01intro_physics_1

426Week 9: Oscillationswhere we used the fact that ∆ =LN x∆ just asxL=N ax.Finally, we rearrange this by putting all of the extra terms together as follows:FA= − keffa∆ LL= − Y∆ LL(897)In words, we defineF/Ato be theStress, ∆L/Lto be theStrain, andY=k /aeff(898)to beYoung’s Modulus. With these definitions, the equation above states that:Compressive or extensive Stress applied to a solid equals Young’s Modulustimes the Strain.Note well that we can rearrange this equation into “Hooke’s Law” as follows:F x= −Y AL∆ =x− K ∆ x(899)which basically states that any given chunk of materialbehaves like a springwhen forces are appliedto stretch or compress it, with a spring constantKthat isproportional to the cross sectionalarea A and inversely proportional to the length L!Young’s modulus is the material-dependentcontribution from the actual geometry and interaction potential energy that holds the materialtogether, but the rest of the dependence isgeneric, and applies to all materials.Thisscalingbehavior of the response of solid (or liquid, or gaseous) matter to applied forcesis the important take-home conceptual lesson of this section. Substances like bone or steel or woodor nearly anything arestronger– respond less to an applied force – as they getthicker, and areweaker– respond more to an applied force – as they getlonger. This intuitive understanding canbe made quantitatively precise to be sure (and for e.g. engineers or orthopedic surgeons willhaveto be quantitatively precise as they craft buildings that won’t fall down or artificial hips or bonesthat mimic natural ones) but is sufficient in and of itself for people to see the world through neweyes, to understand why there are building codes for houses and decks governing the lengths andcross-sectional areas of support beams and joists and so on. We’ll explore a few of these applicationslater, but first let us relativelyquicklyextend the linear extension/compression result tosidewaysforces and understandshear.9.5.4: Shear Forces and the Shear ModulusIf we imagine taking our chunk of model solid pictured in figure 131 above and pushingsidewaysonthe top to the right while the bottom is locked to a table by e.g. static friction on the bottom, we areapplying ashear force– one that bends the material sideways instead of stretching or compressing it.The layers of the material all shift sideways by an amount that distributes the shear force betweenall of the layers as shown and the material deformsat an angleθas shown.A fully microscopic derivation of the response is still quite possible, but not quite as simple asthat for Young’s modulus above. For that reason we will content ourselves with a simple heuristicextension of thescalingideas we learned in the previous section.The displacement layer to layer required to oppose a given shear force is going to be inverselyproportional to the thickness of the materialL, because the longer the material is, the more layersthere are to split the restoring force among. So we expectF sto be proportional to 1/L. The workwe do as the material bends comes primarily from the stretching of the vertical bonds. The numberof vertical bonds is proportional to the cross-sectional areaAof the material at the shear surfaceon the top and bottom. More bonds, a stronger force for a given displacement. We expectF sto

Week 9: Oscillations427be proportional toA. As before, we expect the response to depend on the microscopic propertiesof the material and its structure in some way – we wrap all our ignorance of these properties intoa modulus that we can more easilymeasure in the labthancompute from first principles. We thenassemble thesescalingresults into:F s= −M AsL∆ x(900)whereM sis theshear modulus– the equivalent of Young’s modulus for shear forces. Once againwe can put this in a form involvingshear stressF /Asandshear strain∆x/L:F sA= − M s∆ xL(901)In words:Shear Stress applied to a solid equals the Shear Modulus times the Strain.where note well the minus sign that as always means that the reaction forceopposesthe appliedshear force, trying to make ∆xsmaller in magnitudeno matter what direction the material is beingbent.Shear stress is how springs really work! A “spring” is just a very long piece of material wrappedaround into a coil so that when the ends are pulled, it produces shear stress spread out across theentire coil! So it is not anaccidentthat we discover Hooke’s Law buried within this result – this isa (slighly handwaving)microscopic derivationof Hooke’s Law:F x= −M AsL∆ =x− k∆ x(902)where:k=M AsL(903)From this, we can predict how the strength of springs will vary according to the length of the wirein the coilLand its thicknessA. Pretty cool!9.5.5: Deformation and FractureWhat happens when one stretches or compresses or shears a material by an amount (per bond) thatisnotsmall? Here is a verbal description of what we might expect based on the microscopic modelabove and our own experience.For a range of (relatively small) stresses, the strain remains alinear response– proportionalto the stress, acting in a direction opposed to the stress. As one increases the stress, then, thefirst thing that happens is that the response becomes non-linear. Each additional increment ofstress producesmorethan a linear increment of strain stretching, and quite possiblelessthan alinear increment compressing, as one can understand from the graph of a typical interatomic orintermolecular force above. Materials tend to resist compression better than extension because onecan pull bonded atoms apart but one cannot cause two bonded atoms to interpenetrate with anyreasonable force – they arevery strongly repulsivewhen their electron clouds start to overlap butonly weakly attractive as the electron clouds are pulled apart.At some point, one adds enough energy to the bonds (doingworkas one e.g. stretches thematerial) that atoms or molecules start todislocate– leave their normal place in the lattice orstructure and migrate someplace else. This leaves behind adefect– a missing atom or moleculein the structure – and correspondingly weakens the entire structure. This migration tends to bepermanent– even if one releases the stress on the material, it does not return to its original state but

428Week 9: Oscillationsremains stretched, compressed, or bent out of place. This region is one ofpermanent deformationof the material, as when one bends a paper clip.Finally, if one continues to ratchet up the stress, at some point the number of defects introducedreaches a critical point and each new defect produced weakens the material enough that anotherdefect is produced without further increase in stress, as the number of bonds over which the stressis distributed decreases with each new defect. The number of defects “explodes”, creating afractureplaneand the materialbreaks.|F/A|∆| L/L |ABClinear responsenonlinear responsepermanent deformationfractureFigure 133: The slope of the linear response part of the curve is Young’s Modulus (note well theabsolute magnitude signs).Amarks the point where nonlinear response is first apparent.Bmarksthe boundary where dislocations occur and permanent deformation of the material begins.Cmarksthe point where there is a chain reaction – each defect produced produces on average at least onemore defect at the same stress, so that the material “instantly” fractures.These behaviors – and the critical points where the behavior changes – are summarized on thegraph in figure 133.Materials are not just characterized by their moduli. Moduli describe only the simple linearresponse regime. Furthermore, the moduli themselves depend on how far the atoms or moleculesof the material are spending part of their time away from equilibriumwithoutstress of any sort,because the material has atemperature(basically, a mechanical energy per atom that islargerthanthe minimum associated with sitting at the equilibrium point, at rest). In thermal equilibrium, eachatom isalreadyoscillating back and forth around its equilibrium position and temperature increasesalone are sufficient to drive the system through the same series of states – linear and nonlinearresponse where it can cool back to the original structure, nonlinear response that introduces defectsso that the material “starts to melt” and doesn’t come back to its original state, followed bymeltinginstead offracturewhen the energy per bond no longer keeps atoms localized to a lattice or structureat all. When one mixes heating and stress, one can get many different ranges of physical behavior.One last qualitative measure of interest isbrittlenessortoughness– a measure of how likely amaterial is to bend versus fracture when stresses are applied. Some materials, such as steel, are verytough and not very brittle – they do not easily deform or fracture. Others, like diamond, can bevery hard indeed but are easy to fracture – they arebrittle.Human bone varies tremendously in the range of its brittleness with the lifetime of the humaninvolved, with genetic factors, with dietary factors, and with the history of the person involved.Young people (on average) have bones that bend easily but aren’t very brittle. Old people havebones that become progressively more brittle as they decalcify with age or disease. Normal adultstend to fall in between – bones that are not as flexible but that are also not particularly brittle.

Week 9: Oscillations429In all cases thick bones are stronger than thin ones, long bones are weaker than short bones,allthings being equal(which often is not the case). Still, this section should give you a good chanceof understanding at least semi-quantitatively how bone strength varies and can be described witha few empirical parameters that can be connected (with a fair bit of work) all the way back to theintermolecular bonds within the bone itself and its physical structure.9.6: Human BoneFigure 134: This figure illustrates the principle anatomical features of bone.The bone itself is a composite material made up of a mix of living and dead cells embedded ina mineralized organic matrix. It has significant tensor structure – looking somewhat like a randomhoneycomb structure in cross section but with a laminated microstructure along the length of thebone. Its anatomy is illustrated in figure 134.Bone is layered from the outside in. The very outer hard layer of a bone is called isperios-teum201. In between is compact bone, orosteonthat gives bone much of its strength. Nutrientsflow into living bone tissue through holes in the bone calledforamen, and are distributed up anddown through the osteon throughhaversian canals(not shown) that are basically tubes throughthe bone for blood vessels that runalongthe bone’s length to perfuse it. The periosteum and osteonmake up roughly 80% of the mass of a typical long bone.Inside the osteon is softer inner bone calledendosteum202. The inner bone is made up of a mixof different kinds of bone and other tissue that include spongy bone calledtrabeculaeand bonemarrow (where blood cells are stored and formed). It has only 20% of the bone’s mass, but 90% ofthe bone’s surface area. Much of the spongy bone material is filled with blood, to the point where agood way to characterize the difference between the osteon and the trabeculae is that in the former,bone surrounds blood but in the latter, blood surrounds bone.201Latin for “outer bone”. But it sounds much cooler in Latin, doesn’t it?202Latin for – wait for it – “inner bone”.

430Week 9: OscillationsThe bone matrix itself is made up of a mix of inorganic and organic parts. The inorganic partis formed mostly of calcium hydroxylapatite (a kind of calcium phosphate that is quite rigid). Theorganic part is collagen, a protein that gives bone its toughness and elasticity in much the sameway that tough steels are often a mix of soft iron and hard cementite particles, with the lattercontributing hardness and compressive/extensive strength, the latter reducing the brittleness thatoften accompanies hardness and giving it a broader range of linear response elasticity.There are two types of microscopically distinct bone.Woven bonehas collagen fibers mixedhaphazardly with the inorganic matrix, and is mechanically weak.Lamellar bonehas a regularparallel alignment of collagen fibers intosheets(lamellae) that is mechanically strong. The lattergive the osteon a laminar/layered structure aligned with the bone axis. Woven bone is an earlydevelopmental state of lamellar bone, seen in fetuses developing bones and in adults as the initial softbone that forms in a healing fracture. It serves as a sort of template for the replacement/formationof lamellar bone.Bones are typically connected together with surface layers of cartilage at the joints, augmentedby tough connective tissue and tendons smoothly integrated into muscles that permit mobile bonesto be articulated at the joints. Together, they make an impressive mechanical structure capable ofan extraordinary range of motions and activities while still supporting and protecting softer tissueof our organs and circulatory system. Pretty cool!Bone is quite strong. It fractures under compression at a stress of around 170 MPa in typicalhuman bones, but has a smaller fracture stress under tension at 120 MPa and is relatively ¡I¿easy¡/i¿to fracture with shear stresses of around 52 MPa. This is why it is “easy” (and common!) to breakbones with shear stresses, less common to break them from naked compression or tension – typicallyother parts of the skeletal structure – tendons or cartilage in the joints – fail before the actual bonedoes in these situations.Bone is basically brittle and easy to chip, but does have a significant degree of compressive,tensile, or shear elasticity (represented by e.g. Young’s modulus in the linear regime) contributedprimarily by collagen in the bone tissue. Younger humans still have relatively elastic bones; as oneages one’s bones become first harder and tougher, and then (as repair mechanisms break down withage) weaker and more brittle.Figure 135: Illustration of the alteration of bone tissue accompanying osteoporosis.Figure 135 above shows the changes in bone associated withosteoporosis, the gradual thinning

Week 9: Oscillations431of the bone matrix as the skeleton starts to decalcify. This process is associated with age, especiallyin post-menopausal women, but it can also occur in association with e.g. corticosteroid therapy,cancer, or other diseases or conditions such as Paget’s disease in younger adults.This process significantly weakens the bones of those afflicted to the point where the static shearstresses associated with muscular articulation (for example, standing up) can break bones. A youngperson might fall and break their hip where a person with really significant osteoporosis can actuallybreak their hip (from the stress of standing) and then fall. This is not a medical textbook andshould not be treated as an authoritative guide to the practice of medicine (but rather, as a basis forunderstanding what one might by trying to learnina directed study of medicine), but with that said,osteoporosis can be treated to some extent by things like hormone replacement therapy in women (itseems to go with the reduction of estrogen that occurs in menopause), calcium supplementation tohelp slow the loss of calcium, and certain drugs such asFosamax(Alendronate) that reduce calciumloss and increase bone density (but that have risks and side effects).Example 9.6.1: Scaling of Bones with Animal SizeAn interesting biological example ofscaling lawsin physics – and the reason I emphasize the de-pendence of many physically or physiologically interesting quantities on length and/or area – canbe seen in the scaling of animal bones with the size of the animal203. Let us consider this.We have seen above that the scaling of the “spring constant” of a given material that governs itschange in length or its transverse displacement under compression, tension, or shear is:keff=XAL(904)whereXis the relevant (compression, tension, shear) modulus. Bone strength, including the pointwhere the bone fractures under stresses of these sorts, might very reasonably be expected to beproportional to this constant and to scale similarly.The leg bones of a four-legged animal have to be able to support the weight of that animal undercompressive stress. This enables us to make the following scaling argument:•In general, the weight of any animal is roughly proportional to its volume. Most animals aremostly made of water, and have a density close to that of water, so the volume of the animaltimes the density of water is a decent approximate guess of what its weight should be.•In general, the volume of an animal (and hence its weight) is proportional to any characteristiclength scale that describes the animalcubed. Obviously this won’t work well if one comparesa snake, with one very long length and two very short lengths, to a comparatively roundhippopotamus, but it won’t be crazy comparing mice to dogs to horses to elephants that allhave reasonably similar body proportions. We’ll choose the animal height.•The leg bones of the animal have a strength proportional to the cross-sectional area.We would like to be able to estimate the thickness of an animal’s bone it’s known height andfrom a knowledge of the thickness ofone kindof a “reference” animal’s bone and its height.Our argument then is: The volume, and hence the weight, of an animal increases like the cube ofits characteristic length (e.g. its heightH). The strength of its bones that must support this weightgoes like the square of the diameterDof those bones. Therefore:H 3=CD2(905)203Wikipedia: http://www.wikipedia.org/wiki/On Being the Right Size. This and many other related argumentswere collected by J. B. S. Haldane in an article titledOn Being the Right Size, published in 1926. Collectively theyare referred to as theHaldane Principle. However, the original idea (and 3/2 scaling law discussed below) is due tonone other than Galileo Galilei!

432Week 9: OscillationswhereCis some constant of proportionality. Solving forD H( ) we get:D =1√ CH3 2/(906)This simple equation is approximately satisfied, although notexactlyas given because our modelfor bones breaking does not reflect shear-driven “buckling” and a related need for muscle to scale,for mammals ranging from small rodents through the mighty elephant. Bone thickness does indeedincrease nonlinearly with respect to body size.

Week 9: Oscillations433Homework for Week 9Problem 1.Physics Concepts:Make this week’s physics concepts summary as you work all of the problemsin this week’s assignment. Be sure to cross-reference each concept in the summary to the problem(s)they were key to, and include concepts from previous weeks as necessary. Do the work carefullyenough that you can (after it has been handed in and graded) punch it and add it to a three ringbinder for review and study come finals!Problem 2.Harmonic oscillation is conceptually very important because (as has been remarked in class) manythings that arestablewill oscillate more or less harmonically if perturbed a small distance awayfrom their stable equilibrium point. Draw anenergy diagram, agraphof a “generic” interactionpotential energy with astable equilibrium pointand explain in words and equations where, andwhy, this should be.Problem 3.mLgmoon= gearth/6MoonA pendulum with a string of lengthLsupporting a massmon the earth has a certain periodT. A physicist on the moon, where the acceleration near the surface is aroundg/6, wants to makea pendulum with the same period. What massm mand lengthL mof string could be used toaccomplish this?

434Week 9: OscillationsProblem 4.mmmkk 1k 2k 1k 2effIn the figure above identical massesmare attached to two springsk 1andk 2in “series” – oneconnected to the other end-to-end – and in “parallel” – the two springs side by side. In the thirdpicture another identical massmis shown attached to a single spring with constantkeff.a) What shouldkeffbe in terms ofk 1andk 2for the series combination of springs so that theforce on the mass is the same for any given displacement ∆ from equilibrium?xb) What shouldkeffbe in terms ofk 1andk 2for the parallel combination of springs so that theforce on the mass is the same for any given displacement ∆ from equilibrium?xNote that this problem illustrates the scaling of Young’s modulus with length (series) or area(parallel). It also gives you a head start on how series and parallel addition of resistors and capacitorsworks next semester! It’s therefore well worth puzzling over.

Week 9: Oscillations435Problem 5.P leftPrightP leftP leftP leftPrightPrightPrightcI vI vI vI vabdLrIn this class we usually idealize fluid flow by neglecting resistance (drag) and the viscosity of thefluid as it passes through cylindrical pipes so we can use the Bernoulli equation. As discussed inclass, however, it is often necessary to have at sound conceptual understanding of at least thequal-itative effectsof including the resistance and viscosity. UsePoiseuille’s Lawand the conceptsofresistance, pressure, and flowfrom the online textbook to answer and discuss the follow-ing simple questions. Make sure your conceptual understanding of these concepts isqualitativelysound!a) Is ∆P a= Pleft− Prightgreater than, less than, or equal to zero in figure a) above, where bloodflows at a rateI vhorizontally through a blood vessel with constant radiusrand some lengthLagainst theresistanceof that vessel?b) If the radiusrincreases(while flowI vand lengthLremain the same as in a), does thepressure difference ∆P bincrease, decrease, or remain the same compared to ∆P a ?c) If the lengthincreases(while flowI vand radiusrremains the same as in a), does the pressuredifference ∆P cincrease, decrease, or remain the same compared to ∆P a ?d) If the viscosityµof the bloodincreases(where flowI v, radius , and lengthrLare all un-changed compared to a) do you expect the pressure difference ∆P ddifference across a bloodvessel to increase, decrease, or remain the same compared to ∆P a ?Blood viscosity is chemically related to things like the “stickiness” of platelets and their abundancein the blood. Viscosity and the radiusrof blood vessels can be regulated or altered (within limits)by drugs, disease, diet and exercise – all of which have acompletely understandableeffect uponblood pressure based on the simple ideas above.

436Week 9: OscillationsProblem 6.A2H/3HwaterThis problem will help you learn required concepts such as:•Static equilibrium•Archimedes Principle•Simple Harmonic Oscillationso please review them before you begin.rgb is fishing in still water off of the old dock. He is using acylindricalbobber as shown. Thebobber has a cross sectional area ofAand a length ofH, and is balanced so that it remains verticalin the water. When it is floatingat equilibrium(supporting the weight of the hook and wormdangling underneath) 2H/3 of its length is submerged in the water. You can neglect the volumeof the line, hook and worm (that is, theirbuoyancyis negligible), and neglect all drag/dampingforces. Answer the following questions in terms ofA H ρ,,w(the density of water), :ga) What is the combined massMof the bobber, hook and worm from the data?b) A fish gives a tug on the worm and pulls the bobber straight down anadditional distance(from equilibrium) . What is theynetrestoring force on the bobber as a function of ?yc) Use this force and the calculated massMfrom a) to write Newton’s 2nd Law for the motionof the bobber up and down (in ) and turn it into an equation of motion in standard form forya harmonic oscillator.d) Assume that the bobber is pulled down the specific distancey=H/3 andreleased fromrestat timet= 0. Neglecting the damping effects of the water, write an equation for thedisplacement of the bobber from its equilibrium depth as a function of time, ( ) (the solutiony tto the equation of motion).e) With what frequency does the bobber bob? Evaluate your answer forH= 10 cm,A= 1 cm2(reasonable values for a fishing bobber).

Week 9: Oscillations437Problem 7.LMmax θThis problem will help you learn required concepts such as:•Simple Harmonic Oscillation•Torque•Newton’s Second Law for Rotation•Moments of Inertia•Small Angle Approximationso please review them before you begin.A grandfather clock is constructed with a pendulum that consists of a long, light (assume mass-less) rod and a small, heavy (assume point-like) massMthat can be slid up and down the rodchangingLto “tune” the clock. The clock keeps perfect time when the period of oscillation of thependulum isT= 2 seconds. When the clock is running, the maximum angle the rod makes withthe vertical isθmax= 0 05 radians (a “small angle”)..a)Derivethe equation of motion for the rod when it freely swings and solve for ( ) assuming itθ tstarts atθmaxat = 0. You may use either force or torque.tb) At what distanceLfrom the pivot should the mass be set so that the clock keeps correct time?As always, solve the problem algebraicallyfirstand only then worry about numbers.c) In your answer above, you idealized by assuming the rod to be massless and the ball to bepointlike. In the real world, of course, the rodhas a small massmand the ball isa ball ofradiusr, not a point mass. Will the clock run slow or fast if you set the massexactlywhereyou computed it in part b)? Should you reduceLor increaseLa bit to compensate and getbetter time? Explain.

438Week 9: OscillationsProblem 8.hdθdpivotkkMThis problem will help you learn required concepts such as:•Simple Harmonic Oscillation•Springs (Hooke’s Law)•Small Angle Approximationso please review them before you begin.The pole in the figure has massMand lengthhis supported by two identical springs with springconstantkthat connect its platform to a table as shown. The springs attach to the platform adistancedfrom the base of the pole, which is pivoted on a frictionless hinge so it can rotate only inthe plane of the page as shown. Gravity acts down. Neglect the mass of the platform.a) On a copy of the right-hand figure, indicate the forces acting on the pole and platform whenthe pole tips over as shown, deviating from the vertical by a (small) angle .θb) Using your answer to a), find thetotal torqueacting on the system (pole and platform) aboutthe hinged bottom of the pole, and note its direction. Use the small angle approximationssin( )θ≈ θ; cos( )θ≈ 1.c) Find theminimal valuefor the spring constantkof the two springs such that the pole isstable in the vertical position. This means that a small deviation as above produces a torquethat restores the pole to the vertical.d) ForM= 50 kg,h= 1 0 meter,.d= 0 5 meter, and springs with spring constant.k= 9600N/m, find the angular frequency with which the pole oscillates about the vertical.

Week 9: Oscillations439Advanced Problem 9.PAay0 P0This problem will help you learn required concepts such as:•Bernoulli’s Equation•Torricelli’s Lawso please review them before you begin.In the figure above, a large drum of water is open at the top and filled up to a heightyabove atap at the bottom (which is also open to normal air pressure). The drum has a cross-sectional areaAat the top and the tap has a cross sectional area ofaat the bottom.The questions below help you usecalculusto determine how long it will take for the drumto empty. The first two questions and the last question anybody can answer. The one where youactually have to integrate may be difficult for non-math/physics/engineering students, but feel freeto give it a try!a) Find the speed with which the water emerges from the tap. Assume laminar flow withoutresistance. Compare your answer to the speed a mass has after falling a heightyin a uniformgravitational field (after usingA≫ ato simplify your final answer, Torricelli’s Law).b)Start by guessinghow long it will take water to flow out of the tank byusing dimensionalanalysisand the insight gained from a). That is, think about how youexpectthe time to varywith each quantity that describes the problem and form a simple expression with the relevantparameters that has the right units and goes up where it should go up and down where itshould go down.c) Next, find analgebraic expressionfor the velocity of the top. This isdy/dt.d) Manupulate the expression and integratedyon one side, an expression withdton the otherside, to find the time it takes for the top of the water to reach the bottom of the tank. Compareyour answer to b) to your answer to d) and the time it takes a mass to fall a heightyin auniform gravitational field. Does the correct answer make dimensional and physical sense?You might want to think a bit about Toricelli’s Law here as well...e) Evaluate the answers to a) and d) forA= 0 50 m ,.2a= 0 5 cm , (0) = 100 cm..2y

440Week 10: The Wave EquationOptional ProblemsContinue studying for the final exam!Only three “weeks” (chapters) to go in thistextbook, finals are coming apace!

Week 10: The Wave EquationWave Summary•Wave Equationd y 2dt2− v 2d y 2dx2= 0(907)where for waves on a string:v= ±s Tµ(908)•Superposition Principley x, t() =Ay x, t1 () +By x, t2 ()(909)(sum of solutions is solution). Leads to interference, standing waves.•Travelling Wave Pulsey x, t() = (f x± vt)(910)where ( ) is an arbitrary functional shape or pulsef u•Harmonic Travelling Wavesy x, t() =y 0sin(kx±ωt .)(911)where frequency , wavelength , wave numberfλk= 2π/λand angular frequencyωare relatedtovby:v=fλ= ωk(912)•Stationary Harmonic Wavesy x, t() =y 0cos(kx+ ) cos(δωt+ ) φ(913)where one can selectkandωso that waves on a string of lengthLsatisfy fixed or free boundaryconditions.•Energy (of wave on string)E tot= 12µω A λ22(914)is the total energy in a wavelength of a travelling harmonic wave. The wave transports thepowerP=ET= 12µω A λf22= 12µω A v22(915)past any point on the string.•Reflection/Transmission of Waves441

442Week 10: The Wave Equationa) Light string (medium) to heavy string (medium): Transmitted pulse right side up, re-flected pulse inverted. (A fixed string boundary is the limit of attaching to an “infinitelyheavy string”).b) Heavy string to light string: Transmitted pulse right side up, reflected pulse right sideup. (a free string boundary is the limit of attaching to a “massless string”).10.1: WavesWe have seen how a particle on a spring that creates a restoring force proportional to its displacementfrom an equilibrium position oscillates harmonically in time about that equilibrium. What happensif there aremanyparticles,allconnected by tiny “springs” to one another in an extended way?This is a good metaphor for many, many physical systems. Particles in a solid, a liquid, or a gasboth attract and repel one another with forces that maintain an average particle spacing. Extendedobjects under tension or pressure such as strings have components that can exert forces on oneanother. Even fields (as we shall learn next semester) can interact so that changes in one tinyelement of space create changes in a neighboring element of space.What we observe in all of these cases is that changes in any part of the medium “propagate” toother parts of the medium in a very systematic way. The motion observed in this propagation iscalled awave. We have all observed waves in our daily lives in many contexts. We have watchedwater waves propagate away from boats and raindrops. We listen to sound waves (music) generatedby waves created on stretched strings or from tubes driven by air and transmitted invisibly throughspace by means of radio waves. We read these words by means of light, an electromagnetic wave. Inadvanced physics classes one learns that all matter is a sort of quantum wave, that indeedeverythingis really a manifestation of waves.It therefore seems sensible to make a first pass at understanding waves and how they work ingeneral, so that we can learn and understand more in future classes that go into detail.The concept of a wave is simple – it is anextended structurethat oscillates in bothspaceand intime. We will study two kinds of waves at this point in the course:•Transverse Waves(e.g. waves on a string). The displacement of particles in a transversewave isperpendicularto the direction of the wave itself.•Longitudinal Waves(e.g. sound waves). The displacement of particles in a longitudinalwave is in the same direction that the wave propagates in.Some waves, for example water waves, are simultaneously longitudinal and transverse. Transversewaves are probably the most important waves to understand for the future; light is a transverse wave.We will therefore start by studying transverse waves in a simple context: waves on a stretched string.

Week 10: The Wave Equation443yxvFigure 136: A uniform string is plucked or shaken so as to produce awave pulsethat travels at aspeedvto the right. The circled region is examined in more detail in the next figure.10.2: Waves on a StringSuppose we have a uniformstretchablestring (such as a guitar string) that is pulled at the ends sothat it is under tensionT. The string is characterized by its mass per unit lengthµ– thick guitarstrings have more mass per unit length than thin ones but little else. It is fairly harmless at this pointto imagine that the string is fixed to pegs at the ends that maintain the tension. Our experience ofsuch things leads us to expect that the stretched string will form an almost perfectly straight lineunless we pluck it or otherwise bend some other shape upon it. We will impose coordinates uponthis string such thatxruns along its (undisplaced) equilibrium position andydescribes the verticaldisplacement of any given bit of the string.Now imagine that we have plucked the string somewhere between the end points so that it isdisplaced in the -direction from its equilibrium (straight) stretched position and has some curvedyshape, as portrayed in figure 136. The tensionT, recall, acts all along the string, but because thestring iscurvedthe force exerted on any small bit of string does not balance. This inspires us to tryto write Newton’s Second Law not for the entire string itself, but for just a tiny bit of string, indeedadifferentialbit of string.We thus zoom in on just a small chunk of the string in the region where we have stretched orshaken awave pulseas illustrated by the figure above. If we blow this small segment up, perhapswe can find a way to write the unbalanced forces out in a way we can deal with algebraically.This is shown in figure 137, where I’ve indicated a short segment/chunk of the string of length∆ by cross-hatching it. We would like to write Newton’s 2nd law for that chunk. As you can see,xthe samemagnitudeof tensionTpulls on both ends of the chunk, but the tension pulls in slightlydifferent directions, tangent to the string at the end points.Ifθis small, the components of the tension in the -direction:xF 1 x=− Tcos( )θ 1≈ −TF 2 x=Tcos( )θ 1≈ T(916)are nearly equal and in opposite directions and hence nearly perfectly cancel (where we have usedthe small angle approximation to the Taylor series for cosine:cos( ) = 1θ− θ 22!+ θ 44!− ...≈ 1(917)for smallθ≪ 1).Each bit of string therefore moves more or lessstraight up and down, and a useful solution isdescribed by (y x, t), theydisplacementof the string at positionxand time . The permittedtsolutions must becontinuousif the string does not break. It is worth noting that notallwavesinvolving moving up and down only – this initial example is called atransversewave, but otherkinds of waves exist. Sound waves, for example, arelongitudinalcompression waves, with themotion back and forth along the direction of motion and not at right angles to it. Surface waveson water are a mixture, they involve the waterbothmoving up and downandback and forth. Butfor the moment we’ll stick with these very simple transverse waves on a string because they alreadyexhibit most of the properties of far more general waves you might learn more about later on.

444Week 10: The Wave Equationxyθθ12x∆TTFigure 137: The forces exerted on a small chunk of the string by the tensionTin the string. Notethat we neglect gravity in this treatment, assuming that it is much too small to bend the stretchedstring significantly (as is indeed the case, for a taut guitar string).In the -direction, we find write the force law by considering the total -components of the sumyyof the force exerted by the tension on the ends of the chunk:∆ Fy,tot= F 1 y+ F 2 y= − Tsin( ) +θ 1Tsin( ) = ∆θ 2may= ∆µ xay(918)Here ∆F yis not the “change in the force” but rather the total force acting on the chunk of length∆ . These two quantities will obviously scale together.xWe make the small angle approximation: sin( )θ≈tan( )θ≈ θ :∆ F y= Ttan( )θ 2− Ttan( ) =θ 1T(tan( )θ 2−tan( )) = ∆θ 1µ xa ,y(919)Next, we note that tan( ) =θdydx(the slope of the string is the tangent of the angle the stringmakes with the horizon) and divide out theµ x ∆to isolatea y=d y/dt 22. We end up with thefollowing expression for what might be calledNewton’s Second Law per unit lengthof the string:∆ F y∆ x= T∆(dydx)∆ x= µd y 2dt2(920)In the limit that ∆x→0, this becomes:f y=dFydx= Td y 2dx2= µd y 2dt2(921)where we might considerf y=dF /dxytheforce density(force per unit length) acting at a point onthe string. Finally, we rearrange to get:d y 2dt2−T d yµ dx22= 0(922)The quantityTµhas to have units ofL 2t 2which is a velocity squared.We therefore formulate this as the one dimensionalwave equation:d y 2dt2− v 2d y 2dx2= 0(923)wherev= ±s Tµ(924)

Week 10: The Wave Equation445are the velocity(s) of the wave on the string. This is a second order linear homogeneous differentialequation and has (as one might imagine) well known and well understood solutions.Note well: At the tension in the string increases, so does the wave velocity. As the mass densityof the string increases, the wave velocity decreases. This makesphysical sense. As tension goes upthe restoring force is greater. As mass density goes up one accelerates less for a given tension.10.3: Solutions to the Wave EquationIn one dimension there are at least three distinct solutions to the wave equation that we are interestedin. Two of these solutionspropagatealong the string – energy istransportedfrom one place to anotherby the wave. The third is astationarysolution, in the sense that the wave doesn’t propagate in onedirection or the other (not in the sense that the string doesn’t move). But first:10.3.1: An Important Property of Waves: SuperpositionThe wave equation islinear, and hence it is easy to show that ify x, t1 () solves the wave equationandy x, t2 () (independent ofy 1 )alsosolves the wave equation, then:y x, t() =Ay x, t1 () +By x, t2 ()(925)solves the wave equation for arbitrary (complex)AandB .This property of waves is most powerful and sublime.10.3.2: Arbitrary Waveforms Propagating to the Left or RightThe first solution we can discern by noting that the wave equation equates a second derivative intime to a second derivative in space. Suppose we write the solution as ( ) wheref uuis an unknownfunction ofxandtand substitute it into the differential equation and use the chain rule:d f du 2du2(dt) 2− v 2d f du 2du2(dx) = 0 2(926)ord f 2du2(dudt) 2− v 2 (dudx) 2= 0(927)dudt= ± vdudx(928)with a simple solution:u= x± vt(929)What this tells us is thatanyfunctiony x, t() = (f x± vt)(930)satisfies the wave equation.Anyshape of wave created on the string and propagating to the rightor left is a solution to the wave equation, although not all of these waves will vanish at the ends ofa string.

446Week 10: The Wave Equation10.3.3: Harmonic Waveforms Propagating to the Left or RightAn interesting special case of this solution is the case ofharmonicwaves propagating to the leftor right. Harmonic waves are simply waves that oscillate with a given harmonic frequency. Forexample:y x, t() =y 0sin(x− vt)(931)is one such wave.y 0is called theamplitudeof the harmonic wave. But what sorts of parametersdescribe the wave itself? Are there more than one harmonic waves?This particular wave looks like a sinusoidal wave propagating to the right (positivexdirection).But this is not a very convenient parameterization. To better describe a general harmonic wave, weneed to introduce the following quantities:•Thefrequencyf. This is the number of cycles per second that pass a point or that a pointon the string moves up and down.•Thewavelengthλ. This the distance one has to travel down the string to return to the samepoint in the wave cycle at any given instant in time.To convertx(a distance) into an angle in radians, we need to multiply it by 2πradians perwavelength. We therefore define thewave number:k= 2 πλ(932)and write our harmonic solution as:y x, t()=y 0sin( (k x− vt))(933)=y 0sin(kx−kvt)(934)=y 0sin(kx−ωt)(935)where we have used the following train of algebra in the last step:kv= 2 πλv= 2πf= 2 πT= ω(936)and where we see that we have two ways to write :vv=fλ= ωk(937)As before, you should simplyknowevery relation in this set of algebraic relations betweenλ, k, f, ω, vto save time on tests and quizzes. Of course there is also the harmonic wave travellingto the left as well:y x, t() =y 0sin(kx+ωt .)(938)A final observation about these harmonic waves is that because arbitrary functions can beex-pandedin terms of harmonic functions (e.g. Fourier Series, Fourier Transforms) and because thewave equation is linear and its solutions are superposable, the two solution forms above are not reallydistinct. One can expand the “arbitrary” (f x− vt) in a sum of sin(kx−ωt)’s for special frequenciesand wavelengths. In one dimension this doesn’t give you much, but in two or more dimensions thisprocess helps one compute thedispersionof the wave caused by the wave “spreading out” in multipledimensions and reducing its amplitude.

Week 10: The Wave Equation44710.3.4: Stationary WavesThe third special case of solutions to the wave equation is that ofstanding waves. They are especiallyapropos to waves on a string fixed at one or both ends. There are two ways to find these solutionsfrom the solutions above. A harmonic wave travelling to the right and hitting the end of the string(which is fixed), it has no choice but to reflect. This is because theenergyin the string cannot justdisappear, and if the end point is fixed no work can be done by it on the peg to which it is attached.The reflected wave has to have the same amplitude and frequency as the incoming wave. What doesthesumof the incoming and reflected wave look like in this special case?Suppose one adds two harmonic waves with equal amplitudes, the same wavelengths and fre-quencies, but that are travelling inoppositedirections:y x, t()=y 0(sin(kx−ωt) + sin(kx+ωt))(939)=2 y 0sin(kx) cos(ωt)(940)=Asin(kx) cos(ωt)(941)(where we give the standing wave the arbitrary amplitudeA). Since all the solutions above areindependent of the phase, a second useful way to write stationary waves is:y x, t() =Acos(kx) cos(ωt)(942)Which of these one uses depends on the details of the boundary conditions on the string.In this solution a sinusoidal form oscillates harmonically up and down, but the solution has somevery important new properties. For one, it is alwayszerowhenx= 0 for all possiblelambda:y , t(0 ) = 0(943)For a givenλthere are certain otherxpositions where the wave vanishes at all times. These positionsare callednodesof the wave. We see that there are nodes for anyLsuch that:y L, t() =Asin(kL) cos(ωt) = 0(944)which implies that:kL= 2πLλ=π, π, π, . . .23(945)orλ= 2 Ln(946)forn= 1 2 3, , , ...Only waves with these wavelengthsand their associated frequencies can persist on a string oflengthLfixed at both ends so thaty , t(0 ) = (y L, t) = 0(947)(such as a guitar string or harp string). Superpositions of these waves are what give guitar stringstheir particular tone.It is also possible to stretch a string so that it is fixed at one end but so that theotherend isfree to move– to slide up and down without friction on a greased rod, for example. In this case,instead of having a node at the free end (where the wave itself vanishes), it is pretty easy to seethat theslopeof the wave at the end has to vanish. This is because if the slope were not zero, theterminating rod would be pulling up or down on the string, contradicting our requirement that therod be frictionless and notableto pull the string up or down, only directly to the left or right dueto tension.

448Week 10: The Wave EquationThe slope of a sine wave is zero only when the sine wave itself is a maximum or minimum, so thatthe wave on a string free at an end must have anantinode(maximum magnitude of its amplitude)at the free end. Using the same standing wave form we derived above, we see that:kL= 2πLλ=π/ , π/ , π/ . . .2 32 52(948)for a string fixed atx= 0 and free atx= , or:Lλ=4 L2 n− 1(949)forn= 1 2 3, , , ...There is a second way to obtain the standing wave solutions that particularly exhibits the re-lationship between waves and harmonic oscillators. One assumes that the solution (y x, t) can bewritten as theproductof a fuction inxalone and a second function intalone:y x, t() =X x T t( ) ( )(950)If we substitute this into the differential equation and divide by (y x, t) we get:d y 2dt2=X x( )d T 2dt2=v 2d y 2dx2=v T t 2( )d X 2dx2(951)1T t dt( )d T 22=v 21X x dx( )d X 22(952)=− ω 2(953)where the last line follows because the second line equations a function oft(only) to a function ofx(only) so that both terms must equal a constant. This is then the two equations:d T 2dt2+ω T 2= 0(954)andd X 2dt2+k X 2= 0(955)(where we usek=ω/v).From this we see that:T t( ) =T 0cos(ωt+ ) φ(956)andX x( ) =X 0cos(kx+ ) δ(957)so that the most general stationary solution can be written:y x, t() =y 0cos(kx+ ) cos(δωt+ ) φ(958)10.4: Reflection of WavesWe argued above that waves have to reflect of of the ends of stretched strings because of energyconservation. This is true independent of whether the end is fixed or free – in neither case can thestring do work on the wall or rod to which it is affixed. However, the behavior of the reflected waveis different in the two cases.Suppose a wavepulseis incident on the fixed end of a string. One way to “discover” a wavesolution that apparently conserves energy is to imagine that the stringcontinuesthrough the barrier.

Week 10: The Wave Equation449At the same time the pulse hits the barrier, anidenticalpulse hits the barrier from the other,“imaginary” side.Since the two pulses are identical, energy will clearly be conserved. The one going from left toright will transmit its energy onto the imaginary string beyond at the same rate energy appearsgoing from right to left from the imaginary string.However, we still have two choices to consider. The wave from the imaginary string could beright side upthe same as the incident wave orupside down. Energy is conserved either way!If the right side up wave (left to right) encounters an upside down wave (right to left) theywill always beoppositeat the barrier, and when superposed they willcancelat the barrier. Thiscorresponds to afixed string. On the other hand, if a right side up wave encounters a right side upwave, they willaddat the barrier with oppositeslope. There will be a maximum at the barrier withzero slope – just what is needed for a free string.From this we deduce the general rule that wave pulsesinvertwhen reflected from a fixed boundary(string fixed at one end) and reflect right side up from a free boundary.When two strings of different weight (mass density) are connected, wave pulses on one string areboth transmitted onto the other and are generally partially reflected from the boundary. Computingthe transmitted and reflected waves is straightforward but beyond the scope of this class (it starts toinvolve real math and studies of boundary conditions). However, the following qualitative propertiesof the transmitted and reflected waves should be learned:•Light string (medium) to heavy string (medium): Transmitted pulse right side up, reflectedpulse inverted. (A fixed string boundary is the limit of attaching to an “infinitely heavystring”).•Heavy string to light string: Transmitted pulse right side up, reflected pulse right side up. (afree string boundary is the limit of attaching to a “massless string”).10.5: EnergyClearly a wave can carry energy from one place to another. A cable we are coiling is hung up ona piece of wood. We flip a pulse onto the wire, it runs down to the piece of wood and knocks thewire free. Our lungs and larnyx create sound waves, and those waves trigger neurons in ears faraway. The sun releases nuclear energy, and a few minutes later that energy, propagated to earthas a light wave, creates sugar energy stored inside a plant that is still later released while we playbasketball204. Since moving energy around seems to be important, perhaps we should figure outhow a wave manages it.Let us restrict our attention to a harmonic wave of known angular frequency , although many ofωour results will be more general, because arbitrary wave pulses can be fourier decomposed as notedabove.Consider a small piece of the string of lengthdxand massdm=µdx. This piece of string,displaced to its position (y x, t), will have kinetic energy:dK=12dmdydt2(959)=12A µω 22cos (2kx−ωt dx)(960)204Painfully and badly, in my case. As I’m typing this, my ribs and ankles hurt from participating in the GreatBeaufort 3 on 3 Physics Basketball Tournament, Summer Session 1, 2011! But what the heck, we won and are in thefinals. And it was energy propagated by a wave, stored (in my case) as fat, that helped get us there...

450Week 10: The Wave EquationWe can easily integrate this over any specific interval. Let us pick a particular timet= 0 andintegrate it over a single wavelength:∆ K=Z∆0KdK= Zλ012A µω 22cos (2kx dx)(961)=12 kA µω 22Zλ0cos (2kx kdx)(962)=12 kA µω 22Z2 π0cos ( )2θ dθ(963)=14A µω λ 22(964)(where we have used the easily proven relation:Z2 π0sin ( )2θ dθ= π(965)to do the final form of the integral.) If we assume that the string has lengthL≫ λ, we can averagethis over many wavelengths and get theaveragekinetic energy per unit lengthκ=K/L≈ ∆K/λ= 14A µω 22(966)In the limit of an infinitely long string, or a length of string that contains an integer number ofwavelengths, this expression is exact. Note also that this answer does not depend on time, becauseωtonly corresponds to a different phase and the integral of sin ( ) + ) does not change.2θδThe average potential energy of the string is more difficult. There is a temptation to treat thepotential energy of a chunk of the string of length ∆ as the work done against that force bendingxthe string into its instantaneous shape. This approach will lead to a constant energy per unit lengthfor the string, as it basically treats every string element as an ”independent” mass oscillating as if itis attached to a spring. Unfortunately, this picture doesn’t explain how energy canpropagatein thecase of a travelling wave, nor does it give us any insight into just where the energy is being stored,as thereareno springs.It is better to treat it as the work donestretchingthe local pieces of the string, while maintainthe assumption above that the pieces of string engage in transverse motiononly, that is, the movestraight up or down and not back and forth in the -direction.xThis approach basically treats the string as a very long chain of mass elements, each of whichisa little “spring”. Each mass element has kinetic energy as it moves straight up and down, andhas potential energy to the extent that it is stretched. Note that the stringmuststretch if it is notstraight, because a straight line is the shortest distance between two points and (y x, t) cannot bezero everywhere or it is the boring trivial solution, no actual wave at all.In this approach, the potential energy does not depend on value of (y x, t) at all, but ratheron theslopeof the stringdydxat a point. This is illustrated in figure 138. A small (eventuallydifferentially small) chunk of the string of unstretched length ∆ is seen to be stretched to a newxlength ∆ =ℓp(∆ ) + (∆ ) when it is moved straigth up to the height (x 2y2y x, t) so that it has someslope there. If we assume that this slope is very small (∆y≪∆ ) then:x

Week 10: The Wave Equation451xylx∆y∆∆Figure 138: A small chunk of the string. The chunk that (unstretched at (y x, t) = 0 has length ∆xis stretched by an amount ∆ℓ−∆ .x∆ ℓ− ∆ x=(∆ ) + (∆ )x 2y21 2/− ∆ x=∆ x1 +(∆ )y2(∆ )x 21 2/− ∆ x=∆ x1 +12∆ y∆ x2+ ...!− ∆ x≈∆ x2∆ y∆ x2(967)In the limit that ∆x→0, this becomes:dℓ−dx= 12dydx2dx(968)(where really, the expression should use partial derivatives e.g.dℓ−dx= 12∂y∂x2dxas usual buteither way it is the instantaneous slope of (y x, t)). The workwedo to stretch this small chunk ofstring to its new length is considered to be the potential energy of the chunk:dU=T dℓ(−dx) =T2dydx2dx(969)or the potential energy per unit length of the string is just:dUdx= T2dydx2(970)For thetravelling harmonic wavey x, t() =Asin(kx−ωt) we considered before:dydx=kAcos(kx−ωt)(971)anddU= 12Tk A22cos (2kx−ωt dx)(972)

452Week 10: The Wave Equationso that we can form thetotal energy densityof the stringdE/dxin this formulation:dEdx=12µω A22cos (2kx−ωt) +12Tk A22cos (2kx−ωt)=12µω A22cos (2kx−ωt) + cos (2kx−ωt)=µω A22cos (2kx−ωt)(973)becauseµω2=Tk2fromTµ= ω 2k 2= v 2 .There is apparently exactly as much kinetic energy as there is potential energy in the differentiallysmall chunk of string, and that energy in the chunk isnot a constant. Not only that, but we notethat the energy density is itself a wave propagating from left to right as it has the form (f x− vt).Our interpretation of this is that the string iscarrying energy from left to rightas the wavepropagates, with the peak energy in the string concentrated where (y x, t) = 0, where both the speedof the string and its local stretch (magnitude of its slope) are maximal.It is useful to find the average energy per unit length of this wave. Since the wave is lots of copiesof a single wavelength, we can get the average over a very long string by just evaluating the energyper wavelength from a single wavelength:Eλ= 1λµω A22Zλ0cos (2kx−ωt dx)= 12µω A22(974)(where the integral is considered below, if it isn’t familiar to you).Next, Consider a genericstanding wavey x, t() =Asin(kx) cos(ωt) for a string that (perhaps) isfixed at ends located atx= 0 andx= L. We can easily evaluate the kinetic energy of a chunk oflengthdx:dK= 12µω A22sin (2kx) sin (2ωt dx)(975)and the potential energy of the same chunk:dU=12Tk A22cos (2kx) cos (2ωt dx)=12µω A22cos (2kx) cos (2ωt dx)(976)If we now form the total energy density of this chunk of string as before:dEdx= 12µω A22sin (2kx) sin (2ωt) + cos (2kx) cos (2ωt)(977)we note that the energy density of the string isnot a constant.At first this appears to be a problem, but really it is not. All this tells us is that the nodesand antinodes of the string (where sin(kx) or cos(kx) are zero, respectively) carry the energy of thestringout of phasewith one another. At timesωt= 0, π, π...2, the potential energy is maximal, thekinetic energy is zero, and the potential energy is concentrated at the points where cos(kx) =± 1.These are thenodes, which have maximum magnitude of slope and stretch as the string reaches itsmaximum positive or negative amplitude. At timesωt=π/ , π/ ...2 32the kinetic energy is maximal,the potential energy is zero (indeed, the string is now flat and unstretched at all) and the kineticenergy is concentrated at theantinodes, where the velocity of the string is the greatest.This makes perfect sense!The energy in the string oscillates from being all potential (for theentire string) as it momentarily comes to rest to all kinetic as the string is momentarily straightand moving at maximum speed (for the location) everywhere. The one question we need to answer,though, is: Is thetotalenergy of the string constant? We would be sad if it were not, because weknow that the endpoints of the standing wave are (for example) fixed, and hence no work is donethere. Whatever the energy in the wave on the string, it has nowhere to go.

Week 10: The Wave Equation453We note that no matter what the boundary conditions, the string contains an integer number ofquarter wavelengths. All we have to do is show that the total energy in any quarter wavelength ofthe string is constant in time and we’re good to go. Consider the following integral:Zλ/40sin (2kx dx)= 1kZλ/40sin (2kx kdx)= 1kZπ/20sin θ dθ2( )=λ π2 4π=λ/8 =Zλ/40cos (2kx dx)(978)(the latter from symmetry)From this we can easily find the energy per unit length of any standing wave on a string of lengthL(which as noted has an integer number of quarter waves on it). We’ll express this as the energyper wavelength of string to facilitate comparison with a travelling wave and just remember that itworks just as well for a quarter wave, half wave, etc.Eλ=1 1λ 2µω A22sin (2ωt)Zλ0sin (2kx dx)+ cos (2ωt)Zλ0cos (2kx)!=1 1λ 2µω A22sin (2ωt)λ2+ cos (2ωt)λ2=14µω A22sin (2ωt) + cos (2ωt)=14µω A22(979)If we compare this to the expression for the energy per wavelength of a traveling wave with thesame amplitude, we get:E sλ= 12E tλ(980)We can understand this easily enough. The standing wave consists of two travelling waves of halfof the amplitude going in opposite directions. Their total energy per wavelength are independent –we just add them. Energy per wavelength is proportional to amplitude squared, so each wave withamplitudeA/2 has 1 4 of the energy of a travelling wave with amplitude/A. But there are two ofthem, hence a factor of 1 2./

454Week 10: The Wave EquationHomework for Week 10Problem 1.Physics Concepts:Make this week’s physics concepts summary as you work all of the problemsin this week’s assignment. Be sure to cross-reference each concept in the summary to the problem(s)they were key to, and include concepts from previous weeks as necessary. Do the work carefullyenough that you can (after it has been handed in and graded) punch it and add it to a three ringbinder for review and study come finals!Problem 2.This (and several of the following) problems are fromresonance and damped and/or drivenoscillation, not from waves! Be sure to use the correct concepts on your concept summary!Roman soldiers (like soldiers the world over even today) marched in step at a constant frequency –except when crossing wooden bridges, when they broke their march and walked over with randompacing. Why? What might have happened (and originally did sometimes happen) if they marchedacross with a collective periodic step?

Week 10: The Wave Equation455Problem 3.123456789101112131415161718192012345678910omegaP(omega)(a)(b)(c)In the figure above, threeresonance curvesare drawn showing the power delivered to a steady-state driven oscillator,P ω( ). In all three cases the resonance frequencyω 0is the same. Put downanestimateof theQ-value of each oscillator by looking at the graph. It may help for you to putdown the definition ofQmost relevant to the process of estimation on the page.a)b)c)

456Week 10: The Wave EquationProblem 4.kb mA massmis attached to a spring with spring constantkand immersed in a medium with dampingcoefficient . (Gravity, if present at all, is irrelevant as shown in class). The net force on the massbwhen displaced byxfrom equilibrium and moving with velocityv xis thus:F x=max= − kx− bvx(in one dimension).a) Convert this equation (Newton’s second law for the mass/spring/damping fluid arrangement)into the equation of motion for the system, a “second order linear homogeneous differentialequation” as done in class.b) Optionallysolvethis equation, finding in particular the exponential damping rate of the solu-tion (the real part of the exponential time constant) and the shifted frequencyω ′, assumingthat the motion is underdamped. You can put down any form you like for the answer; theeasiest is probably a sum of exponential forms. However, you may also simplyput down thesolutionderived in class if you plan tojustmemorize this solution instead of learn to deriveand understand it.c) Using your answer forω ′from part b), write down the criteria for damped, underdamped, andcritically damped oscillation.d) Drawthreequalitatively correct graphs of ( ) if the oscillator is pulled to a positionx tx 0andreleased at rest at time = 0, one for each damping. Note that you should be able to do thistpart even if you cannot derive the curves that you draw orω ′. Clearly label each curve.

Week 10: The Wave Equation457Problem 5.m4m9mabcThree strings of lengthL(not shown) with the same mass per unit lengthµare suspendedvertically and blocks of massm , 4mand 9mare hung from them. The total mass of each stringµL≪ m(the strings aremuchlighter than the masses hanging from them). If the speed of a wavepulse on the first string isv 0, fill in the following table with entries for b) and c):a)v 0b)c)

458Week 10: The Wave EquationProblem 6.4T4Tabcdµ2µ3µ4µIn the figure above, the neck of a stringed instrument is schematized. Four strings of differentthickness(and hence differentµas shown) and the same length are stretched in such a way that thetension in each is about the same ( ) in each string. This produces a total of 4TTbetween the endbridges – if this were not so, the neck of the guitar or ukelele or violin would tend to bow towardsthe side with the greater tension.If the speed of a wave pulse on the first (lightest) string isv 0, fill in the following table for thespeed of a wave pulse for the other three:a)v 0b)c)d)

Week 10: The Wave Equation459Problem 7.abvavbTwo combinations of two strings with different mass densities are drawn above that are connectedin the middle. In both cases the string with the greatest mass density is drawn darker and thickerthan the lighter one, and the strings have the same tensionTin both a and b. A wave pulse isgenerated on the string pairs that is travelling from left to right as shown. The wave pulse will arriveat the junction between the strings at timet a(for a) andt b(for b). Sketch reasonableestimatesfor thetransmitted and reflected wave pulsesonto (copies of) the a and b figures at time 2t aand 2t brespectively. Your sketch should correctly represent things like therelative speed of thereflected and transmitted waveand any changes you might reasonably expect for theamplitudeand appearanceof the pulses.

460Week 10: The Wave EquationProblem 8.T,µyx0LA string of mass densityµis stretched to a tensionTand isfixed at bothx= 0andx= L .The transverse string displacement is measured in theydirection. All answers should be given interms of these quantities or new quantities you define in terms of these quantities.a) Following the text, derive the wave equation (the equation of motion) for waves on a stringand identify the wave velocity squared in terms ofTand . This one derivation suffices forµthis and the next problem.b) Write down the equationy x, tn () for a genericstanding waveon this string with mode indexn, assuming that the string is maximally displaced at = 0. Verify that it is a solution to thetODE in a). Remember that the string is fixed at both ends!c) Findk , ω , f , λnnnnfor thefirst three modes supported by the string. Sketch them inon the axes below, labelling nodes and antinodes. Note that you should be able to draw themodes and find at least the wavelengths from the pictures alone.yx0Lyx0Lyx0L

Week 10: The Wave Equation461Problem 9.T,µyx0LA string of mass densityµis stretched to a tensionTand isfixed atx= 0and free (fric-tionless loop) atx= L. The transverse string displacement is measured in theydirection. Allanswers should be given in terms of these quantities or new quantities you define in terms of thesequantities.a) Write down the equationy x, tn () for a genericstanding waveon this string with mode indexn, assuming that the string is maximally displaced at = 0. Verify that it is a solution to thetODE in a). Remember that the string is free at one end!b) Findk , ω , f , λnnnnfor thefirst three modes supported by the string. Sketch them inon the axes below, labelling nodes and antinodes. Note that you should be able to draw themodes and find at least the wavelengths from the pictures alone.yx0Lyx0Lyx0L

462Week 10: The Wave EquationProblem 10.LThis problem will help you learn required concepts such as:•Speed of Wave on String•Static Equilibrium•Relationship betweeen Distance, Velocity, and Timeso please review them before you begin.A string of total lengthLwith a mass densityµis shown hanging from the ceiling above.a) Find the tensionT y( ) in the string as a function of , the distanceyupfrom its bottom end.Note that the string is not massless, so each small bit of string must be instatic equilibrium.b) Find the velocity ( ) of a small wave pulse cast into the string at the bottom that is travellingv yupward.c) Find the amount of time it will take this pulse to reach the top of the string, reflect, and returnto the bottom. Neglect the size (width in ) of the pulse relative to the length of the string.yHint for last part. Setv=dy/dt, rearrange to get all the -dependent parts on one side andydtandsome constants on the other side, then integrate both sides, theypart from 0 toL, thetpart from0 tot 0. Solve fort 0, double it. This is what calculus isfor!

Week 10: The Wave Equation463Problem 11.0LA string of total massMand total lengthLis fixed at both ends, stretched so that the speed ofwaves on the string is . It is plucked so that it harmonically vibrates in itsvn= 4 mode:y x, t() =Asin k x cos ω t .(4)(4)Find the instantaneous total energy in the string in terms ofM L n,,= 4,vandA(although itwill simplify matters to useλ 4andω 4once you define them in terms of the givens).If you are aphysics or math major, the word ”find” should be interpreted as “derive”,following the derivation presented in the book. All others can get by from remembering the generalway the total energyscales with the givensin understandable ways to lead to the formula derivedin the book above.For those who attempt the derivation, remember (or FYI):Znπ0sin ( )2u du= Znπ0cos ( )2u du=nπ2

464Week 11: SoundOptional ProblemsContinue studying for the final exam!Onlyone more week of class(and onechapter) to go in this textbook! Don’t wait until the last moment to start!

Week 11: SoundSound Summary•Speed of Soundin a fluidv=s Bρ(981)whereBis the bulk modulus of the fluid andρis the density. These quantities vary withpressure and temperature.•Speed of Soundin air isv a≈340 m/sec.•Travelling Sound waves:Plane (displacement) waves (in the -direction):xs x, t() =s 0sin(kx−ωt)Spherical waves:s r, t() =s 0R( 4 ) √π rsin(kr−ωt)whereRis a reference length needed to make the units right corresponding physically to the“size of the source” (e.g. the smallest ball that can be drawn that completely contains thesource). The√4 is needed so that the intensity has the right functional form for a sphericalπwave (see below).•Pressure Waves:The pressure waves that correspond to these two displacement waves are:P x, t() =P 0cos(kx−ωt)andP r, t() =P 0R( 4 ) √π rcos(kr−ωt)whereP 0=v ρωsa0=Zωs0withZ=v ρa= √Bρ(a conversion factor that scales microscopicdisplacement to pressure). Note that:P x, t() =Z ddts x, t()or, the displacement wave is a scaled derivative of the pressure wave.The pressure waves represent the oscillation of the pressurearoundthe baseline ambient pres-sureP a, e.g. 1 atmosphere. The total pressure is reallyP a+P x, t() orP a+P r, t() (and wecould easily put a “∆” in front of e.g.P r, t() to emphasize this point but don’t so as to notconfuse variation around a baseline with a derivative).465

466Week 11: Sound•Sound Intensity:The intensity of sound waves can be written:I= 12v ρω sa2 20orI=1 12v ρaP 20=12 Z P 20•Spherical Waves:The intensity of spherical sound waves drops off like 1/r2(as can be seenfrom the previous two points). It is usually convenient to express it in terms of thetotalpower emitted by the sourcePtotas:I=Ptot4πr2This is “the total power emitted divided by the area of the sphere of radiusrthrough whichall the power must symmetrically pass” and hence itmakes sense!One can, with some effort,take the intensity at some reference radiusrand relate it toP 0and tos 0, and one can easilyrelate it to the intensity at other radii.•Decibels:Audible sound waves span some 20 orders of magnitude in intensity. Indeed, theear is barely sensitive to adoublingof intensity – this is the smallest change that registers asa change in audible intensity. This motivate the use ofsound intensity levelmeasured indecibels:β= 10 log10II 0(dB)where the reference intensityI 0= 10− 12watts/meter is the2threshold of hearing, theweakest sound that is audible to a “normal” human ear.Important reference intensities to keep in mind are:–60 dB: Normal conversation at 1 m.–85 dB: Intensity where long term continuous exposuremaycause gradual hearing loss.–120 dB: Hearing loss islikelyfor anything more than brief and highly intermittent ex-posures at this level.–130 dB: Threshold ofpain. Pain is bad.–140 dB: Hearing loss isimmediate and certain– you are actively losing your hearingduring any sort of prolonged exposure at this level and above.–194.094 dB: The upper limit of undistorted sound (overpressure equal to one atmosphere).This loud a sound will instantly rupture human eardrums 50% of the time.•Doppler Shift: Moving Sourcef ′=f 0(1∓ v sv a)(982)wheref 0is the unshifted frequency of the sound wave for receding (+) and approaching (-)source, wherev sis the speed of the source andv ais the speed of sound in the medium (air).•Doppler Shift: Moving Receiverf ′= f 0(1± v rv a)(983)wheref 0is the unshifted frequency of the sound wave for receding (-) and approaching (+)receiver, wherev ris the speed of the source andv ais the speed of sound in the medium (air).

Week 11: Sound467•Stationary Harmonic Wavesy x, t() =y 0sin(kx) cos(ωt)(984)for displacement waves in a pipe of lengthLclosed at one or both ends. This solution has anode atx= 0 (the closed end). The permitted resonant frequencies are determined by:kL=nπ(985)forn= 1 2, ...(both ends closed, nodes at both ends) or:kL= 2 n− 12π(986)forn= 1 2, , ...(one end closed, nodes at the closed end).•BeatsIf two sound waves of equal amplitude and slightly different frequency are added:s x, t()=s 0sin(k x0−ω t0) +s 0sin(k x1−ω t1)(987)=2 s 0sin(k 0+ k 12x− ω 0+ ω 12t) cos(k 0− k 12x− ω 0− ω 12t )(988)which describes a wave with the average frequency and twice the amplitude modulated so thatit “beats” (goes to zero) at the difference of the frequenciesδf= |f 1− f 0 | .11.1: Sound Waves in a FluidWaves propagate in a fluid much in the same way that a disturbance propagates down a closed hallcrowded with people. If one shoves a person so that they knock into their neighbor, the neighborfalls againsttheirneighbor (and shoves back), and their neighbor shoves against their still furtherneighbor and so on.Such a wave differs from the transverse waves we studied on a string in that the displacement ofthe medium (the air molecules) isin the same directionas the direction of propagation of the wave.This kind of wave is called alongitudinalwave.Although different, sound waves can be related to waves on a string in many ways. Most of thesimilarities and differences can be traced to one thing: a string is a one dimensional medium and ischaracterized only by length; a fluid is typically a three dimensional medium and is characterizedby a volume.Air (a typical fluid that supports sound waves) does not support “tension”, it is under pressure.When air is compressed its molecules are shoved closer together, altering its density and occupiedvolume. For small changes in volume the pressure alters approximatelylinearlywith a coefficientcalled the “bulk modulus”Bdescribing the way the pressure increases as the fractional volumedecreases. Air does not have a mass per unit length , rather it has a mass per unit volume, .µρThe velocity of waves in air is given byv a=s Bρ≈343m sec/(989)The “approximately” here is fairly serious. The actual speed varies according to things like theair pressure (which varies significantly with altitude and with the weather at any given altitudeas low and high pressure areas move around on the earth’s surface) and the temperature (hottermolecules push each other apart more strongly at any given density). The speed of sound can varyby a few percent from the approximate value given above.

468Week 11: Sound11.2: Sound Wave SolutionsSound waves can be characterized one of two ways: as organized fluctuations in thepositionofthe molecules of the fluid as they oscillate around an equilibrium displacement or as organizedfluctuations in thepressureof the fluid as molecules are crammed closer together or are divenfarther apart than they are on average in the quiescent fluid.Sound waves propagate in one direction (out of three) at any given point in space. This meansthat in the directionperpendicularto propagation, the wave is spread out to form a “wave front”.The wave front can be nearly arbitrary in shape initially; thereafter it evolves according to themathematics of the wave equation in three dimensions (which is similar to but a bit more complicatedthan the wave equation in one dimension).To avoid this complication and focus on general properties that are commonly encountered, wewill concentrate on two particular kinds of solutions:a)Plane Wavesolutions. In these solutions, the entire wave moves in one direction (say thexdirection) and the wave front is a 2-D plane perpendicular to the direction of propagation.These (displacement) solutions can be written as (e.g.):s x, t() =s 0sin(kx−ωt)(990)wheres 0is the maximum displacement in the travelling wave (which moves in thexdirection)and where all molecules in the entireplaneat positionxare displaced by the same amount.Waves far away from the sources that created them are best described as plane waves. So arewaves propagating down a constrained environment such as a tube that permits waves to onlytravel in “one direction”.b)Spherical Wavesolutions. Sound is often emitted from a source that is highly localized (suchas a hammer hitting a nail, or a loudspeaker). If the sound is emitted equally in all directionsfrom the source, a spherical wavefront is formed. Even if it is not emitted equally in alldirections, sound from a localized source will generally form a spherically curved wavefront asit travels away from the point with constant speed. The displacement of a spherical wavefrontdecreasesas one moves further away from the source because theenergyin the wavefront isspread out on larger and larger surfaces. Its form is given by:s r, t() =s 0rsin(kr−ωt)(991)whereris the radial distance away from the point-like source.11.3: Sound Wave IntensityThe energy density of sound waves is given by:dEdV= 12ρω s2 2(992)(again, very similar in form to the energy density of a wave on a string). However, this energy perunitvolumeis propagated in a single direction. It is therefore spread out so that it crosses anarea,not a single point. Just how much energy an object receives therefore depends on how muchareaitintersects in the incoming sound wave, not just on the energy density of the sound wave itself.For this reason the energy carried by sound waves is best measured byintensity: the energy perunit time per unit area perpendicular to the direction of wave propagation. Imagine a box with sides

Week 11: Sound469given by ∆A(perpendicular to the direction of the wave’s propagation) and ∆ (in the directionv tof the wave’s propagation. All the energy in this box crosses through ∆Ain time ∆ . That is:t∆ E= (12ρω s2 2)∆Av t∆(993)orI=∆ E∆ ∆A t= 12ρω s v2 2(994)which looks very much like thepowercarried by a wave on a string. In the case of a plane wavepropagating down a narrow tube, it is very similar – the power of the wave is the intensity timesthe tube’s cross section.However, consider a spherical wave. For a spherical wave, the intensity looks something like:I r, t() =12ρω2 0 s 2sin (2kr−ωt)r 2v(995)which can be written as:I r, t() =Ptot4πr2(996)wherePtotis the total power in the wave.This makes sense from the point of view of energy conservation and symmetry. If a source emitsa powerPtot, that energy has to cross each successive spherical surface that surrounds the source.Those surfaces have an area equal toA= 4πr2. Thus the surface atr= 2r 0has 4 times the area ofone atr= r 0, but thesametotal power has to go through both surfaces. Consequently, the intensityat ther= 2r 0surface has to be 1 4 the intensity at the/r= r 0surface.It is important to remember this argument, simple as it is. Think back to Newton’s law ofgravitation. Remember that gravitational field diminishes as 1/r2with the distance from the source.Electrostatic field also diminishes as 1/r2. There seems to be a shared connection between symmetricpropagation and spherical geometry; this will form the basis forGauss’s Lawin electrostatics andmuch beautiful math.11.3.1: Sound Displacement and Intensity In Terms of PressureThe pressure in a sound wave (as noted) oscillates around the mean/baseline ambient pressure of theair (or water, or whatever). The pressurewavein sound is thus the time varying pressuredifference– the amplitude of the pressure oscillation around themeanof normal atmospheric pressure.As always, we will be interested in writing the pressureasa harmonic wave (where we can) andhence will use the peak pressure difference as the amplitude of the wave. We will call this pressurethe (peak) “overpressure”:P 0= Pmax− P a(997)whereP ais the baseline atmospheric pressure. It is easy enough to express the pressure wave interms of the displacement wave (and vice versa). The amplitudes are related by:P 0=v ρωsa0=Zωs0(998)whereZ=v ρa, the product of the speed of sound in air and the air density. The pressure anddisplacement waves areπ/2out of phase!with the pressure wave leading the displacement wave:P t( ) =P 0cos(kx−ωt)(999)(for a one-dimensional “plane wave”, usekrand put it over 1/rto make a spherical wave as before).The displacement wave is (Ztimes) the time derivative of the pressure wave, note well.

470Week 11: SoundThe intensity of a sound wave can also be expressed in terms ofpressure(rather than displace-ment). The expression for the intensity is then very simple (although not so simple to derive):I= P 202 Z=P 202v ρa(1000)

Week 11: Sound47111.3.2: Sound Pressure and DecibelsSource of SoundP(Pa)IdBAuditory threshold at 1 kHz2×10− 510− 120Light leaf rustling, calm breathing6 32.×10− 510− 1110Very calm room3 56.×10− 43 16.×10− 1225A Whisper2×10− 310− 840Washing machine, dish washer6 32.×10− 310− 750Normal conversation at 1 m2×10− 210− 660Normal (Ambient) Sound6 32.×10− 210− 570“Loud” Passenger Car at 10 m2×10− 110− 480Hearing DamagePossible0.3563 16.×10− 485Traffic On Busy Roadway at 10 m0.3563 16.×10− 485Jack Hammer at 1 m210− 2100Normal Stereo at Max Volume210− 2100Jet Engine at 100 m6.32-20010− 1to 103110-140Hearing DamageLikely201120Vuvuzela Horn201120Rock Concert (“The Who” 1982) at 32 m201120Threshold ofPain63.210130Marching Band (100-200 members, in front)63.210130“Very Loud” Car Stereo11231.6135Hearing DamageImmediate, Certain200102140Jet Engine at 30 m632103150Rock Concert (“The Who” 1982) atspeakers63210315030-06 Rifle 1 m to side6,320105170Stun Grenades20,000106180Limit of Undistorted Sound101,3252 51.×107194.094(human eardrums rupture 50% of time)(1 atm)Table 5: Table of (approximate)P 0and sound pressure levels in decibels relative to the threshold ofhuman hearing at 10− 12watts/m . Note that ordinary sounds only extend to a peak overpressure2ofP 0= 1 atmosphere, as one cannot oscillate symmetrically to underpressures pressureslessthana vacuum.The one real problem with the very simple description of sound intensity given above is one ofscale. The human ear isroutinely exposed to and sensitive tosounds that vary bytwentyorders of magnitude– from sounds so faint that they barely can move our eardrums to soundsso very loud that they immediatelyrupture them!Even this isn’t the full range of sounds out there– microphones and amplifiers allow us to detect even weaker sounds – as much as eight orders ofmagnitude weaker – and much stronger “sounds” calledshock waves, produced bysupersoniceventssuch as explosions. The strongest supersonic “sounds” are some 32 orders of magnitude “louder”than the weakest sounds the human ear can detect, although even the weakest shock waves arealmost strong enough to kill people.It is very inconvenient to have to describe sound intensity in scientific notion across this wide arange – basically 40 orders of magnitude if not more (the limits of technology not being well definedat the low end of the scale). Also the humanminddoes not respond to sounds linearly. We do notpsychologically perceive of sounds twice as intense as being twice as loud – in fact, a doubling ofintensity is barely perceptible. Both of these motivate our using a different scale to represent soundintensities a relativelogarithmicscale, calleddecibels205.205Wikipedia: http://www.wikipedia.org/wiki/Decibel. Note well that the term “decibel” is not restricted to sound

472Week 11: SoundThe definition of a sound decibel is:β= 10 log10II 0dB(1001)where “dB” is the abbreviation for decibels (tenths of “bels”, the same unit without the factor of10). Note that log10is logbase 10, not the natural log, in this expression. Also in this expression,I 0= 10− 12watts/meter2(1002)is the reference intensity, called thethreshold of hearing. It is, by definition, the “faintest soundthe human ear can hear” although naturally Your Mileage May Vary here – the faintest soundmyrelatively old and deaf ears is very likely much louder than the faintest sound a young child canhear, and there obviously some normal variation (a few dB) from person to person at any given age.The smallest increments of sound that the human ear can differentiate as being “louder” aretypicallytwo decibel increases– more likely 3. Let’s see what adoubling the intensitydoes tothe decibel level of the sound.∆ β=10 log102 II 0−log10II 0=10 log102I II I00=10 log10(2)=3 01. dB(1003)In other words,doubling the sound intensity from any value corresponds to an increasein sound intensity level of 3 dB!This is such a simple rule that it is religiously learned as arule of thumb by engineers, physicists and others who have reason to need to work with intensitiesof any sort on a log (decibel) scale. 3 dB per factor of two in intensity can carry you a long, longway! Note well that we used one of the magic properties of logarithms/exponentials in this algebra(in case you are confused):log( )A−log( ) = logBAB(1004)You should remember this; it will be very useful next year.Table 5 presents a number of fairly common sounds, sounds you are likely to have directly orindirectly heard (if only from far away). Each sound is cross-referenced with theapproximatepeakoverpressureP 0in the sound pressure wave (in pascals), the sound intensity (in watts/meter ),2and the sound intensity level relative to the threshold of hearing in decibels.The overpressure is the pressureoverthe background of (a presumed) 1 atm in the sinusoidalpressure wave. The actual peak pressure would then bePmax= P a+ P 0while the minimum pressurewould bePmin= P a− P 0 .Pmin, however,cannot be negativeas the lowest possible pressure is avacuum,P= 0! At overpressures greater than 1 atm, then, it is no longer possible to have a puresinusoidal sound wave. Waves in this category are a train of highly compressed peak amplitudes thatdrop off to (near) vacuum troughs in between, and are given their own special name:shock waves.Shock waves are typically generated by very powerful phenomena and often travel faster than thespeed of sound in a medium. Examples of shock waves are the sonic boom of a jet that has brokenthe sound barrier and the compression waves produced by sufficiently powerful explosives (close tothe site of the explosion).Shock waves as table 6 below clearly indicates, are capable of tearing the human body apartand accompany some of the most destructive phenomena in nature and human affairs – explodingvolcanoes and colliding asteroids, conventional and nuclear bombs.– it is rather a way of transforming any quantity that varies over a very large (many orders of magnitude) range into alog scale. Other logarithmic scales you are likely to encounter include, for example, the Richter scale (for earthquakes)and the F-scale for tornadoes.

Week 11: Sound473Source of SoundP 0(Pa)dBAll sounds beyond this pointare nonlinearshock wavesHuman Death from Shock Wave Alone200,0002001 Ton of TNT632,000210Largest Conventional Bombs2,000,0002201000 Tons of TNT6,320,00023020 Kiloton Nuclear Bomb63,200,00025057 Megaton (Largest) Nuclear Bomb2,000,000,000280Krakatoa Volcanic Explosion (1883 C.E.)63,200,000,000310Tambora Volcanic Explosion (1815 C.E.)200,000,000,000320Table 6: Table of (approximate)P 0and sound pressure levels in decibels relative to the thresholdof human hearing at 10− 12watts/m of2shock waves, events that produce distorted overpressuresgreater than one atmosphere. These “sounds” can be quite extreme! The Krakatoa explosion crackeda 1 foot thick concrete wall 300 miles away, was heard 3100 miles away, ejected 4 cubic miles of theearth, and created an audible pressure antinode on the opposite side of the earth. Tambora ejected36 cubic miles of the earth and was equivalent to a14 gigatonnuclear explosion (14,000 1 megatonnuclear bombs)!11.4: Doppler ShiftEverybody has heard the doppler shift in action. It is the rise (or fall) in frequency observed when asource/receiver pair approach (or recede) from one another. In this section we will derive expressionsfor the doppler shift for moving source and moving receiver.11.4.1: Moving Sourcev Tsλ ’SourceReceivers vλ0Figure 139: Waves from a source moving towards a stationary receiver have a foreshortened wave-length because the source movesinto the wave it produces. Thekeyto getting the frequency shiftis to recognize that the new (shifted)wavelengthisλ ′= λ 0−v TswhereTis the unshifted periodof the source.Suppose your receiver (ear) is stationary, while a source of harmonic sound waves at fixed fre-quencyf 0is approaching you. As the waves are emitted by the source they have a fixed wavelengthλ 0=v /fa0=v Taand expand spherically from the point where the source was at the time thewavefront was emitted.However, that point moves in the direction of the receiver. In the time between wavefronts (oneperiodT) the source moves a distancev Ts. The shifted distance between successive wavefronts inthe direction of motion ( ) can easily be determined from an examination of figure 139 above:λ ′λ ′= λ 0−v Ts(1005)We would really like thefrequencyof the doppler shifted sound. We can easily find this by using

474Week 11: Soundλ ′=v /fa′andλ=v /fa. We substitute and usef= 1/T:v af ′= v a− v sf 0(1006)then we factor to get:f ′=f 01− v sv a(1007)If the source is moving away from the receiver, everything is the same except now the wavelengthis shifted to be bigger and the frequency smaller (as one would expect from changing the sign onthe velocity):f ′=f 01 +v sv a(1008)11.4.2: Moving ReceiverSourcevv T’rv T’arλ 0ReceiverFigure 140: Waves from a stationary source are picked up by a moving receiver. They have ashortenedperiodbecause the receiver doesn’t wait for the next wavefront to reach it, at receivesit when it has only moved part of a wavelength forward. Thekeyto getting the frequency shift isto recognize that the sum of the distance travelled by the wave and the receiver in a new periodT ′must equal the original unshifted wavelength.Now imagine that the source of waves at frequencyf 0is stationary but the receiver is movingtowards the source. The source is thus surrounded by spherical wavefronts a distanceλ 0=v Taapart.Att= 0 the receiver crosses one of them. At a timeT ′later, it has moved a distanced=v Tr′inthe direction of the source, and the wave from the source has moved a distanceD =v Ta′towardthe receiver, and the receiver encounters the next wave front.This can be visualized in figure 140 above. From it we can easily get:λ 0=d+ D(1009)=v Tr′+v Ta′(1010)=( v r+v Ta )′(1011)v Ta=( v r+v Ta )′(1012)We usef 0= 1/T f,′= 1/T′(whereT ′is the apparent time between wavefronts to the receiver)and rearrange this into:f ′= f 0(1 +v rv a)(1013)Again, if the receiver is moving away from the source, everything is the same but the sign ofv r ,so one gets:f ′= f 0(1− v rv a)(1014)

Week 11: Sound47511.4.3: Moving Source and Moving ReceiverThis result is just the product of the two above – moving source causes one shift and moving receivercauses another to get:f ′= f 01∓ v rv a1± v sv a(1015)where in both casesrelativeapproach shifts the frequency up andrelativerecession shifts the fre-quency down.I donotrecommend memorizing these equations – I don’t have them memorized myself. It isveryeasy to confuse the forms for source and receiver, and the derivations take a few seconds andare likely worth points in and of themselves. If you’re going to memorize anything, memorize thederivation(a process I call “learning”, as opposed to “memorizing”). In fact, this is excellent advicefor 90% of the material you learn in this course!11.5: Standing Waves in PipesEverybody has created a stationary resonant harmonic sound wave by whistling or blowing over abeer bottle or by swinging a garden hose or by playing the organ. In this section we will see how tocompute the harmonics of a given (simple) pipe geometry for an imaginary organ pipe that is openor closed at one or both ends.The way we proceed is straightforward. Air cannot penetrate a closed pipe end. The air moleculesat the very end are therefore “fixed” – they cannot displace into the closed end. Theclosedend ofthe pipe is thus adisplacement node. In ordernotto displace air the closed pipe end has to exert aforce on the molecules by means of pressure, so that the closed end is a pressure antinode.At an open pipe end the argument is inverted. The pipe is open to the air (at fixed back-ground/equilibrium pressure) so that there must be a pressure node at the open end. Pressure anddisplacement areπ/2 out of phase, so that theopenend is also adisplacement antinode.Actually, the air pressure in the standing wave doesn’t instantly equalize with the backgroundpressure at an open end – it sort of “bulges” out of the pipe a bit. The displacement antinode istherefore justoutsidethe pipe end, notatthe pipe end. You may still draw a displacement antinode(or pressure node) as if they occur at the open pipe end; just remember that the distance from theopen end to the first displacement node isnota very accurate measure of a quarter wavelength andthat open organ pipes are a bit “longer” than they appear from the point of view of computing theirresonant harmonics.Once we understand the boundary conditions at the ends of the pipes, it is pretty easy to writedown expressions for the standing waves and to deduce their harmonic frequencies.11.5.1: Pipe Closed at Both EndsAs noted above, we expect adisplacement node(and hencepressure antinodeat the closedend of a pipe, as air molecules cannot move through a solid surface. For a pipe closed at both ends,then, there are displacement nodes at both ends, as pictured above in figure 141. This is just like astring fixed at both ends, and the solutions thus have the same functional form:s x, t() =s 0sin(k xm) cos(ω tm )(1016)This has a node atx= 0 for all . To get a node at the other end, we require (as we did for thekstring):sin(k Lm) = 0(1017)