Acoustics and Sound Measurement Ch 1 PDF

| 12.09.22 - 17:44 1 Acoustics and Sound Measurement We begin our study of audiology by reviewing the of MKS units because this is the internationally ac- nature of sound because, after all, sound is wh...

| 12.09.22 - 17:44 1 Acoustics and Sound Measurement We begin our study of audiology by reviewing the of MKS units because this is the internationally ac- nature of sound because, after all, sound is what we cepted standard in the scientific community, known hear. The science of sound is called acoustics, which as the Système International d’Unites (SI). Equiva- is a branch of physics, and relies on several basic lent cgs values will often be given as well because physical principles. Many useful sources are avail- the audiology profession has traditionally worked in able for students wishing to pursue the areas cov- cgs units, and the death of old habits is slow and ered in this chapter in greater detail (e.g., Peterson & labored. These quantities are summarized with Gross 1972; Hewitt 1974; Kinsler, Frey, Coppens, & equivalent values in MKS and cgs units in Table 1.1. Sanders 1982; Sears, Zemansky, & Young 1982; In addition, the correspondence between scientific Beranek 1986; Gelfand 2018). notation and conventional numbers, and the mean- ings of prefixes used to describe the sizes of metric ▪Physical Quantities The basic physical quantities are mass, time, and units are shown for convenience and ready reference in Table 1.2 and Table 1.3. Quantities may be scalars or vectors. A scalar can length (or distance). All other physical quantities are be fully described by its magnitude (amount or size), derived by combining these three basic ones, as well but a vector has both direction and magnitude. For as other derived quantities, in a variety of ways. example, length is a scalar because an object that is The principal basic and derived quantities are one meter long is always one meter long. However, summarized in Table 1.1. These basic quantities are we are dealing with a vector when we measure the expressed in terms of conventional units that are distance between two coins that are one meter apart measurable and repeatable. The unit of mass (M) is because their relationship has both magnitude and the kilogram (kg) or the gram (g); the unit of direction (from point x1 to point x2). This quantity is length (L) is the meter (m) or the centimeter (cm); called displacement (x). Derived quantities will be and the unit of time (t) is the second (s). Mass is not vectors if they have one or more components that really synonymous with weight even though we ex- are vectors; for example, velocity is a vector because press its magnitude in kilograms. The mass of a body it is derived from displacement, and acceleration is a is related to its density, but its weight is related to vector because it involves velocity. We distinguish the force of gravity. If two objects are the same size, between scalars and vectors because they are the one with greater density will weigh more. How- handled differently when calculations are being ever, even though an object’s mass would be identi- made. cal on the earth and the moon, it would weigh less Velocity Everyone knows that “55 miles per on the moon, where there is less gravity. hour” refers to the speed of a car that causes it to When we express mass in kilograms and length travel a distance of 55 miles in a one-hour period of in meters, we are using the meter-kilogram-second time. This is an example of velocity (v), which is or MKS system. Expressing mass in grams and equal to the amount of displacement (x) that occurs length in centimeters constitutes the centimeter- over time (t): gram-second or cgs system. These two systems also x have different derived quantities. For example, the v¼ t units of force and work are called newtons and joules in the MKS system and dynes and ergs in the Displacement is measured in meters and time is cgs system, respectively. We will emphasize the use measured in seconds (s); thus, velocity is expressed 1 | 12.09.22 - 17:44 2 1 Acoustics and Sound Measurement Table 1.1 Principal physical quantities Quantity Formula MKS (SI) units cgs units Comments Mass (M) M kilogram (kg) gram (g) 1 kg = 103 g Time (t) t second (s) s Area (A) A m2 cm2 1 m2 = 104 cm2 Displacement (x) x meter (m) centimeter (cm) 1 m = 102 cm Velocity (v) v = x/t m/s cm/s 1 m/s =102 cm/s Acceleration (a) a = v/t m/s2 cm/s2 1 m/s2 = 102 cm/s2 Force (F) F = Ma kg · m/s2 g · cm/s2 1 N = 105 dyne = Mv/t newton (N) dyne Pressure (p) p = F/A N/m2 dyne/cm2 2 × 10−5 N/m2 or 20 μPa Pascal (Pa) microbar (μbar) (reference value) 2 × 10−4 dyne/cm2 or μbar (reference value) Work (W) W = Fx N·m dyne · cm 1 J = 107 erg joule (J) erg Power (P) P = W/t joule/s erg/s 1 W = 1 J/s = Fx/t watt (W) watt (W) 1 W = 107 erg/s = Fv Intensity (I) I = P/A W/m2 W/cm2 10−12 W/m2 (reference value) 10−16 W/cm2 (reference value) in meters per second (m/s). Velocity is the vector math-minded, it refers to the velocity when the dis- equivalent of speed because it is based on displace- placement and time between one point and the next ment, which has both magnitude and direction. one approach zero, that is, the derivative of displace- When we take a trip we usually figure out the dis- ment with respect to time: tance traveled by making a mental note of the start- ing odometer reading and then subtracting it from dx v¼ the odometer reading at the destination (e.g., if we dt start at 10,422 miles and arrive at 10,443 miles, then Acceleration Driving experience has taught us all the distance must have been 10,443 – 10,422 = 21 that a car increases its speed to get onto a highway, miles). We do the same thing to calculate the time it slows down when exiting, and also slows down took to make the trip (e.g., if we left at 1:30 and while making a turn. “Speeding up” and “slowing arrived at 2:10, then the trip must have taken 2:10 – down” mean that the velocity is changing over 1:30 = 40 minutes). Physical calculations involve the time. The change of velocity over time is accelera- same straightforward approach. When an object is tion (a). Suppose a body is moving between two displaced, it starts at point x1 and time t1 and arrives points. Its velocity at the first point is v1, and the at point x2 and time t2. Its average velocity is simply time at that point is t1. Similarly, its velocity at the the distance traveled (x2 – x1) divided by the time it second point is v2 and the time at that point is t2. took to make the trip (t2 – t1): Average acceleration is the difference between the x 2 " x1 two velocities (v2 – v1) divided by the time interval by v¼ t2 " t1 the time interval (t2 – t1): The term instantaneous velocity describes the veloc- v2 " v1 ity of a body at a particular moment in time. For the a¼ t2 " t1 | 12.09.22 - 17:44 1 Acoustics and Sound Measurement 3 Table 1.2 Expressing numbers in standard notation In more general terms, acceleration is written sim- and scientific notation ply as v Standard notation Scientific notation a¼ t 0.000001 10−6 Because velocity is the same as displacement div- 0.00001 10−5 ided by time, we can replace v with x/t, so that 0.0001 10−4 x=t a¼ t 0.001 10–3 which can be simplified to 0.01 10–2 x a¼ 0.1 10–1 t2 1 100 Consequently, acceleration is expressed in units of meters per second squared (m/s2) in the MKS system. 10 101 When measurements are made in cgs units, acceler- ation is expressed in centimeters per second 100 102 squared (cm/s2). 1000 103 Acceleration at a given moment is called instan- taneous acceleration, and quantitatively oriented 10,000 104 readers should note it is equal to the derivative of 100,000 105 velocity with respect to time, or 1,000,000 106 dv a¼ dt 3600 3.6 × 103 Because velocity is the first derivative of displace- 0.036 3.6 × 10−2 ment, we find that acceleration is the second deriva- tive of displacement: 0.0002 2 × 10−4 2 0.00002 2 × 10–5 d x a¼ dt 2 Table 1.3 Examples of prefixes used to express metric units Multiply by Prefix Symbol Definition Standard notation Scientific notation micro μ millionths 1/1,000,000 or 0.000001 10−6 milli m thousandths 1/1000 or 0.001 10–3 centi c hundredths 1/100 or 0.01 10−2 deci d tenths 1/10 or 0.1 10–1 deka da tens 10 101 hecto h hundreds 100 102 kilo k thousands 1000 103 mega M millions 1,000,000 106 | 12.09.22 - 17:44 4 1 Acoustics and Sound Measurement Force An object that is sitting still will not move in the same direction, then the net force would be unless some outside influence causes it to do so, and the sum of those two forces. Conversely, if a 4 N an object that is moving will continue moving at the force pushes an object toward the right at the same same speed unless some outside influence does time that a 9 N force pushes it toward the left, then something to change it. This commonsense state- the net force is 9 – 4 = 5 N toward the left. Thus, if ment is Newton’s first law of motion. It describes the two forces push an object in opposite directions, attribute of inertia, which is the property of mass to then the net force is the difference between the two continue doing what it is already doing. The “outside opposing forces, and it causes the object to acceler- influence” that makes a stationary object move, or ate in the direction of the greater force. If two equal causes a moving object to change its speed or forces push in opposite directions, then the net force direction, is called force (F). Notice that force causes is zero. Because the net force is zero it will not cause the moving object to change velocity or the motion- the motion of the object to change. The situation in less object to move, which is also a change in velocity which net force is zero is called equilibrium. In this (from zero to some amount). Recall that a change of case, a moving object will continue moving and an velocity is acceleration. Hence, force is that influence object that is at rest (i.e., not moving) will continue (conceptually a “push” or “pull”) that causes a mass to remain still. to be accelerated. In effect, the amount of “push” or Friction When an object is moving in the real “pull” needed depends on how much mass you want world, it tends to slow down and eventually comes to influence and the amount of acceleration you are to a halt. This happens because anything that is trying to produce. In other words, force is equal to moving in the real world is always in contact with the product of mass times acceleration: other bodies or mediums. The sliding of one body on the other constitutes a force that opposes the mo- F ¼ Ma tion, called resistance or friction. Since acceleration is velocity over time (v/t), we can The opposing force of friction or resistance de- also specify force in the form pends on two parameters. The first factor is that the amount of friction depends on the nature of the ma- Mv F¼ terials that are sliding on one another. Simply stated, t the amount of friction between two given objects is The quantity Mv is called momentum, so we may greater for “rough” materials than for “smooth” or also say that force equals momentum over time. “slick” ones, and is expressed as a quantity called The amount of force is measured in kg · m/s2 the coefficient of friction. The second factor that de- because force is equal to the product of mass (meas- termines how much friction occurs is easily appreci- ured in kg) and acceleration (measured in m/s2). The ated by rubbing the palms of your hands back and unit of force is the newton (N), where one newton is forth on each other. First rub slowly and then more the amount of force needed to cause a 1 kg mass to rapidly. The rubbing will produce heat, which occurs be accelerated by 1 m/s2; hence, 1 N = 1 kg · 1 m/s2. because friction causes some of the mechanical en- (This might seem very technical, but it really simpli- ergy to be converted into heat. This notion will be fies matters; after all, it is easier to say “one newton” revisited later, but for now we will use the amount than “one kg · m/s2.”) It would take a 2 N force to of heat as an indicator of the amount of resistance. cause a 1 kg mass to be accelerated by 2 m/s2, or a Your hands become hotter when they are rubbed to- 2 kg mass to be accelerated by 1 m/s2. A 4 N force is gether more quickly. This illustrates the notion that needed to accelerate a 2 kg mass by 2 m/s2, and a 63 the amount of friction depends on the velocity of N force is needed to accelerate a 9 kg mass by 7 m/s2. motion. In quantitative terms, In the cgs system, the unit of force is called the dyne, Ff ¼ Rv which is the force needed to accelerate a 1 g mass by 1 cm/s2; that is, 1 dyne = 1 g · cm/s2. It takes 105 where Ff is the force of friction, R is the coefficient of dynes to equal 1 N. friction between the materials, and v is the velocity Many different forces are usually acting upon an of the motion. object at the same time. Hence, the force we have Elasticity and restoring force It takes some effort been referring to so far is actually the net or resul- (an outside force) to compress or expand a spring; and tant force, that is, the “bottom line” effect of all the the compressed or expanded spring will bounce back forces that act upon an object. If a force of 3 N is to its original shape after it is released. Compressing or pushing an object toward the right and a second expanding the spring is an example of deforming an force of 8 N is also pushing that object toward the object. The spring bouncing back to its prior shape is right, then the net force would be 3 + 8 = 11 N toward an example of elasticity. More formally, we can say the right. In other words, if two forces push a body that elasticity is the property whereby a deformed | 12.09.22 - 17:44 1 Acoustics and Sound Measurement 5 object returns to its original form. Notice the dis- Because force is measured in newtons and displace- tinction between deformation and elasticity. A rub- ment is measured in meters, work itself is quanti- ber band and saltwater taffy can both be stretched fied in newton-meters (N · m). For example, if a (deformed), but only the rubber band bounces back. force of 2 N displaces a body by 3 m, then the In other words, what makes a rubber band elastic is amount of work is 2 × 3 = 6 N. There can only be not that it stretches, but rather that it bounces back. work if there is displacement. There cannot be The more readily a deformed object returns to its work if there is no displacement (i.e., if x = 0) original form, the more elastic (or stiff) it is. because work is the product of force and displace- We know from common experiences, such as ment, and zero times anything is zero. The MKS using simple exercise equipment, that it is rela- unit of work is the joule (J). One joule is the tively easy to begin compressing a spring (e.g., a amount of work that occurs when one newton of “grip exerciser”), but that it gets progressively force effects one meter of displacement, or 1 J = 1 harder to continue compressing it. Similarly, it is N · m. In the cgs system, the unit of work is called easier to begin expanding a spring (e.g., pulling the erg, where 1 erg = 1 dyne · cm. One joule corre- apart the springs on a “chest exerciser”) than it is sponds to 107 erg. to continue expanding it. In other words, the more Energy is usually defined as the capability to a spring-like material (an elastic element) is de- do work. The energy of a body at rest is potential formed, the more it opposes the applied force. The energy and the energy of an object that is in mo- force that opposes the deformation of an elastic or tion is kinetic energy. The total energy of a body is spring-like material is known as the restoring the sum of its potential energy plus its kinetic force. If we think of deformation in terms of how energy, and work corresponds to the exchange be- far the spring has been compressed or expanded tween these two forms of energy. In other words, from its original position, we could also say that energy is not consumed when work is accom- the restoring force increases with displacement. plished; it is converted from one form to the other. Quantitatively, then, restoring force (FR) depends This principle is illustrated by the simple example on the stiffness (S) of the material and the amount of a swinging pendulum. The pendulum’s poten- of its displacement as follows: tial energy is greatest when it reaches the extreme of its swing, where its motion is momentarily F R ¼ Sx zero. On the other hand, the pendulum’s kinetic Pressure Very few people can push a straight pin into energy is greatest when it passes through the mid- a piece of wood, yet almost anyone can push a thumb- point of its swing because this is where it is mov- tack into the same piece of wood. This is possible ing the fastest. Between these two extremes, because a thumbtack is really a simple machine that energy is being converted from potential to kinetic concentrates the amount of force being exerted over a as the pendulum speeds up (on each down swing), larger area (the head) down to a very tiny area (the and from kinetic to potential as the pendulum point). In other words, force is affected by the size of slows down (on each up swing). the area over which it is applied in a way that consti- Power The rate at which work is done is called tutes a new quantity. This quantity, which is equal to power (P), so that power can be defined as work force divided by area (A), is called pressure (p), so divided by time, F W p¼ P¼ A t Because force is measured in newtons and area is The unit of power is called the watt (W). One unit of measured in square meters in MKS units, pressure is power corresponds to one unit of work divided by measured in newtons per square meter (N/m2). one unit of time. Hence, one watt is equal to one The unit of pressure is the pascal (Pa), so that 1 joule divided by one second, or 1 W = 1 J/s. Power is Pa = 1 N/m2. In the cgs system, pressure is measured also expressed in watts in the cgs system, where in dynes per square centimeter (dynes/cm2), occa- work is measured in ergs. Since 1 J = 107 erg, we can sionally referred to as microbars (μbars). also say that 1 W = 107 erg/s. Work and energy As a physical concept, work Power can also be expressed in other terms. For (W) occurs when the force applied to a body results example, because W = Fx, we can substitute Fx for W in its displacement, and the amount of work is equal in the power formula, to arrive at to the product of the force and the displacement, or Fx P¼ W ¼ Fx t | 12.09.22 - 17:44 6 1 Acoustics and Sound Measurement We know that v = x/t, so we can substitute v for x/t fixed amount of oil in the prior example. Conse- and rewrite this formula as quently, the sound power will be divided over the ever-expanding surface as distance increases from P ¼ Fv the source, analogous to the thinning out of the wid- In other words, power is also equal to force times ening oil slick. This notion is represented in the fig- velocity. ure by the thinning of the lines at greater distances Intensity Consider a hypothetical demonstration from the source. Suppose we measure how much in which one tablespoonful of oil is placed on the power registers on a certain fixed amount of surface surface of a still pond. At that instant the entire area (e.g., a square inch). As a result, a progressively amount of oil will occupy the space of a tablespoon. smaller proportion of the original power falls onto a As time passes, the oil spreads out over an square inch as the distance from the source in- expanding area on the surface of the pond, and it creases, represented in the figure by the lighter therefore also thins out so that much less than all shading of the same-size ovals at increasing the oil will occupy the space of a tablespoon. The distances from the source. wider the oil spreads the more it thins out, and the The examples just described reveal that a new proportion of the oil covering any given area gets quantity, called intensity (I), develops when power smaller and smaller, even though the total amount is distributed over area. Specifically, intensity is of oil is the same. Clearly, there is a difference be- equal to power per unit area, or power divided by tween the amount of oil, per se, and the concentra- area, or tion of the oil as it is distributed across (i.e., divided P by) the surface area of the pond. I¼ An analogous phenomenon occurs with sound. It A is common knowledge that sound radiates outward Because power is measured in watts and area is meas- in every direction from its source, constituting a ured in square meters in the MKS system, intensity is sphere that gets bigger and bigger with increasing expressed in watts per square meter (W/m2). Intensity distance from the source, as illustrated by the con- is expressed in watts per square centimeter (W/cm2) centric circles in Fig. 1.1. Let us imagine that the in the cgs system. sound source is a tiny pulsating object (at the center Intensity decreases with increasing distance of the concentric circles in the figure) and that it from a sound source according to a rule called the produces a finite amount of power, analogous to the inverse square law. It states that the amount of Fig. 1.1 Intensity (power divided by area) decreases with distance from the sound source because a fixed amount of power is spread over an increasing area, represented by the thinning of the lines. Proportionately less power falls on the same unit area (represented by the lighter shading of the ovals) with increasing distance from the source. | 12.09.22 - 17:44 1 Acoustics and Sound Measurement 7 Fig. 1.2 Illustrations of the inverse square law. (a) Doubling of distance: The intensity at 10 m away from a loud- speaker is one quarter of its intensity at 5 m because 1/22 = 1/4. (b) Tripling of distance: The intensity at 15 m away from the sources is one-ninth of its intensity at 5 m because 1/32 = 1/9. intensity drops by 1 over the square of the change in a washing machine, a guitar string, a tuning fork distance. Two examples are illustrated in Fig. 1.2. prong, and air molecules. The vibration is usually Fig. 1.2a shows that when the distance from a loud- called sound when it is transferred from air particle speaker is doubled from 5 m to 10 m, the amount of to air particle (we will see how this happens later). intensity at 10 m will be one quarter of what it was The vibration of air particles might have a simple pat- at 5 m (because 1/22 = 1/4). Similarly, Fig. 1.2b shows tern such as the tone produced by a tuning fork, or a that tripling the distance from 5 m to 15 m causes very complex pattern such as the din heard in a the intensity to fall to one-ninth of its value at the school cafeteria. Most naturally occurring sounds are closer point because 1/32 = 1/9. very complex, but the easiest way to understand An important relationship to be aware of is that sound is to concentrate on the simplest ones. power is equal to pressure squared, P ¼ p2 Simple Harmonic Motion and pressure is equal to the square root of power, A vibrating tuning fork1 is illustrated in Fig. 1.3. The p!!! initial force that was applied by striking the tuning p¼ P fork is represented by the green arrow in frame 1. The progression of the drawings represents the mo- In addition, intensity is proportional to pressure tion of the prongs at selected points in time after squared, the fork has been activated. The two prongs vibrate I / p2 as mirror images of each other, so that we can de- scribe what is happening in terms of just one prong. and pressure is proportional to the square root of The circular insert highlights the motion of the right intensity, prong. Here the center position is where the prong p!! would be at rest. When the fork is struck the prong is p/ I forced inward as shown by arrow a. After reaching the This simple relationship makes it easy to convert leftmost position it bounces back (arrow b), acceler- between sound intensity and sound pressure. ating along the way. The rapidly moving prong over- shoots the center and continues rightward (arrow c). ▪The Nature of Sound Sound is often defined as a form of vibration that It slows down along the way until it stops for an instant at the extreme right, where it reverses direction again and starts moving toward the left propagates through the air in the form of a wave. (arrow d) at an ever-increasing speed. It overshoots Vibration is nothing more than the to-and-fro motion (oscillation) of an object. Some examples include a 1 See Russell (2020) for detailed review of the characteristics and history playground swing, a pendulum, the floorboards under of tuning forks. | 12.09.22 - 17:44 8 1 Acoustics and Sound Measurement resting position. In fact, the prong is moving at its maximum speed as it passes through the resting position. The force of inertia causes the prong to overshoot the center and continue moving right- ward. The deformation process begins again once the prong overshoots its resting position. As a re- sult, opposing elastic restoring forces start building up again, now in the leftward direction. Just as be- fore, the increasing (leftward) restoring force even- tually overcomes the rightward inertial force, thereby stopping the prong’s displacement at the rightmost point, and causing a reversal in the di- rection of its movement. Hence, the same course of events is repeated again, this time in the leftward direction; then rightward, then leftward, etc., over and over again. This kind of vibration is called simple harmonic motion because the oscillations repeat themselves at the same rate over and over again. We know from experience that the oscillations just described do not continue forever. Instead, they dissipate over time and eventually die out com- pletely. The dying out of vibrations over time is called damping, and it occurs due to resistance or Fig. 1.3 After being struck, a tuning fork vibrates or friction. Resistance occurs because the vibrating oscillates with a simple pattern that repeats itself over prong is always in contact with the surrounding air. time. One replication (cycle) of this motion is illustrated going from frames 1 to 5. The arrows in the insert highlight As a result, there will be friction between the oscil- the motion of one of the prongs. lating metal and the surrounding air molecules. This friction causes some of the mechanical energy that has been supporting the motion of the tuning fork the center again, and as before, the prong now to be converted into heat. In turn, the energy that follows arrow a, slowing down until it stops has been converted into heat is no longer available momentarily at the extreme left. Here it reverses to maintain the vibration of the tuning fork. Conse- direction again and repeats the same process over quently, the sizes of the oscillations dissipate and and over again. One complete round trip (or repli- eventually die out altogether. cation) of an oscillating motion is called a cycle. A diagram summarizing the concepts just de- The number of cycles that occur in one second is scribed is shown in Fig. 1.4. The curve in the figure called frequency. represents the tuning fork’s motion. The amount of This form of motion occurs when a force is ap- displacement of the tuning fork prong around its plied to an object having the properties of inertia resting (or center) position is represented by dis- and elasticity. Due to its elasticity, the deformation tance above and below the horizontal line. These of the fork caused by the applied force is opposed by events are occurring over time, which is represented a restoring force. In the figure the initial leftward by horizontal distance (from left to right). The initial force is opposed by a restoring force in the opposite displacement of the prong due to the original ap- direction, that is, toward the right. The rightward plied force is represented by the dotted segment of restoring force increases as the prong is pushed the curve. Inertial forces due to the prong’s mass progressively toward the left. As a result, the move- and elastic restoring forces due to the elasticity of ment of the prong slows down and eventually the prong are represented by labeled arrows. Damp- stops. Under the influence of its elasticity the ing of the oscillations due to friction is shown by the prong now reverses direction and starts moving decline in the displacement of the curve as time rightward. As the restoring force brings the prong goes on. The curve in this diagram is an example of a back toward the center, we must also consider its waveform, which is a graph that shows displace- mass. Because the prong has mass, inertia causes it ment (or another measure of magnitude) as a func- to accelerate as it moves back toward its center tion of time. | 12.09.22 - 17:44 1 Acoustics and Sound Measurement 9 Fig. 1.4 Diagrammatic representation of tuning fork oscillations over time. Vertical displacement represents the amount of the tuning fork prong dis- placement around its resting position. Distance from left to right represents the progression of time. (Adapted from Gelfand 2018, courtesy of CRC Press.) Sound Waves decreases and is momentarily zero as they pass through their resting positions. As they continue to Tuning fork vibrations produce sound because the move to the left of their resting positions, the par- oscillations of the prongs are transmitted to the ticles are now becoming increasingly farther from surrounding air particles. When the tuning fork the molecules to their right (compared with when prong moves to the right, it displaces air molecules they are in their resting positions). We now say that to its right in the same direction. These molecules the air particles are rarefied compared with their are thus displaced to the right of their own resting resting states, so that the air pressure is now below positions. Displacing air molecules toward the right pushes them closer to the air particles to their atmospheric pressure. This state of lower than ambi- right. The pressure that exists among air molecules ent pressure is called rarefaction. When the air par- that are not being disturbed by a driving force (like ticles reach the leftmost position they are maximally the tuning fork) is known as ambient or atmospher- rarefied, which means that the pressure is maxi- ic pressure. We can say that the rightward motion mally negative. At this point, the restoring force in- of the tuning fork prong exerts a force on the air stigates a rightward movement of the air molecules. molecules that pushes them together relative to This movement is enhanced by the push of the tun- their undisturbed, resting situation. In other words, ing fork prongs that have also reversed direction. forcing the air molecules together causes an increase The air molecules now accelerate rightward (so that in air pressure relative to the ambient pressure that the amount of rarefaction decreases), overshoot existed among the undisturbed molecules. This state their resting positions, and continue to the right, of positive air pressure is called compression. The and so on. The tuning fork vibrations have now been amount of compression increases as the prong con- transmitted to the surrounding particles, which are tinues displacing the air molecules rightward. A now also oscillating in simple harmonic motion. maximum amount of positive pressure occurs when Sounds that are associated with simple harmonic the prong and air molecules reach their greatest motion are called pure tones. rightward displacement. Let us consider one of the air molecules that has The tuning fork prong then reverses direction, already been set into harmonic motion by the tun- overshoots its resting position, and proceeds to its ing fork. This air particle now vibrates to-and-fro in leftmost position. The compressed air molecules the same direction that was originally imposed by reverse direction along with the prong. This occurs the vibrating prong. When this particle moves to- because air is an elastic medium, so the particles ward its right it will cause a similar displacement of compressed to the right develop a leftward restoring the particle that is located there. The subsequent force. Small as they are, air particles do have mass. leftward motion is also transmitted to the next par- Therefore, inertia causes the rebounding air particles ticle, etc. Thus, the oscillations of one air particle are to overshoot their resting positions and to continue transmitted to the molecule next to it. The second toward their extreme leftward positions. As the par- particle is therefore set into oscillation, which in ticles move leftward, the amount of compression turn initiates oscillation of the next one, and so forth | 12.09.22 - 17:44 10 1 Acoustics and Sound Measurement down the line. In other words, each particle vibrates This propagation of vibratory motion from par- back and forth around its own resting point, and ticle to particle constitutes the sound wave. This wave causes successive molecules to vibrate back and appears as alternating compressions and rarefactions forth around their own resting points, as shown radiating from the sound source in all directions, as schematically in Fig. 1.5. Notice that each molecule already suggested in Fig. 1.1. The transmission of par- vibrates “in place” around its own average position; ticle motion along with the resulting variations in air it is the vibratory pattern that is transmitted pressure with distance from the source are repre- through the air. sented in Fig. 1.6. Most people are more familiar with the kinds of waves that develop on the surface of a pond when a pebble is dropped into the water. These are transverse waves because the particles are mov- ing at right angles to the direction that the wave is propagating. That is, the water particles oscillate up and down (vertically) even though the wave moves out horizontally from the spot where the pebble hit the water. This principle can be demonstrated by floating a cork in a pool, and then dropping a pebble in the water to start a wave. The floating cork reflects the motions of the water particles. The wave will move out horizontally, but the floating cork bobs up and down (vertically) at right angles to the wave. In contrast, sound waves are longitudinal waves because each air particle oscillates in the same direction in which the wave is propagating (Fig. 1.6). Although sound waves are longitudinal, it is more convenient to draw them with a transverse represen- tation, as in Fig. 1.6. In such a diagram, the vertical dimension represents some measure of the size of the signal (e.g., displacement, pressure, etc.), and left to Fig. 1.5 Sound is initiated by transmitting the vibratory right distance represents time (or distance). For ex- pattern of the sound source to nearby air particles, and then the vibratory pattern is passed from particle to ample, the waveform in Fig. 1.6 shows the amount of particle as a wave. Notice how it is the pattern of vibration positive pressure (compression) above the baseline, that is being transmitted, whereas each particle oscillates negative pressure (rarefaction) below the baseline, around its own average location. and distance horizontally going from left to right. Fig. 1.6 Longitudinal and transverse representations of a sound wave. Wave- length (λ) is the distance covered by one replication (cycle) of a wave, and is most easily visualized as the distance from one peak to the next. | 12.09.22 - 17:44 1 Acoustics and Sound Measurement 11 The Sinusoidal Function that we now have a right triangle in the circle, where r is the hypotenuse, θ is an angle, and d is Simple harmonic motion is also known as sinusoi- the leg opposite that angle. Recall from high school dal motion, and has a waveform that is called a math that the sine of an angle equals the length of sinusoidal wave or a sine wave. Let us see why. the opposite leg over the length of the hypotenuse. Fig. 1.7 shows one cycle of a sine wave in the center, Here, sin θ = d/r. If we conveniently assume that the surrounded by circles labeled to correspond to length of r is 1, then displacement d becomes the points on the wave. Each circle shows a horizontal sine of angle θ, which happens to be 0.707. In other line corresponding to the horizontal baseline on words, displacement is determined by the sine of the sine wave, as well as a radius line (r) that will the phase angle, and displacement at any point on move around the circle at a fixed speed, much like the sine wave corresponds to the sine of θ. This is a clock hand but in a counterclockwise direction. why it is called a sine wave. Point a on the waveform in the center of the fig- The peak labeled c on the sine wave corre- ure can be viewed as the “starting point” of the sponds to circle c, where the rotating radius has cycle. The displacement here is zero because this reached the straight-up position. We are now one- point is on the horizontal line. The radius appears as fourth of the way into the wave and one-fourth of shown in circle b when it reaches 45° of rotation, the way around the circle. Here, θ = 90° and the dis- which corresponds to point b on the sine wave. The placement is the largest it can be (notice that d = r angle between the radius and the horizontal is on the circle). Continuing the counterclockwise ro- called the phase angle (θ) and is a handy way to tell tation of r causes the amount of displacement from location going around the circle and on the sine wave. the horizontal to decrease, exemplified by point d In other words, we consider one cycle (one “round on the sine wave and circle d, where θ is 135°. The trip” of oscillation) to be the same as going around a oscillating air particle has already reversed direc- circle one time. Just as a circle has 360°, we also con- tion and is now moving back toward the resting po- sider one cycle to have 360°. Since 45/360 = 1/8, a sition. When it reaches the resting position there is phase angle (θ) of 45° is the same as one-eighth of again no displacement, as shown by point e on the the way around a circle or one-eighth of the way into sine wave and by the fact that r is now superim- a sine wave. Returning to the circle, the vertical dis- posed on the horizontal at θ = 180° in circle e. No- placement from the horizontal to the point where r tice that 180° is one half of the 360° round trip, so intersects the circle is represented by a vertical line we are now halfway around the circle and halfway labeled d. This vertical line corresponds to the dis- into the cycle of simple harmonic motion. In addi- placement of point b on the sine wave, where the tion, displacement is zero at this location (180°). displacement of the air particle is represented by Continuing the rotation of r places it in the lower the height of the point above the baseline. Notice left quadrant of circle f, corresponding to point f on Fig. 1.7 Sinusoidal motion (θ, phase angle; d, displacement). (Adapted from Gelfand 2018, courtesy of CRC Press.) | 12.09.22 - 17:44 12 1 Acoustics and Sound Measurement the wave, where θ = 225°. The oscillating particle has rotation. This is why the horizontal axis in Fig. 1.8 overshot its resting position and the displacement is can be labeled in terms of phase. As such, the phase now increasing in the other direction, so that we are of the wave at each of the points indicated in Fig. 1.7 in the rarefaction part of the wave. Hence, displace- is 0° at a, 45° at b, 90° at c, 135° at d, 180° at e, 225° ment is now drawn in the negative direction, indicat- at f, 270° at g, 315° at h, and 360° at i, which is also ing rarefaction. The largest negative displacement is 0° for the next cycle. reached at point g on the wave, where θ = 270°, corre- Phase is often used to express relationships be- sponding to circle g, in which r points straight down. tween two waves that are displaced relative to each The air particle now begins moving in the positive other, as in Fig. 1.8. Each frame in the figure shows direction again on its way back toward the resting two waves that are identical to each other except position. At point h and circle h, the displacement in that they do not line up exactly along the horizontal the negative direction has become smaller as the (time) axis. The top panel shows two waves that rotating radius passes through the point where are 45° apart. Here, the wave represented by the 0 = 315° (point h on the wave and circle h). The air thicker line is at 45° at the same time that the particle is again passing through its resting position at other wave (shown by the thinner line) is at 0°. point i, having completed one round trip or 360° of The phase displacement is highlighted by the rotation. Here, displacement is again zero. Having shaded area and the dotted vertical guideline in completed exactly one cycle, 360° corresponds to 0°, the figure. This is analogous to two radii that are and circle i is the same one previously used as circle a. always 45° apart as they move around a circle. In Recall that r rotates around the circle at a fixed other words, these two waves are 45° apart or out- speed. Hence, how fast r is moving will determine of-phase. The second panel shows the two waves dis- how many degrees are covered in a given amount of placed from each another by 90°, so that one wave is time. For example, if one complete cycle of rotation at 90° when other one is at 0°. Hence, these waves takes 1 second, then 360° is covered in 1 second; are 90° out-of-phase, analogous to two radii that are 180° is covered in 1/2 second; 90° takes 1/4 second; always 90° apart as they move around a circle. The 270° takes 3/4 second, etc. Hence, the phase angle third panel shows two waves that are 180° out-of- also reflects the elapsed time from the onset of phase. Here, one wave is at its 90° (positive) peak Fig. 1.8 Each panel shows two waves that are identical in every way except they are displaced from one another in terms of phase, highlighted by the shaded areas and the dotted vertical guidelines. Analogous examples of two radii moving around a circle are shown to the left of the waveforms. Top panel: Two waves that are 45° out-of-phase, analo- gous to two radii that are always 45° apart as they move around a circle. Second panel: Waves that are 90° out-of- phase, analogous to two radii moving around a circle 90° apart. Third panel: Waves that are 180° out-of phase, analogous to two radii that are always 180° apart (pointing in opposite direc- tions) moving around a circle. Bottom panel: Two waves (and analogous radii moving around a circle) that are 270° out-of-phase. | 12.09.22 - 17:44 1 Acoustics and Sound Measurement 13 at the same time that the other one is at its 270° kind of periodic wave because simple harmonic (negative) peak, which is analogous to two radii motion is the simplest form of vibration. Later we that are always 180° apart as they move around a will address complex periodic waves. circle. Notice that these two otherwise identical The duration of one cycle is called its period. The waves are exact mirror images of each other when period is expressed in time (t) because it refers to they are 180° out-of-phase, just as the two radii the amount of time that it takes to complete one are always pointing in opposite directions. The last cycle (i.e., how long it takes for one round trip). For example in the bottom panel shows the two waves example, a periodic wave that repeats itself every 270° out-of-phase. one-hundredth of a second has a period of 1/100 seconds, or t = 0.01 seconds. One-hundredth of a Parameters of Sound Waves second is also 10 thousandths of a second (millisec- onds), so we could also say that the period of this We already know that a cycle is one complete repli- wave is 10 milliseconds. cation of a vibratory pattern. For example, two Similarly, a wave that repeats itself every one- cycles are shown for each sine wave in the upper thousandth of a second has a period of 1 millisecond frame of Fig. 1.9, and four cycles are shown for each or 0.001 seconds; and the period would be 2 milli- sine wave in the lower frame. Each of the sine waves seconds or 0.002 seconds if the duration of one cycle in this figure is said to be periodic because it repeats is two-thousandths of a second. itself exactly over time. Sine waves are the simplest The number of times a waveform repeats itself in one second is its frequency (f), or the number of cycles per second (cps). We could say that fre- quency is the number of cycles that can fit into one second. Frequency is expressed in units called hertz (Hz), which means the same thing as cycles per sec- ond. For example, a wave that is repeated 100 times per second has a frequency of 100 Hz; the frequency of a wave that has 500 cycles per second is 500 Hz; and a 1000 Hz wave has 1000 cycles in one second. If frequency is the number of cycles that occur each second, and period is how much time it takes to complete one cycle, then frequency and period must be related in a very straightforward way. Let us consider the three examples that were just used to illustrate the relationship of period and frequency: A period of 0.01 seconds corresponds to a frequency (f) of 100 Hz. A period of 0.002 seconds corresponds to a frequency of 500 Hz. A period of 0.001 seconds corresponds to a frequency of 1000 Hz. Now, notice the following relationships among these numbers: 1/100 = 0.01 and 1/0.01 = 100. 1/500 = 0.002 and 1/0.002 = 500. 1/1000 = 0.001 and 1/0.001 = 1000. In each case, the period corresponds to 1 over the frequency, and the frequency corresponds to 1 over the period. In formal terms, frequency equals the re- ciprocal of period, and period equals the reciprocal Fig. 1.9 Each frame shows two sine waves that have the of frequency, same frequency but different amplitudes. Compared with the upper frame, twice as many cycles occur in the same amount of time in the lower frame; thus, the period is half as long and 1 f ¼ the frequency is twice as high. t | 12.09.22 - 17:44 14 1 Acoustics and Sound Measurement Each wave in the upper frame of Fig. 1.9 contains average of the positive and negative instantaneous two cycles in 4 milliseconds, and each wave in the amplitudes will not work because it will always be lower frame contains four cycles in the 4 millisec- equal to zero. A different kind of overall measure is onds. If two cycles in the upper frame last 4 millisec- therefore used, called the root-mean-square (RMS) onds, then the duration of one cycle is 2 milliseconds. amplitude. Even though measurement instruments Hence, the period of each wave in the upper frame provide us with RMS amplitudes automatically, we is 2 milliseconds (t = 0.002 s), and the frequency can understand RMS by briefly reviewing the steps is 1/0.002, or 500 Hz. Similarly, if four cycles last that would be used to calculate it manually: (1) All 4 milliseconds in the lower frame, then one cycle of the positive and negative values on the wave are has a period of 1 millisecond (t = 0.001 s), and the squared, so that all values are positive (or zero for frequency is 1/0.001, or 1000 Hz. values on the resting position itself). (2) A mean Fig. 1.9 also illustrates differences in the ampli- (average) is calculated for the squared values. (3) tude between waves. Amplitude denotes size or This average of the squared values is then rescaled magnitude, such as the amount of displacement, back to the “right size” by taking its square root. power, pressure, etc. The larger the amplitude at This is the RMS value. The RMS amplitude is nu- some point along the horizontal (time) axis, the merically equal to 70.7% of (or 0.707 times) the peak greater its distance from zero on the vertical axis. amplitude (or 0.354 times the peak-to-peak ampli- With respect to the figure, each frame shows one tude). Even though these values technically apply wave that has a smaller amplitude and an other- only to sinusoids, for practical purposes RMS values wise identical wave that has a larger amplitude. are used with all kinds of waveforms. As illustrated in Fig. 1.10, the peak-to-peak Referring back to Fig. 1.6, we see that the dis- amplitude of a wave is the total vertical distance tance covered by one cycle of a propagating wave is between its negative and positive peaks, and peak called its wavelength (λ). We have all seen water amplitude is the distance from the baseline to one waves, which literally appear as alternating crests peak. However, neither of these values reflects the and troughs on the surface of the water. Using this overall, ongoing size of the wave because the ampli- common experience as an example, wavelength is tude is constantly changing. At any instant an oscil- simply the distance between the crest of one wave lating particle may be at its most positive or most and the crest of the next one. For sound, wavelength negative displacement from the resting position, or is the distance between one compression peak and anywhere between these two extremes, including the next one, or one rarefaction peak and the next the resting position itself, where the displacement one, that is, the distance between any two succes- is zero. The magnitude of a sound at a given instant sive peaks in Fig. 1.6. It is just as correct to use any (instantaneous amplitude) is applicable only for other point, as long as we measure the distance be- that moment, and will be different at the next mo- tween the same point on two successive replications ment. Yet we are usually interested in a kind of of the wave. The formula for wavelength is “overall average” amplitude that reveals the magni- c tude of a sound wave throughout its cycles. A simple λ¼ f where f is the frequency of the sound and c is the speed of sound (~ 344 m/s in air). This formula indi- cates that wavelength is inversely proportional to frequency. Similarly, frequency is inversely propor- tional to wavelength: c f ¼ λ These formulas show that wavelength and frequency are inversely proportional to each other. In other words, low frequencies have long wavelengths and high frequencies have short wavelengths. Complex Waves When two or more pure tones are combined, the re- sult is called a complex wave. Complex waves may Fig. 1.10 Peak, root-mean-square (RMS), and peak-to- contain any number of frequencies from as few as peak amplitude. two up to an infinite number of them. Complex | 12.09.22 - 17:44 1 Acoustics and Sound Measurement 15 periodic waves have waveforms that repeat them- selves over time. If the waveform does not repeat itself over time, then it is an aperiodic wave. Combining Sinusoids The manner in which waveforms combine into more complex waveforms involves algebraically adding the amplitudes of the two waves at every point along the horizontal (time) axis. Consider two sine waves that are to be added. Imagine that they are drawn one above the other on a piece of graph paper so that the gridlines can be used to identify similar moments in time (horizontally) for the two waves, and their am- plitudes can be determined by simply counting boxes vertically. The following exercise is done at every point along the horizontal time axis: (1) Determine the amplitude of each wave at that point by counting the boxes in the positive and/or negative direction. (2) Add these two amplitudes algebraically (e.g., + 2 plus + 2 is + 4; –3 plus –3 is –6; and + 4 plus – 1 is + 3, etc.). (3) Plot the algebraic sum just obtained on the graph paper at the same point along the hori- zontal time axis. After doing this for many points, drawing a smooth line through these points will reveal the combined wave. Several examples of combining two sinusoids are illustrated in Fig. 1.11. This figure shows what occurs when two sinusoids being combined have exactly the same frequencies and amplitudes. The two sinusoids being combined in Fig. 1.11a are in Fig. 1.11 Combining sinusoids that have the same fre- phase with each other. Here, the combined wave quency and amplitude when they are (a) in-phase looks like the two identical components, but has an (showing complete reinforcement); (b) 180° out-of-phase amplitude twice as large. This case is often called (showing cancellation); and (c) 90° out-of-phase. complete reinforcement for obvious reasons. The addition of two otherwise identical waves that are 180° out-of-phase is illustrated in Fig. 1.11b. In this those just described for two similar waves: Their case, the first wave is equal and opposite to the sec- amplitudes are algebraically summed on a point-by- ond wave at every moment in time, so that algebraic point basis along the horizontal (time) axis, addition causes the resulting amplitude to be zero at regardless of their individual frequencies and ampli- every point along the horizontal (time) axis. The tudes or their phase relationships. However, combin- result is complete cancellation. ing unequal frequencies will not produce a sinusoidal When the sinusoids being combined are identi- result. Instead, the combined waveform depends on cal but have a phase relationship that is any value the nature of the particular sounds being combined. other than 0° (in-phase) or 180° (opposite phase), For example, consider the three different sine waves then the appearance of the resulting wave will de- labeled f1, f2, and f3 in Fig. 1.12. Wave f1 has a fre- pend on how the two components happen to line up quency of 1000 Hz, f2 is 2000 Hz, and f3 is 3000 Hz. in time. Fig. 1.11c shows what happens when the The lower waveforms show various combinations of two otherwise identical sinusoids are 90° out-of- these sinusoids. The combined waves (f1 + f2, f1 + f3, phase. The result is a sinusoid with the same fre- and f1 + f2 + f3) are no longer sinusoids, but they are quency as the two (similar) original waves but that periodic because they repeat themselves at regular differs in phase and amplitude. intervals over time. In other words, they are all com- plex periodic waves. Notice that the periods of the complex periodic Complex Periodic Waves waves in Fig. 1.12 are the same as the period of f1, The principles used to combine any number of simi- which is the lowest frequency component for each lar or dissimilar waves are basically the same as of them. The lowest frequency component of a | 12.09.22 - 17:44 16 1 Acoustics and Sound Measurement Fig. 1.12 Waveforms (left) and corre- sponding spectra (right) for three har- monically related sine waves (pure tones) of frequencies f1, f2, and f3; and com- plex periodic waves resulting from the in-phase addition of f1 + f2, f1 + f3, and f1 + f2 + f3. Notice that the fundamental frequency is f1 for all of the complex waves. Also notice that each pure tone spectrum has one vertical line, while the spectrum of each complex periodic sound has a separate vertical line for each of its components. complex periodic wave is called its fundamental determined by examining its waveform. In fact, frequency. The fundamental frequency of each of the same frequencies can result in dramatically the complex periodic waves in the figure is 1000 Hz different-looking complex waveforms if their phase because f1 is the lowest component in each of relationships are changed. Hence, another kind of them. The period (or the time needed for one com- graph is needed when we want to know what fre- plete replication) of a complex periodic wave is the quencies are present. This kind of graph is a spec- same as the period of its fundamental frequency. trum, which shows amplitude on the y-axis as a Harmonics are whole number or integral multi- function of frequency along the x-axis. Several ex- ples of the fundamental frequency. In other words, amples are given in Fig. 1.12 and Fig. 1.13. The fre- the fundamental is the largest whole number quency of the pure tone is given by the location of a common denominator of a wave’s harmonics, and vertical line along the horizontal (frequency) axis, the harmonics are integral multiples of the funda- and the amplitude of the tone is represented by the mental frequency. In fact, the fundamental is also height of the line. According to Fourier’s theorem, a harmonic because it is equal to 1 times itself. In complex sounds can be mathematically dissected the case of wave f1 + f2 + f3, 1000 Hz is the funda- into their constituent pure tone components. The mental (first harmonic), 2000 Hz is the second process of doing so is called Fourier analysis, which harmonic, and 3000 Hz is the third harmonic. results in the information needed to plot the spec- Another example of combining sinusoids into a trum of a complex sound. The spectrum of a com- complex periodic wave is given in Fig. 1.13. Here, plex periodic sound has as many vertical lines as the sine waves being added are odd harmonics of there are component frequencies. The locations of the fundamental (1000 Hz, 3000 Hz, 5000 Hz, etc.), the lines show their frequencies, and their heights and their amplitudes get smaller with increasing show their amplitudes, as illustrated in Fig. 1.13. frequency. The resulting complex periodic wave becomes squared off as the number of odd har- monics is increased, and is called a square wave Aperiodic Waves for this reason. Aperiodic sounds are made up of components that Waveforms show how amplitude changes with