by a greater amount than High Frequency (Treble) sounds.High Frequency sounds suffering less diffraction are said to be much more “directional” or “line of sight”.This is one of reasons we can hear the low frequency (bass) sounds but not hear the high frequency (treble) sounds coming from a “party” a few streets away on a Saturday night.
48Question 17A Speaker system uses a single, wide-frequency response speaker. Explain why the quality (fidelity) will deteriorate as the listener moves off the centreline. Hence explain why a multiple-loudspeaker system, as shown in Figure 1, would be more satisfactory.The amount of diffraction depends on the ratio λ/w.For a single speaker, the high frequencies would not diffract away from the centre line as much as the low frequencies.Using different speaker sizes for different frequency ranges would ensure that comparable spreading will occur for all frequencies.
49Chapter 5 Topics Covered: Superposition Standing Waves Standing Waves on Strings.Standing Waves in Open and Closed Pipes.Overtones & Harmonics.Resonance.
50Constructive Superposition Destructive Superposition Two or more waves occupying the same space will interact to form a single, composite or total wave which reflects the size and orientation of the individual waves making it up.This addition process is called SUPERPOSITION.Superposition is a VECTOR addition process, so wave orientation as well as amplitude are importantConstructive SuperpositionDestructive SuperpositionDESTRUCTIVE SUPERPOSITION occurs when two waves with opposite orientations add.CONSTRUCTIVE SUPERPOSITION occurs when two waves with similar orientations add.
515.1 Standing WavesStanding waves are produced when two identical wave trains travelling in opposite directions interact with one another (undergo Superposition) to produce a so called “Standing Wave Pattern”.The whole “pattern” remains fixed in space as long as the frequencies of the two travelling wave trains remain constant.The “picture” representing the standing wave is, in fact, an “envelope” showing that in some areas large “motions” occur (the antinodes) and in other areas no “motions” occur (the nodes).½Standing Wave“Envelope”TravellingWaveTravellingWaveAN¼Adjacent Nodes [N] in a Standing Wave Pattern are ½ apart.Similarly Antinodes [A] are also ½ apart.This then makes an adjacent Node and Antinode ¼ apart.
52Two speakers facing each other are connected to the same signal generator/amplifier and are producing 340 Hz.Question 18Assume the speed of sound is 340 ms-1.Mary stands in the centre, equidistant to speakers A and B. She then moves towards speaker B and experiences a sequence of loud and quiet regions. She stops at the second region of quietness experienced. How far is she from speaker B? Explain your reasoning.Being connected to a common source, the speakers will be ‘in phase’, ensuring the midpoint is an antinode or a region of loudness.The distance between successive antinodes is λ/2, and λ/4 between an antinode and an adjacent node.Accordingly, Mary has moved 3λ/4 from the central antinode to the second node.Since λ = 1.0 m, she has moved 0.75 m toward B and therefore is 5.0 – 0.75 or 4.25 m from B
535.2 Standing Waves - Strings 1. Fundamental Frequency2. First OvertoneLFor a string FIXED AT BOTH ENDS:Simplest Standing Wave Pattern for a string fixed at both endsLength of string L = /2Thus = 2L and from v = f we get: f = v/2LThis is the Fundamental Frequency (f1)Now, L = λThus, f = v/L = 2f1and 1st overtone = 2f11. All OVERTONES exist2. Nth Overtone =(N + 1) x Fundamental Freq.3. The Fundamental and ALL Overtones can exist on the string AT THE SAME TIME.4. Third Overtone3. Second OvertoneNow L = 3λ/2Thus, f = 3V/2L = 3f1So, 2nd Overtone = 3f1L = 2λThus, f = 2V/L = 4f13rd Overtone = 4f1
545.3 Standing Waves - Open Pipes FUNDAMENTAL FREQUENCY2nd OVERTONEThus, f = 3V/2L = 3f12nd Overtone = 3 x Fundamental FrequencyLL = 3λ/2Open Pipe, Length = LNASimplest Standing Wave in an open pipe.Pressure node at each end, pressure antinode in middle.Length of pipe, L = /2Thus = 2L, and from v = f we get: f = v/2LThis is called The Fundamental Frequency (f1) for this pipe.Thus f = 2V/L = 4f13rd Overtone = 4 x Fundamental Frequency3rd OVERTONELL = 2λLNA1st OVERTONEThus f = v/L = 2f11st overtone = 2 x Fundamental FrequencyL = λFor a pipe OPEN AT BOTH ENDS:1. All OVERTONES exist2. Nth Overtone = (N + 1) x Fundamental Freq.3. The Fundamental and ALL Overtones can exist in the pipe AT THE SAME TIME.
55Question 19The fundamental of a bugle is 88 Hz. Other notes easily produced by a bugle have frequencies of 264 Hz, 352 Hz and 440 Hz. Should the bugle be modelled as an open or a closed pipe? Justify your answer. Take the speed of sound in air as 340 ms-1 .The fundamental frequency for the bugle f1 = 88Hz.The other easily produced notes are 264Hz = 3f1, 352Hz =4f1 440Hz =5 f1and hence the bugle is an open pipe since there is an even harmonic produced.Question 20Calculate the length of the air column in the bugle. You may neglect any end correction.For the fundamental freq L = λ/2λ = v/f = 340/88 = 3.86 mL = λ/2 = 3.86/2 = 1.93 mF
56Sarah is planning to buy some plastic pipe from a hardware store Sarah is planning to buy some plastic pipe from a hardware store. To measure the length of the pipe, she intends to blow across one end of the pipe and measure the frequency of the resonance produced.The shop owner questions this method, but in the end agrees to let her perform the measurements. Sarah takes a section of pipe open at both ends, and performs the measurements. A clear resonance of 200 Hz can be heard.Question 21Use this information to determine the length of the pipe. Show your working/reasoning. (speed of sound 340 ms-1)v = fλ λ = v/f = 340/200 = 1.7 mAn open/open pipe has L = λ/2 L = 1.7/2 = 0.85 m
57Question 22At which one or more of the following frequencies could Sarah’s pipe also resonate?A. 300 HzB. 400 HzC. 500 HzD. 600 HzQuestion 23Briefly explain resonance in terms of the behaviour of the sound waves in a tube open at both ends.Since the pipe was open at both ends, all harmonics were possible.Resonance is the matching of frequency between the natural frequency of the tube and the frequency within the source of the excitation, the blowing across the tube. (1)At an open end there is a pressure node, the distance between adjacent nodes is half a wavelength, (1) so this determines the natural frequency of the tube. (1).
585.4 Standing Waves - Closed Pipes FUNDAMENTAL FREQUENCYNA2nd OVERTONEThus, f = 5V/4L = 5f12nd Overtone = 5 x Fundamental FrequencyLL = 5λ/4Simplest Standing Wave pattern in a Closed Pipe. Pressure antinode at closed end, node at open end.Length of pipe, L = /4Thus = 4L and from v = f we get: f = v/4LThis is called the Fundamental Frequency (f1) for this pipe.Thus f = 7V/4L = 7f13rd Overtone =7 x Fundamental Frequency3rd OVERTONELL = 7λ/41st OVERTONEThus f = 3v/4L = 3f11st overtone = 3 x Fundamental FrequencyLANL = 3λ/4FOR A PIPE CLOSED AT ONE END.1. Only ODD overtones exist.2. The Nth Overtone = (2N + 1) x Fund. Freq.3. The Fundamental and all ODD OVERTONES can exist in the pipe AT THE SAME TIME.
59Roger, an instrument maker, is constructing and testing pipes for a pipe organ. The pipes can be considered to be uniform tubes open at one end and closed at the other.Roger tests the pipe by placing a loudspeaker attached to a very precise audio signal generator at the open end of the pipe, and gradually increases the frequency.He finds that in addition to the resonance at 256 Hz, there is a higher resonance (the second harmonic).Question 24At which one of the following frequencies will this second harmonic be observed?A. 128 HzB. 512 HzC. 768 HzD Hz
60Question 25Which one of the following statements best describes how Roger was able to identify this second harmonic?A. At the frequency of this second harmonic a standing wave is set up in the tube. This absorbs sound energy, hence the volume heard by Roger decreases.B. At the frequency of this second harmonic the first harmonic is also heard, so when Roger hears this as well, he knows the signal generator is at a harmonic.C. At the frequency of this second harmonic a standing wave is set up in the tube. This causes the volume heard by Roger to increase.D. At the frequency of this second harmonic the fidelity (quality) of the note changes, and Roger is able to identify this.Question 26Roger is later designing a different pipe to give a wavelength of m.Which one of the following lengths should Roger make the pipe?A mB mC mD m
615.5 Standing Waves Overtones vs Harmonics Overtones and Harmonics are two terms used to describe the same effect - standing wave patterns which are whole number multiples of a Fundamental Frequency.Overtones are numbered AFTER the Fundamental Frequency.Harmonics are numbered to INCLUDE the Fundamental FrequencyOvertones on a String.Fixed at both ends.Let Fundamental Frequency = 256 HzThen1st overtone = 2 x 256 = 512 Hz2nd overtone = 3 x 256 = 768 Hz3rd overtone = 4 x 256 = 1024 HzHarmonics on a String.Fixed at both ends.Let fundamental frequency = 256 HzThen:1st Harmonic = 256 Hz2nd Harmonic = 512 Hz3rd Harmonic = 768 Hz4th Harmonic = 1024 Hz
625.6 TimbreSo an accordion, playing the same note, has a different timbre toMusic, played on any musical instrument, has a depth and resonance beyond the simple note or tone played.Not only that, but the same note can be played on different instruments and sound completely different.Instruments all have very different qualities to their sounds which make them distinctive; these qualities are often referred to as the instrument's timbre.a trumpetThe timbre for an instruments is determined by its fundamental frequency and overtones/harmonics all of which exist at the same time.It is the unique combination of fundamental and harmonics that give each musical instrument its unique sound.
635.7 ResonanceEvery known object has a so called “Natural Frequency of Vibration”.This is the frequency with which the object would wish to vibrate if it has to.If the object is attached to an external source which vibrates at the object’s natural frequency of vibration, the object will vibrate with an amplitude much larger than expected.This is because a “standing wave pattern” is set up within the object producing the large amplitudes observed.This is known as RESONANCE.Examples of this phenomenon are:The opera singer breaking the glass.and the Tacoma Narrows Bridge collapse in the USA in 1940.
645.8 Tacoma NarrowsThe bridge earned the nickname "Galloping Gertie" from its rolling, undulating behaviour.Motorists crossing the 2,800-foot centre span sometimes felt as though they were travelling on a giant roller coaster, watching the cars ahead disappear completely for a few moments as if they had been dropped into the trough of a large wave.The original, 5,939 ft (1810 m) long Tacoma Narrows Bridge opened to traffic on July 1, 1940 after two years of construction, linking Tacoma and Gig Harbour in Washington State, USA.It collapsed just four months later during a 68 km/h wind storm on Nov. 7, 1940.
666.0 MicrophonesA microphone is a device that converts sound into an electrical signal.Microphones are used in many applications such as telephones, tape recorders, hearing aids, motion picture production and in radio and television broadcasting.In all microphones, sound waves (sound pressure) are translated into mechanical vibrations in a thin, flexible diaphragm.These sound vibrations are then converted by various methods into an electrical signal which varies in voltage amplitude and frequency in an analogue of the original sound.DynamicRibbonCondenserCrystalElectretFor this reason, a microphone is an acoustic wave to voltage modulation transducer.Many types of microphones exist as can be seen
676.1 Electret MicrophonesAn electret is a dielectric material that has been permanently electrically charged or polarised.Capacitor microphones can be expensive and require a power supply, but give a high-quality sound signal and are used in laboratory and studio recording applications.In a capacitor microphone, also known as an electret - condenser microphone, the diaphragm acts as one plate of a capacitor, and the distance changing sound vibrations produce changes in a voltage maintained across the capacitor plates.Sound information exists as patterns of air pressure; the microphone changes this information into patterns of electric current.Fidelity is the term used for the accuracy of this transformation, and a “flat response” is the most sought after property of a microphone.
68Question 27Which one of the following (A to D below) best describes the physical operating principle of the electret-condensor microphone?A. electromagnetic inductionB. piezo-electric effectC. capacitanceD. electrical resistanceThe diagram represents a particular microphone.Question 28Identify the type of microphone this diagram represents.electret-condensor microphone?Question 29Describe how this microphone detects the wave and produces the signal output.The sound wave vibrates the diaphragm which is permanently charged (polarised).The diaphragm is effectively one side of a capacitor.As the capacitance changes, a tiny current is induced and hence a signal voltage appears across the resistor.
696.2 Crystal MicrophonesThese microphones utilises the ‘piezoelectric effect’. Piezo (Greek for Push) electric solids produce a voltage between surfaces when a mechanical stress is applied.Sound waves cause the diaphragm to move which in turn communicates the resulting vibration to an attached piezo electric crystal. Charges and hence voltages are proportional to the crystal’s bending.Crystal microphones tend to be used for low quality audio applications such as telephone handsets since they don’t require phantom powering or amplification and are cheap to produce.The frequency response of crystal microphones is often limited to a relatively narrow band restricting their application.
706.3 Dynamic MicrophonesIn the dynamic microphone a small movable induction coil, positioned in the magnetic field of a permanent magnet, is attached to the diaphragm.When the diaphragm vibrates, the coil moves in the magnetic field, producing a varying current in the coil.Dynamic microphones are robust and relatively inexpensive, and are used in a wide variety of applications.However the inertia of the coil reduces high frequency response.It is important to remember that current is produced by the motion of the diaphragm, and that the amount of current is determined by the speed of that motion.So this type of microphone is also known as a velocity microphoneHence they are NOT best suited to studio applications where quality and subtlety are important such as high quality vocal recording or acoustic instrument micking.
71Question 30Explain the operation of a dynamic microphone.A dynamic microphone consists of a coil of wire inside a magnetic field. (1) The coil is made to vibrate backwards and forwards by sound waves. This movement within the magnetic field produces an induced voltage that matches the variation in sound pressure. (1)Figure 2 shows the frequency response curve for a dynamic microphone.Question 31From the data on the graph, what makes this microphone particularly suitable for use by a singer?The microphone is suitable because it has a linear response over the frequency range of the singer.
726.4 Ribbon MicrophonesRibbon microphones employ electromagnetic induction to convert sound to voltage.A long thin strip of conductive foil moves within a magnetic field to generate a current hence voltage.The foil’s lower mass when compared to a moving coil gives it a smoother and higher frequency response.However the relatively low output requires a step up transformer.Ribbon microphones are good for quality studio recording of acoustic instruments though can be delicate, for instance you wouldn’t want to put one in front of a bass cabinet.
73Three types of microphone are • electret-condenser• crystal• dynamic.The physical properties on which the operation of these microphones depend are listed below (not in order).Question 32Which one of the boxes correctly matches the microphone type to the relevant physical property?
746.4 LoudspeakersThe loudspeakers are almost always the limiting element on the quality of a reproduced sound.A loudspeaker without an enclosure does a very poor job of producing sounds whose wavelengths are longer than the diameter of the loudspeaker.Once you have chosen a good loudspeaker from a reputable manufacturer and paid good money for it, you might presume that you would get good sound reproduction from it.But you won't -- not without a good enclosure or cabinet.The enclosure is an essential part of sound reproduction.These wavelengths are prone to generate interference patterns (through the process shown below) which particularly affect the lower frequency or bass aspects of the musicSound wave from back of cone tends to cancel wave from frontPressure wave from front of conePressure wave from back of coneBesides the severe bass loss, the overall efficiency of such loudspeakers is low, about 3-5% compared to 25-50% for well designed horn type loudspeakers.
756.5 Speaker EnclosuresvariousspeakerenclosuresThe enclosure increases the effective size of the loudspeaker.The ideal (but impractical) mount for a loudspeaker would be a flat board (flat baffle) of infinite size with infinite space behind it.Thus the rear sound waves cannot cancel the front sound waves.The bass-reflex enclosure (baffle) makes use of a tuned port which projects some of the sound energy from the back of the loudspeaker, energy which is lost in a sealed enclosure.But care must be taken to avoid the back-to-front cancellation of low frequencies which characterizes unenclosed loudspeakers.
76Question 33In the paragraph below, options to complete each sentence are given within the brackets. Circle the correct option in each case.A loudspeaker is removed from its enclosure box. When an audiosignal is connected, the loudspeaker produces sound waves atboth its front and rear surfaces. The sound waves from the frontof the loudspeaker are [ in phase with / out of phase with /of much higher intensity than ] the waves from the rear. For alistener in front of the speaker the waves from the front[ interfere constructively / interfere destructively / diffractdestructively ] with those generated from the rear surface. Thisaffects the [ frequency / intensity / directional spread ]of the resulting sound.
77A high fidelity loudspeaker system comprising individual speakers mounted on a baffle board is shown in thediagram in Figure 1.Question 34Explain the role of the baffle board in improving the performance of the loudspeaker system above.Sound from the back and front of the speaker was out of phase. The baffle prevented these from meeting and interfering.
786.6 Multiple SpeakersEven with a good enclosure, a single loudspeaker cannot be expected to deliver optimally balanced sound over the full audible sound spectrum.For the production of high frequencies, the driving element should be small and light to be able to respond rapidly to the applied signal.Such high frequency speakers are called "tweeters".On the other hand, a bass speaker should be large to efficiently impedance match to the air.Such speakers, called "woofers", must also be supplied with more power since the signal must drive a larger mass.
796.7 Enclosure ProblemsEnclosures play a significant role in the sound production, adding resonances, diffraction, and other unwanted effects.Problems with resonance are usually reduced by increasing enclosure rigidity, added internal damping and increasing the enclosure mass.Diffraction problems are addressed in the shape of the enclosure; avoiding sharp corners on the front of the enclosure for instance.Sometimes the differences in reaction time of the different size drivers (speakers) is addressed by setting the smaller drivers further back in the enclosure, so that the resulting wavefront from all drivers is coherent when it reaches the listener.
806.8 Frequency ResponseFrequency Response attempts to describe the range of frequencies or musical tones a speaker can reproduce, measured in Hertz.The range of human hearing is generally regarded as being from 20Hz, through to 20kHz.Presumably a speaker that could reproduce that range would sound lifelike.Alas, it is no guarantee.The most important determinant of a speaker's frequency performance is not its width or range, but whether it's capable of reproducing all the audible frequencies at the same volume at which they were recorded.Frequency (Hz)Sound Intensity Level (dB)The relatively flat line on the graph indicates that the speaker is "flat”.This means that it will treat all sounds equally.It won't impose its will on the music but will allow you to hear the music as it was recorded.Flat is good.Flat response means that the speaker reproduces sound accurately.Remember that the ear is barely able to discern changes of 3 dB in SIL’s, so flat ± 3 dB is flat to our ear.
81The response curve for a loudspeaker is shown opposite: Question 35State the frequency range where the speaker performs well.50 Hz to 1 kHzThe speaker is used in a speaker box with crossover circuits to supply separate speakers within the box.Question 36What specific application would this speaker perform within the system?This speaker is a low frequency speaker (or woofer or bass speaker).
82The frequency response curve for one of the speakers in the system shown in Figure 1 is shown in Figure 2 above.Question 37Which type of speaker is most likely to have a response curve similar to that shown in Figure 2?A. sub-wooferB. wooferC. mid-range speakerD. tweeter
836.8 Crossover NetworksMost loudspeakers use multiple drivers and employ crossover networks to route the appropriate frequency ranges to the different drivers.A two speaker system (tweeter and woofer)are called a two way crossoverA three speaker system (tweeter, mid range, woofer) is called a three way crossover