### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# Special Topics in Astrophysics ASTRON 250

GPA 3.9

### View Full Document

## 21

## 0

## Popular in Course

## Popular in Astronomy

This 60 page Class Notes was uploaded by Lester Reinger on Thursday October 22, 2015. The Class Notes belongs to ASTRON 250 at University of California - Berkeley taught by Staff in Fall. Since its upload, it has received 21 views. For similar materials see /class/226675/astron-250-university-of-california-berkeley in Astronomy at University of California - Berkeley.

## Similar to ASTRON 250 at

## Popular in Astronomy

## Reviews for Special Topics in Astrophysics

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 10/22/15

UC Berkeley7 Astro 2507 Order of Magnitude Physics Eugene Chiang transcribed by Aaron Parsons Spring 2006 Lecture 1 Everyday estimation problems 0 How many plumbers are there in Berkeley Population of Berkeley few 105 breakable xtures per person few xture breakage rate 1 per 10 years so the demand is few 104 xtures per year If a year is 250 work days and a plumber can x 3 xtures per workday then it takes 50 plumbers to meet the demand of Berkeley The yellow pages has 200 listings 0 How much lipstick has Brittany Spears eaten Substitute chapstick for lipstick Then think in bulk Eugene uses 1 stick every 2 months Then take data chapstick is 5 grams Brittany Spears is 25 and has probably been using lipstick for 15 years So at 30 grams per year for 15 years this is about 1 lb of worn lipstick How much of this did she eat Eugene says the ef ciency factor is 01 So BS has probably eaten 05 lbs Amount of money in an armored car The natural length scale of the problem is that of a human so we ll say this car is 6x6x12 ft A bill is performing a measurement 5x3 inches What about thickness Thinking in bulk a 500 page book is about 2 inches This comes to 1 107 bills in a truck If these are 20 s we have 2 108 dollars At 1g per bill this comes to 10 metric tons and that s a little full We ll say you can only ll a truck to 3 which means there are 70 million dollars in a full armored car Lecture 2 Material Properties look at Tabor Gases Liquids and Solids Density and Cohesive Energy Enemy of an atom in a amce 71 E4 soheswe enemy LU mean mtEVammK Spatmg Bohx Atom The Coulomb one keeps an election in ocbit Mound a pioton my 4 N oo and angulax momentum comes quantized mama h 2 72 2A 7 i This tells us that ii and ev axe the natuiai units oi atoms 1ev 16 1042279 t H but election shielding makes size quite insensitive to z aitei hthmm imilazly eneigies ox most atoms not H don t change much aitei Lithium wheie h Densities Amx F7 as The iatio oi z pioton numbex to A atomic numbex is about It tuins is rough independent oi applied pressure ox P lt Mbm This thxeshold comes hom meme zanwatwn In the axth s cote the pzessuze is about a s Mbax and these s 121 We can estimate the pzessuxe ionization oiH pxessuze is an enexgy density so out that density 13 62V Pm 1 3 10mm 2 These conditions can be met in the interiors of large planets like Jupiter and Saturn and H undergoes a phase transition from molecular to metallic After metallicizing H the pressuredensity relation depends on whether electrons are rela tivistic or not For non relativistic electrons the pressure goes as E P x p3 p2 This is because energy goes as E ZWE and p g from the deBroglie wavelength Electrons l are fermions so the won t be squeezed more densely than their wavelength so 713 N Since 715 X p we have 2 P x pE x p715 x 03 For relativistic electrons we have the same derivation except the energy per particle doesn t goes as p2 but rather op In a white dwarf which is on the border of being relativistic because neutron stars de nitely are 1MD 1R art h N 106 p white N diuan So the threshold for relativistic electrons is at a density of 106 Cohesive energies Latent heats of vaporization Vaporization is the breaking bonds there are several kinds lonic like NaCl NaF HCl have 60 z 25V Co valent like H20 7 H20 60 z 055V or diamond 60 z 45V Metallic when many atoms come together they become metallic and the outermost electrons decide they are free and go running around making a sea of electrons In this case the energy to take out an electron is simply working against Coulomb forces and 60 N 35V Van der Waals dipoledipole bonds like long chain hydrocarbons wax or solid methane or solid rare gases Kr Ar Ne Permanent dipoles u ed If x is the distance between the 2 dipoles the force between them is F g which you an get just from dimensional analysis This is in contradiction of Tabor Anyway this gives a binding energy of 2 p2 52 d E 7 N i 7 T133 13 1 V mowm m was Emmy dzzumm ltlt1 Permanent induced dipole The electric eld from a dipole causes electrons to be displaced and create an inverse dipole and Minduced O Eimposed and o e s the polarizability and depends only on the volume ol an atom Thus the one becomes Item the pxevlous one equatloh ezdzo 22 d 2 F m 7 ltpgt a9 a9 9 v 27 1 quotDispersionquot attzactloh Symmetuc atoms m lsolatloh have mstahtaheous dlpole momeh s The cutlcal sepazatloh we can estlmate as e Tl whex Tl ls the ocbltal pexlod ol the electcoh hls ls saymg that the mlozmatloh about you mstahtaheous dlpole hatuxe plopagates at t e speed ol hght and you dlpole hatuze ls c ahgmg on t e tlmescale ol the ocbltal pexlod ol the electzoh Th A 4 3whele Fox e i m example HZ molecules m blown dwaxfs have a blndlng enexgy ol E 1 as o The take home message s that these enexgles axe all 1732V except 02 Wn dex Waals whlch 10 a 104w We can estlmate the mass ol lood pet person pet day Combustloh 051375 7502 6002 2H20 We bleak s c bonds at say 2 ev pet bond 7 12 ev Each sullen bums s c and s H 80Mg Thls gwes us 15 10111l Conveztmg mto Caloues 1 Cal 1000 cal A000 1 we get A Calg A human needs 2000 Cal a day so w eat 1 lb ol lood a day Lateht heat of upouzallon ls about 1 ev pet molecule The latent heat ol lusloh llqulds axe sohds whlch have lost them lorlgrzange oxdez and has about 116 to 1 the heat of upouzallon Lecture 3 H Recall that we louhd that 02 dehse thmgs packed at a lew A on a slde the coheswe enexgles u c e c To W Law which opposes the dipole eld We can therefore write 3 e lt dinduced gt I Eimposed d 2 induced 7 Edinduced 7 E a3 04 where u is the induced dipole7 and 04 is the polarizability We also made the assertion that the ratio of fusion heat to vaporization heat is 110 to 130 However7 there are counter examples such as mercury 0615 and gold 1081 Mechanical Properties 0 Moduli The Young Modulus Tensile stress is the ratio of the force applied to the cross sectional area of a cylinder Fapplied ACTOSS Young s modulus relates the strain 6 to the tensile stress T A6 T MYoungE MYoungT The Shear Modulus is the change in orientation with pressure 7 E Mshea39re and the Bulk Modulus is the volume change under pressure AV P MlmlkV compressibility is the inverse of the bulk modulus Strain is dimensionless7 so the units of moduli are stress7 forcearea7 energyvolume7 etc For most solids ionic7 covalent7 metals7 van der Waals7 these moduli are of the same order Mlmlk N Mshea39r N MYoung There are exceptions to this ruloloer7 jello7 and it has something to do with long chain molecules Neutron star crusts are also an exception 0 Estimating the bulk modulus Mbulk 7V Now the change in internal energy of a solid is dU iPdV so this gives us the relation dZU M V7 bulk de A cavalier approach to simplifying this equation is to say this is of order the internal energy density We know that bond energies are of order 3eV7 and the volumes of these energies axe ol ozdez 33 Thls Wes us an estlmate ol 10114395 0 MW Thls ls hot so a off except oz covalent sl steel and dlamohd and van dex Waals sohds A less cavallez method ls to xeduce the above equatloh to consldex ohh a small cell and then to say that Vest Ueau N 5e 7 aunmm gsscd lndex ol a powex law The lesson hexe ls that eve shoxt mtemls you can model auythmg as a powex law two pomts ls a lme and that powex law wlll ohh mtzoduoe a ado ol a lew when taken the deuvatlve ol em 72 ls some Elastlo Wave Speed We can talk about Shea oz compzesslve mves The speed ol a wave depehds ol the tehsloh vs the mass ol the medlum T A we substltute M oz T and a fox A Fox most thmgs we have shown that M N hd a N aim and thls glves us A Thls seems to hold oz most mateuals glass to e Fox sohds 0 SMbM a wood lock e Falluze stxess weld sness fallme Sness E Umdueotlohal stxess wlll cause a mateual to undezgo buttle lalluxe Genezalh lalluxe stxess SN 102 gt 10 s 4h the dlsplaoemeht Falluze ocouzs bee e applled stxess gets magm ed heal omks lmpexfechons m the mat lal Ahheale teel ox e p e allo ed these lmpexleotlohs to settle out and has sN 18 101 compaxed to nozmal steel at sN 5 10a The ls a wlde xange m Ialluxe stzesses Comets have s N 10 stone ls Mound 108 and non and bone axe Mound s N 109 If a material is under hydrostatic isotropic pressure then failure requires that a gt S P Generally yield stresses are around Y N Yield stress is the point at which the stress strain relation goes non linear 0 Surface Tension 39y measures the surface energy of an object and it has units of It takes energy to break to the surface if there are intermolecular forces which you have to break to pull a molecule up to increase the surface area Surface energy arise from unsaturated bonds molecular bonds which are unlinked with other molecules Surface tension is a 2D tension or a tension per length F 71 c We can estimate the surface tension as 39y N 5162 The 6 is for the 6 faces of a cube of liquid Putting in 05eV for H20 and the quintessential 3A for a we get 39y N 150 The actual 39yHZO 470 Thermal Properties 0 Speci c Heat in Thermodynamic Equilibrium the contribution to the mean energy is kT per quadratic term in the system s Hamiltonian For solids and liquids there are 6 such terms 3 for three dimensions in potential and 3 for three dimensions in kinetic energy Thus Usolid NatOWSBkT Speci c heat is 1 8U 07137 7 3k Natoms To the extent that things don t puff up like gases do op N cw For atomic solids 1 x Cu z 37k and for molecular solids op x Cu z where 7 is the number of atoms in 3 71 H mass of molecule7 the molecule Lecture 4 More Thermal Properties 3 molecular mas a bit larger than what we measure at 300 The reason for this is that this temperature is below the Debye Temperature which is another way of saying that not all the Vibrational modes of Al are populated We can calculate the mode frequency of a cube of Al length L on a side as the wave speed over the wavelength If we are in resonance this is Last time we said that cW If you work this out for Al A13 we get something Ce Vmode i where cg is the elastic wave speed Now we say the amplitude of the mode is equivalent to the energy in the mode which is the number of phonons 5 having frequency 1 where Ce 1 7 X n 2L of phonon is Ps olt 2 22 m Then Ps 55 17 at and so the numbex of phonon is gwen by 25 50 1 727 2671 Now these modes wxll lend to be well populated when 5 N1 IL weH exuted L T Hun In 3D 7 r xepxesemmg a spheucal sulface 2amp5 ngj g so the above gxaph mu become a 3D plot wnh a thxeshold minimum based on none spam anespunds to Q The minimum waveiehgth oi a phonon set by the iattioe spacing sets a maximum 7 which is the We can estimate this tempeiatuie as T 7 Me My 7 M Using o N 323 and e N 3 we get AUUK So the takeaway point is that Debye temperaturas a ilar to mom temperature Thermal Di usivity Conductivity in insulators ah ihsuiatoi because oi inhomogeneities iii the iattioe and because oi phononrphonon 1 down ihteiieiehoe phohohs aie ooiiieioha vavnnR dm H vde an get somewheze by go hg howheie This is the diuhhaide mlk you have a 5050 ohahoe oi moving ohe diieotioh oi ahothei You say that a step width is a and sub t is taken in time 7 ii you have a bunch oi diuhhaide then you have the piopeity that e but 02 7e 0 in iaot we can wiite 05 1 N NZ e me 1 H 12mm 1 c1 63 is a term which is always positive and the 3rd term above averages to 0 in the limit of large N Now we ll say that m E and this gives us 205 W where D is the diffusivity The diffusivity is all you care about it can be 2 or 126 or 122739 0 You can get a ux from diffusion if you have more drunkards on one side of a surface than on the other We can write this as Net 7 7Nm 6 7 Nz Fl a At AT where A is the area of the surface Multiplying the top and bottom by 62 and de ning C E as a concentration we can write the above as a derivative of concentration times the ratio of 6 i T Flux this is Fick s Law Alternately Flum so we could write this we ll simplify this equation to say that N Di 0 Back to the phonon random walking through an insulator we can say 1T 1T 7 7 dt dmz where a is the thermal diffusivity D in our digression Using that a N 612 we ll say u N 05 and 6 N Amfp and this is typically a few lattice spacings at the Debye temperature We ll say N 3a simlOA So a N 1 where we ve divided our naive 612 by 3 for 3 dimensions 8 which slows our naive linear diffusion down 0 We can estimate the time it takes to cook a turkey We ll just by dimensional analysis say 2 2 T N rs For a 20 cm bird and a 104 and so we get something around 10 hrs Lecture 5 More Diffusion o How long does it take for perfume to diffuse in air We can estimate the mean free path of air as Amfp N where mm N 3 10190m 3 and a N 321 x 2 x 2 where the 2 twos 1 air 10 are because perfume is a heavy molecule7 and because cross sections are doubled for air air collisions Doing this7 we can get a diffusivity of D N 005 Plugging this into if N we get a time which is a little over a year much slower than experience suggests This tells us that diffusion is not what is wafting perfume across the roomA More Thermal Conductivity 0 Heat conducts by the equation F ichT where kc is the thermal conductivity Comparing this to the diffusion of thermal energy density we see F ichT 7kVU 7kVpcT 7 kpcp VT 5 kc Estimating a value7 we ll say p N 17 GP N gig and k N 10 27 for water7 so we have kc N 10 2 w Water ice is actually 5 10 37 wood is like 3 1047 and feathers are 85 10 5 cmsK39 o Conductivity should have a thermal dependence7 because of the Debye temperature For T gt TDezyye7 we have K cx Aphmm This is because the mean free path depends on the number of phonons7 which depends on the expected number of the highest energy phonons shymm This number is given by 1 SmMD W e M 7 1 7 T TDebye TDebye TDebye i i i For T g T K cx AZth N 3a 5 2T This is because the phonons which are most important for scattering other phonons are those having wavelength N 2017 independently of Thermal Expansion 0 Why do things expand when you heat them up It is because of the anharmonicity of the potential Anhavmamz pmennal avevage Spamg mueases W m empevamve We can expxess the change In the mean latttce spacing as a unctton of tempexatuxe as Aw 7 aTAT a wheze w s the coef cient of thermal expansion H out guxe above 15 to scale then W N g and so H N E N 5T 5 We estnnate that QT N g N 5 10 Coppex s Mound 16 10 s and quaxtz IS 0 05 10 my Summev Sme H I 9 ottsemnn mermat Expansmn m penanta Properties of Metals Metals have me electxons 1 hee ox every positive Ion 12 You d guess that electrons should just have energy kT multiplied by the degrees of freedom but this is wrong because electrons have way more energy than ng apiece The electrons in most solids are degenerate they are degenerate if they have less energy than the Fermi Energy Ef The Fermi Energy is given by 2 P E 7 f 2mg where p m Thus7 the Fermi energy is the energy of con nement who hzn E N 7 f 2mg We can estimate the con nement of electrons in the blackboard by saying 715 N 117 and getting a Fermi Energy of 105V We then estimate that kT at room temperature is 035V Thus7 we know that blackboards are degenerate Another way of saying this is that the mean deBroglie wavelength is greater than the mean spacing The wavefunction of electrons in a box is given by Q i6ina mnyynz2 L3 and these electrons have energy 2 2 2 2 E WW t 712 t W 8 At T O7 electrons ll the lowest energy states In the limit of nf gtgt 1 4 37w 2 N where N is the total number of free electrons in the cube The energy of an electron which lives at the top of the Fermi sea is h 3N g 2m5L2 E and thus the number of electrons per energy is EF dN 87r g L 3 A Ede 32mg2 E Eg EF and so dN MM of Eleumns ham enemy belween E and Bag ehh the top ol the Fezml sea can gam enexgy to leave Remembez that oz ozdmary mateuals kt ltlt E p Because the numbex ol electmhs whlch can any the heat ls AN N 375 IcT N E the thexmal conductmty horn electmhs ls much less than what you d expect so the Ions m metals cany the hon shale ol the specl c NOW cv ktul BMW cvatlmhoras heat Thus at zoom tempezatuze metals and msulatozs have the same heat capauty to ozdez ol magmtude Lecture 6 Thermal Conductivity The thexmal ux m metals ls cauled by mobile hee electmhs and thls makes them much bettez heat cenduetozs than the msulalozs we have dlscussed pxevloush F WUttmex Thus the thermal conductivity is kc E We can estimate some thermal conductiv ities We ll write a as A3 with the factor of 3 reducing the diffusivity for 3 dimensions it takes longer to randomly get somewhere than in 1D Thus the important factors are 2118 kT 3ki F 4718 In comparison to insulators 4 is the same 715 mom 3 is lower by 77 N 2 U5 is 2E k m N 100077 This makes 2 higher by about a factor of 300 Finally 1 the mean free path of an electron in a solid must be much larger than the mfp of a phonon as we know by the fact that metals are excellent thermal conductors If a is the spacing between ions Kittel pg 302 304 tells us that 2 AdeBroglie fa d2 where d2 is the rms thermal displacement of an ion their jiggling around If all the ions were on a perfect lattice a plane wave of electrons would never scatter An ad hoc rational for the above equation is as follows each step in a changes Adagmglie by id Thus the number of a lot bigger than in conductors 125 N A5 N 2 steps for Adagmglie to change by order unity is W We can estimate this factor as 2 l 7153 2 AdeBroglie N T 1 3 12 5mman 12 N ng Now LUZ 2713921mam 27r2 Cir and CB N M23 and Mzmlk N so we end up with AgeBroglie N 2702E 2 3 kT This means that the mean free path of an electron propagating through a lattice is N 600a several hundred lattice spacings Thus 1 is higher by a factor of 200 72 cal kansulato r N kgmetal N 2 And as it turns out Cu and Al have kc 1 and Fe has kc 02 On the other hand liquid Hg has kc 01 because liquids have their long range order broken scattering electrons before thermal wiggling does 1 Notice also that though 5 X T 3 includes a factor of T so thermal conductivity should be temperatureindependent which it is until at low energies there aren t any phonons to scatter electrons and then instead scatter off of the same impurities that scatter phonons in insulators Electrical Conductivity Resistivity o The current through a wire of cross section A length L and voltage V is VA I 705 where as is the electrical conductivity as E E is the electric eld and J USE i Now J is the current density 0 Electrons encounter resistance from ions I E A J A 715 5 vimft The drift velocity is much smaller than the average forward velocity vimft N atfree N A5 2 T as 75 n5 5 8 1017 1 meUF L7 Thus We womer 39 39 ls 9 1090hm m so p5 N 1 1078071771471 This estimate is a little low Al is 3 10 80hm m Cu is 2 Fe is 10 and liquid Hg is 100 Magnetic Diffusivity 0 Magnetic elds decay because currents decay This decay is governed by Ohmic diffusion Ampere s law says vx C and Ohm says J 05E and this gives us that e e e 4 e e VxVxBampVxE C Using Faraday s law and the fact that there are no magnetic monopoles a 2e 3 5 B and ram E 4 is the magnetic diffusivity 0 Thus the Ohmic decay time is of order L2 tdecay where L is the length scale over which E changes Fluid Mechanics o Navlez stokes ea a7 7 VP 2 E EH V r 7 7mglzw Well assume a ls constant v 17 0 and that we axe movlng subrsonlcally Lecture 7 More Fluid Mechanics o The Vlscous Texm Mulezules al boundary at Hum elemems lmnstsy momentum whlch says that the vlscous ux ol momentum ls nelateol to u the dlf fuswlty We can then nd the nate ol change ol the momentum We may also wnte ths tenm as a viscous stress a one pex umt anea ol constant op Newtoman uld an o pi F7 83 Viscoelastlc meoha behave llke ulds on some long tlme scales and llke spungs elastlc on othen shont tlme scales sllh puny ls agoool example pull lt last and lt bxeaks pull lt slowly and lt stnetches sheans molehmtely The eanth s also hlce thls We would llke to know oh oz an anlutnany uld The klnematlc Vlscoslty oz gas ls slmple because the pantlcles axe the unlque beanens ol momentum 1 27722 V9 EhwnT N 0 2T 17 Note that 7 is the kinematic Viscosity and u u is the dynamic viscosity Thus it N 02 104 2 10quotcgs Foi liquids Be as WWW 3 our2 and as it comes out mm N wipes but this estimation is very coaise and in pmtice wuss Ethyiaicohoi 08 15102 G asoiine 0 7 A s 102 theiine 1a 12 101 Hg 136 12 102 Skin niction Diag Boundary Layeis tumuignt iagims X J3 haigtii Hf isaveioniiy cunmuis Juli disianze aieng pints It takes time ioi the negative momentum iiom the piate to diffuse upwaid and begin sioWing ieam at some height a This time goes as 32 tauuse N 7 The advection tiavei time is iust 5 so thickness oi the boundary iayei as a iunction oi a9 is It is impoitant to note that this curve does not descube a stieamline but iathei Just a locus oi constant it is is ssenhalh a s aied a9 cooidinate ts ould be noted that you typically have to go so iiom a plate beioie you see youi oiiginai veiocity Anothei way oi wnting is is in teims oi the Reynolds number We can calculate the total drag force on the plate as FD adm dz waxz Bu 7 i i where a 7 puay N pu cx Then we have 1 3 1 FD N pl Uiziz This gives us that CD 2 Re If you go out far enough the laminar boundary slowing down generated by the plate stops going as z and starts going as x because turbulence starts kicking off momentum swirls upward How far do you have to go down the line before this turbulence occurs Turbulence occurs when R55 2 103 where R55 E g The number 103Z comes from the fact that if you introduce perturbations on top of u they are either ampli ed or attenuated based on the Reynolds number However there is a range of Reynolds numbers where ampli cation occurs and if you exceed this range you will get attenuated This is why you can get transient turbulence in cases where the Reynolds Number is increasing as you go down the ow The number 1000 is where ampli cation rst can happen A Note that R55 Reg How does the displacement and stress due to turbulence scale with z We ll do a simple minded poor man s mixing length theory We have z 6N Ila where this time 1 is a turbulent dil lFusivity77 and is the product ofa length scale and a velocity scale The length scale must be 6 and the velocity scale must be U Thus 1 6 x U x Fudge 6 N A MUEFWW udgez We can solve for Fudge by matching 6m aming at Rem 135th 106 Then 5mm 61Mquot z A x 73 Fudgec39rit merit Lecture 8 Turbulent Drag To detennlne the dug one ln the luxbulenl xeglme an Fates N ma Atee N puma Atee U vaginalrm that N Rated2W And ths gwes us a dag coef clem ln the luxbulenl xeglme el CD N 2mm N 0 002 hub N FwdW swim Howevez CD can change by an oxdex ol magnltude dependmg on the suxlace toughness ol the plate e The takeaway message hexe s 1 C W WNW X N emstum D twwmt bolmdmy tenet Blunt Bodies Not Streamlined uD at Stagnanun um R vlsmus boundary lave The pzessuxe neat the stagnatlon pelnt ls gzeatex because uld elements had to slow down as they appxoach the sphexe Thus the Bemoulll value ox the stzeamhne BPpUZ 20 H the pzessuxe at lnhnlty ls Po then a the stagnatlon polnt all ol the enexgy gets moved lnto pzessuxe U s 0 As the uld elennent zounds the bend ol the sphexe P PorngZ 212 the pxessuxe is less t an at the stagnatlon polnt and the uld elements accelexatlon Mound the bend Hyou lgnoxe the boundaxy layex then stxeannhnes would be able to c Ive the ball and the ow would look symmetno beloxe and altex the b ll low P axound Wall P lugb P V advevsa byessuye uyumsnt u buundaw layer l hypmhetlzal lnzamm m absenze ufvlsmsltv The boundary lays teaes off because ol vlscous dug ol the boundary lays on uld elennents next to lt It doesn t mane how snnall wsooslty s the ow wlll always look llke thls raghue L Pea WESSqu mag F I F q l v Note that u you mcxease the suxface mughness though the boundary lays gets thickex 1t hugs the suxface of the ban 2mm and mgiucas the pzessuxe dug on the ban Ultunately the dug one is Fwy P P57FRZ N 2 42sz In genexal on nonsphexes the pressure drag s 1 2 FM yaw A722 when an s a dag coef cient that depends on the shape of the object How much 5km tnctmn does the ban suffe in F N A N 7A new a m w it If Hch u U N WEA N W Then the who of skm to pzessuxe dug 1s FNW pm 1 Fumesquot UZRZ REE In a hxghh viscous uxd ReR lt1we have ow whxch s the solutmn to 72 1in 0 p 22 with 17 0 at the boundary 17 U at in nity and P P0 at in nity For a highly viscous uid the ball has an effective cross section radius of 2B you have to go out 2 radii before ow is unaffected by the ball Stokes solved this problem and found that although skin friction does win out in this scenario it does so only by a factor of 2 That is to say there is a small component of pressure drag which scales as 1 the viscosity The effective Stokes drag is FD 67rpzUR 47139 skin 27139 pressure and we can de ne a drag coef cient in the Stokes regime C 7 67rpzUR 7 121 1 D 7 pU2R27r 7 UR X Re 0 The reason why the boundary layer separates from the ball is because of an adverse pressure gradient E behind the ball Now VP N so increasing L can help prevent boundary layer separation Lecture 9 Just a note if you use linearized uid equations then Cartesian shear ow in boundary layers is stable all disturbances decay However if you include non linear terms you can nd instabilities like turbulent boundary layers Also remember that while laminar ow may be completely 2D turbulence is inherently 3D If you want to order of magnitude drag equations where there is a free stream velocity U molecular viscosity 1 and a sphere of radius R then you compute your Reynolds number U R 3537 1 and if Re lt 1 then you have Stokes drag mostly skin friction with 81 F N D puay Area N 67rpzUR For Reg gt 1 there is mostly pressure drag due to boundary layer separation with 1 FD N EpCDU27rR2 where CD the coef cient of drag is N 1 for most non streamlined bluff objects More speci cally CD N 1 if the boundary layer is laminar and 013 if the boundary layer is turbulent which is when Reg Reg N 105 a 106 o In other drag regimes U Supersonic compressible ow where U gt cs N UT there is the Mach number Ma E a and the coef cient of drag is CD1 23 nee moleculai wheze AM gt R we have 2 techniques You take the Stokes Ieglme with 2 N and we get FD N p 5U7FRZ so you substitute a e5 ion a U Note that Am gt R and e5 ltlt U We can descube this as the momentum ol molecules hitting the liont lace ol the ball pU sUlsRZ less the mo entum ol molecules hitting the lean lace ol the ball p25 s Ue5 s URZ Thus the net loice on the ball is pucst ll you mm to see some awesome uld moVies go to the National Committee olFluiol Me chanics Films http www mth uea ac ukhUU7MTH3DA1backdoox on maybe oliiectly at http web mit eolulluidswwwshapiionclml html Energy Dissipation How hot to pipes get when you put a uld thiough them7 A Fluid Paynel uJup In this case the uld pucel exeits a loice on the bottom pm mbotbm ai d d may 9 c uoat 0n the bottom Woxk is alone by the paicel on the suiiounoling pei time On the top Woxk is alone on the paicel by the suiiounoling pei time dam21140 op an V p 8y t The the powei iniecteol into the paicel is memo bo bm an an F V7 em 7 V7 F7 apr by F7 83 wheie g is the same ewiluateol at the top and bottom Thus a P N pyid d olbp r moat an al N 7d d 7d Fway a 383 3 2A Thus the enexgy chsslpatlcn ale pexvolume the txanslczmatlcn cl the bulk lnnetlc enexgy lnto xandom klnetlc enexgy ls an 2 E 7 W lt 53 Mom genexalhx E s a stxess tunes the Shea tale whlch s E N 03 8y c Hyou put a at plate cl axea A on a llulcl and want to know how much powex lt xequlxes to m 12 that plate along at constant velcclty v you can calculate thls as 2 Pmua39r F walaczty W37 1 N WA n N WA when L A L K L a A Altexnatebx lulu heatlng la e Z Z 7 an 1 2 plA39u P E Velmepp 8 AD pl 5 AD T Thls ls all equwalent to solvlng the NaVlenStckes equallon whlch ln thls case looks llke win 0 Lecture 10 Velocity Pro les of Imbulent Boundary Layer Fust xecall when we wexe tallnng about the velcclty pxo le cl a ow ovex a llat plate we lound the Blaslus lamlnax pzo le mu Enu lemmer Blew zu s H wevex when things get tuxbulent thexe s a lamina sublayez whxch looks hke Blasms but then velocity goes as t 3 LI ammav snhtaysr In the viscous subrlayex 13 2 J u s 7 p1 v9 jump Mum at meompxesslble uxd Thus we have a a n Nata 15 gt Us VVZ39Uz We ll thxow out the rst tecnn because uy y 0 We then xelate the tecnns whxch xemam W2 Vii e 1 My algae atsue 62s We axgue P whxch by t t h w nnust be small small In this case nneans less than 1 But the second tecnn s laxge gt 10 way t s all comes to show that VVZue s much btggez than h A t e othex side of the equation we stated thh so 1t tuzns out the equation we axe mlezesled 1s m We then nd that pug camtant so we 1y 22 26 We know that u no at y the and u 0 at y 0 so we have the Law or the Wall Ths Jusll es what we wexe Saying the veloolty pxo le ol the sublayex ls lmeax Next we want to know about the tuxbulent lays plVZ39F 0 but V m ths case ls oz tuxbulent and Want N ay gtlt May v V y a Fund says that whezevez you axe away mm the wall the eddy whlch In uences you the most ls the one whlch has length scale you ohstahoe hom a mll v vellew dlffevenze LLvl x U R Ths leaves us wlth the equatloh an t t W 720715 on 5 yyay But m wllh L L tzahsltloh mm the lamina sublayex to the tuzbulent xeg lme Ths allows us to say that the constant above is 311 puffs Thus laminar y 8y 619 digdu 615 y Vu mi lt1 1 Elsi 615 UT 1 in may be interpreted as the velocity at the boundary of the viscous sublayer Where we use the boundary conditions u in at y 615 The result is the Logarithmic Law which is valid for y gt 615 c We expect 6151 is a constant and is the magic number which sets the transition between the laminar sublayer and the turbulent layer above This allows us to write the logarithmic law as u 7 1 constant X lni UT 619 1 constant X ln gm 1 0 Comparison with observations The linear law is perfect 7 yuf and is good for y g 107 For the logarithmic law 3 5 24ln in 1 has some weird constants and is only good for y 2 10 Don t forget 14 E A if and awau N 311 N pug For order of magnitude u U at y 6th so i5241n 9 V lncredibly this theory is still good inside pipes where VP 7 O contrary to our previous assumption For fully developed turbulence inside pipes 6m is just the radius of the pipe Lecture 1 1 Flying v gt v min pavzel navel fastev F l 5 WWW an s gt gt a lil LL e hum lennih Ritual aimiils ave Symmemz yelymn an mdmamm my h The ll loioe geneieleol by a nile Wing is Whexe A is lhe planform area and choxd lenglh 2L is ol oiolei 1 in oiolei lo y actually 2L N 27rsm o This diomes lhsl o 1 410 1 Wing lheie is a lill loioe FL whlch opposes lhe weighl w ol lhe plane Thexe is also a dag loioe FD whlch has oonliibiilions lionn ski liiolion and piessnie dug il lheie is oundary lays sepmlion To oppose lhis dug you nniisl pxo llde lhinsl T When ymg sl oonslsnl velooily FL m3 and FD T We can osloiilsle lhsl L UZ m g A And so lhe speed you need lo y at depends on lhe wing loading Wilh U olt 7 Selfrslmllaz isonneliio llieis m L e K olt olt L olt m3 Thus U cxm and we know that 2mg 2 MIA Fol a seagull m 3709 A 0 115M and CL N 1 so U 10 20mph 1 8 5 anti we get hls ls a llttle lowel than ola mNa gand so UN 77 leallty and thls ls beoause ms ily hlgh whele the denslty ol all ls about g that ol sea level Fowex Requllelnents We ale lightlng 2 lolnns ol parasitic dlag plessule dlag and slnn lllotlon Pralasatae FD U inept2A U2 ch that A ls the planlolln alea the Wlng plus the blldlseye View ol the alea olt e body Note the m N 01ooe iolentoldlag ln the above equatlon ls not to be oonlused Wlth en whele convention is h Howevel thele ls anothel induced drag which ls the extla powel slphoned oil to oleate Wlngrhp voltlces manned Dvag m mafmm ufvumces m lbshonlulew of a plans These voltloes ale caused by the plessule dlilelentlal moss the Wlng at lts tlps and these voltloes spllal oil behlnd the plane as lt illes lolwald It tulns out that 15an N og N h We can estllnate 71x the hollzontal veloolty lnduced along the ads ol the Wlng as a lesult ol the plessule at the end ol the Wing being gleatel than that on t e top on N E x t F7 whele t N 5 and VP N whele AP be N mg so EL m9 quot3 be 2pr 2mm N n The klnetlo enelgy ln the voltex ls XE gmmswimue N pbzwiats 30 Hwe spawn 1 voltex every twee we get the pcwel needed to gehelate the Wlhgetlp voltlces l 2 2 2 75 cwmtee m3 Psalm N 2f N y ZpUbZ The algumeht lcl b lh the equatlch ls that the dlstahce cvel whlch plessllle decays ls b and hem a chle we mtched lh class the veltlcal scale cl vcltlces seems to be the same as the hcllzchtal scale T ele ls a second algumeht lcl why thele ls lhdllced dlag It ls the extla pcwel leqlllle to push all down whlch you must do lh cldel to stay up m3 mawe 2 ll Md 2 l1 length stale P A mlwtlldl vaSSmE dezays J J vemzaHy 0 IT Thus the mass cl all belhg them down ls mm N pU eb and lt lllst tllllls cut that e and thls glves us that WV ZpUbZ Plus as N Lecture 1 2 m you may W t o minimize the c E E72279th fly WaghtXDzstMLca t p pavasmc may um 7th manned mag 1N UK M To minimize the pews dissipated m ght them 1s a magic speed at which you axe most ef cient Planes m holahng pattems y at ths speed H wevex thexe 1s anothez quantity which n as or transport w s 7 me T MW 7 MU 221 MU 1 Z a 9 U m NW 9de The mmlmum powex is called the ghde Ths quantity s minimized when s52quot whoquot 01 ne Em Ema We can solve for the angle of inclination as my sini de9 mg cosi Flift amp Flift tan i Um Fd39rag Thus we have degum Fliftuz Which says that the power dissipated due to horizontal motions assuming ixl is equal to the power dissipated due to vertical motion The glide ratio is solved for by Emin Ep l EI 2P1 Umg N 2Fpa39ra NQEPN BioMechanics 0 Rules of thumb for energy budgets Land bound creatures through exertion burn calories at a rate up to 10 20 times their basal metabolic rate BMR This maximum can only be maintained temporarily Airborne creatures have proportionately larger lungs and can burn up to 20 times their BMR Maximum efficiency is 25 in converting chemical energy to mechanical energy The basal metabolic rate is proportional to M075 in warm blooded animals mice A elephantsl You d expect Mg because of surfacearea scaling and this is the case within a species but across species it is Lecture 14 Land Transport We can calculate the length of an animals heart rate At N Al X mhea39rt heartbeat V l X m4 And in fact the total number of heartbeats in a lifetime is independent of mass 15 109 This tells us that 1 tufe 0lt m 33 Thls ls onh lcl mammal s We also have the elastlc slmllallty whlch says that the ladlus cl the bone cl a cleatule ls lelated to the length cl the cleatule tcxel Thls gwes us Then uslng that muscle stlaln goes as 5N N 8 27Ay 72 And muscle stless ls We also use whele TG ls the txotrgallop tlansltlcn We can also use Assummg when we ale Iunmng cul leet ale actually Iollmg wlthcut sllpplng cvel the glcund we have cm x me x tm oltm Lecture 15 Struts We wlll study lclce balance ln stluts as applled to tlees and thls wlll be uselul lcl a wlde vallety clslt atlcns We stalt by cbsemng a plogt cl the helght cl a tlee vs lts dla Thele ls a blt cl scattel but thele ls a nctlceable t olt Li Thls can be explalned by the lcllcwlng analysls F fume qum ed m de ect RLr malul urnlnallun and we assume that Ag lt e As the stlut ls bent then lelatlve to the centexrllne cl the stlut 3A one slde s m compxesslon havmg a length L 7 AL anol the othex m extenslo L AL consldex the base ol the stxut to be the ougm anol we balance tozques a n n N we ound thls polnt then we wute o FL N T7 x 2 2 o F N T L ab T N otmsate x 2 X 2 AL ab T N T Whexe T ls the tenslon that xesults mm the compxesslo dlzectlons n and extenslon lt polnts In two opposlng ox each ol these We may wnte Ag m texms ol the xadlus ol curvatuze L2 2 Ag Hag LZ 7amp4 s 2R N 2R4 We may also say that anol puttlng these togethez we have and llkewlse o In the above dlaglam the bucklmg tozque mm gxavlty must be balanceol by a xestoung tozque Bucklzng N mgAy N pabLgAy 2 he bAy Rastmzng N To N W o N we say that than s an evolutlonauh xed salety zatlo 7 by whlch the zestoxmg toxque exceeds the bucklmg toxque L2 a2 7 olt anol thls suggests that Cooling Mechanisms Without cooling you overheat Given an input power of BMW 100W 7 dMch 7 dt 3k 1 l I We can est1mate the heat capac1ty 0U N molecule W etc and using M N 70kg we get for humans dT 4 K dt hr If we wait for radiative cooling to stop this we could say L anA Plugging in Boltzmann s constant but this is failing to account for the ambient temperature and radiation of the environment So Lmt an 7 aT A but only if we make the assumption that the photon temperature from the environment is the environmental temperature This assumption is called local thermal equilibrium We may then write AT T8 L N 8 7 W 5 H 288K So for normal circumstances radiative cooling can account for our cooling needs Conductive Advective collisional cooling Assuming laminar ow the boundary layer around us has a width 1 6 1 E U We ll assume the momentum diffusivity is the same as the thermal diffusivity for gases this is true but not always for liquidsithis is the Peclet number 6 1 206m 1 R N7 104 5 2 02 Thus we are not turbulent We can calculate the energy per time being removed by the wind AT A Lamde EhermalA Then using the thermal conductivity is related to the thermal diffusivity kc N pcpr N pepmfva N 0ch 7 5k 28mH7 erg cmsK7 1 AT 1 5 Lamducti39ue N 5 W Thus wind is competitive with probably dominant over radiative cooling and so Then op N and we get kc N 2 103 36 Wmd Chxll We Just need to zelate the addxtxonal coohng mtzoduced by wmd to an effectwe xadxatwe coohng Lb L LWJ eAT bAT The pluggmg Ln e and t we have ATWW M ATlt1 lt Lecture 16 Ewpoxatwe coohng The latent heat oprouzahon Ls 2 101 oxwatex Tth means that you get nd of a lot of heat by sweating magnew Bouncing Balls den lluckuess Heztz Law of Cousmns e magnitude of the denth gwen by R1rcos R1cos Nlt1717gt New L L L n e t om d displacement changes the most We would naweh guess that tth Ls R but than s a shone distance ovex thch the displacement changes 7 Thus the max sham Ls a t e N e N 7 t 23 The maxnnum stzess s 0 N Me N 22a The xestoung one Ls then x F N 072 N M The tune of contut us At N g so to stop the ball FAt N pRH t a M77 N R R 1 quot 37 The maxxmum sham 1s then We can wnte M N peg the bulk modulus can be est1mated as a density t1mes a speed squaxed so phsno 1 p tux N efm toesz and 1we plug m the maximum e at which thmgs bxeak we 0Z Th t means that th1s theory We got that 35 a get that fox a matenal thh as N 7 t at 1 a N s onh apph able at qmte smau speeds So why don t n Because they as made o1 matenals whxch don t hactuze zubbex ox example These have It 1s mtezestmg to compaxe the contact t1me o1 a bounomg ban to the sound speed ozoss 1 e5 mg t me Hextz Law guaxantees the czossmg t1me 1s gxeatez oz 1 lt Magnus one the Aezodnamxo de ection o1 a ban due to spin The dug one on the bottom oi the b 11s 1 1 1 FD N P 7 PoA N EozZAPo Epal 7 RV PM 540 Slmxlaxhx oz the top 1 Po Emil MRZ Ptop The Magnus force 1s A FM N Pop 7 Bat 38 so we get This gwes ah ucelexahon ot FM in W N i N r W Wall 2 The de eotuoh ovez the mam path d s 1 lt d gt2 1 pas 12 s N Euquot TM U E m t t7 mt mm vto tuzbuleht ow because spmmng forwaxd mcxeases tuzbulehoe whxch causes the boundary lays to stay at tached longez oausmg the wak of the ban to Mail off to one side and ozeatmg a pxessuze ushmg m the oppostte duechon ML t t a 85m Lecture 18 We now ohsouss suxfue waves We have 6 physical muables wtkthtetpw T no TV 6733 mdepehdeht dimensionless pzoduots Fust we wul try the sumplest possible oombmatuohs when k s the mvenumbex and h s the depth of the watez Secondhx we wm considex dx ezenl a1 her we axe akmg we do t sally can about the suzface teh o we ehmmate muables whxch do not mane m ths case suxfue tehsuoh 7 Then we have when o A and y as numbers Usmg dimensional nahsis we have 3 0 77 T La Tm L37 g 0 Thus So to get an ot M L and T to cancel we have Wm n2 m gn 39 lnally we need to consldex anotheu physuoal case wheue we c dumensuonal nahms to mot a unut ol length Slnce small waves the type whlch oaue about39y don t oaue about the depth ol the wateu we should use lc Thus motuvates the nal Bucklngham pu as about o and v Then we can use uvate the lncluslon olo and The x gume wheue only 1T2 matteus deep wateu no suulaoe tensuon umplues the lollowung dlspezslon uelatuon o l39lZ glc Thus us the case ol suulaoe guavutv waves on deep wateu A shozlrcommg ol Buokungham Fl us that we don t know what n2 us but ut tuuns out that un thus case H2 1 o Thexe should also be a case whe e matteus but h and 9 do not wateu globules lloatung un spaoe lou example In thus case we have the dlspezslon uelatuon And ut tuuns out that He 1 We can comblne thus dlspezslon uelatuon wuth the pxevlous one to say that In the nal ease h and 9 matteu butv cannot matteu I th FHlnz w n us case we have a lunotuon eonstont Theze aue seveual ways to comblne these vauuables un a dumensuonless b Thus l t t l 8y t untuutuon to select between the optuons y L ll we mauntaun that 5 e o lou we but not lou oy so that uld elements move unlfozmly togetheu along the x axus then we can wuute 811 h at and uslng that p p90 5 we have We then use continuity to say that the mass surface density a E ph E obeys 8 87 V av 0 Note that 1 here is Um assuming Um is constant with y Then we are then left with 0 1 8 870 pet 8m wam 95 8 85 7 h 7 7 0 p875 p 5 8m 1QO We then throw away terms which are not rst order in perturbation quantities so that 85 8 3t 8m 8 5 3t 8m 825 a at w T T a 8x 3 8111 7 Th3 8t These equations have the solutions 5 06iltwt7kz wz 97449 This is the dispersion relation for gravity waves in shallow water Furthermore we can solve for W By continuity we can say 8 81121 i 0 8m 3y 7 Then knowing Um cx ewm m we can say 12y fyeikm t Solving the above equation we nd that f Ag C lmposing the boundary conditions that W 0 at y O we have that f Ay so that we have uid particles traveling on increasingly attened ellipses as we go down 0 Dan Stevenson had a COM derivation of the above in Physics Today 18 1 1amp5 5A 38 pAm p g ElmV 41 The velocity in the X direction we can estimate as Um NamTNgi T where T is the wave period For velocity in the y direction we estimate We can then use continuity to say that the mass uX in the y direction must equal the mass uX in the X direction because uid parcels that get shoved down into the bottom need to spread out in the X direction and in the z direction pvyoz pwh W h 7 N 7 ltlt 1 11 i i 2 A2 i Plugging in for 11 and 12y we have T N E so An OOM derivation for deep water is as follows Firstly we need to know how far down we have to go before you don t feel the disturbance on the surface of the water This point must be of order the wavelength of the disturbances on the surface because as we go farther down we see more of the surface of the water and then the peaks and troughs start canceling De ning y to increase downward then P p9y5 and the X acceleration is given by l LP 3 38 ppgaz 9 am and the y acceleration is 1 5 5 5 5 E ay P8y9 ltp98ygt9 98y 9y Setting the X acceleration equal to the y acceleration which must be true by continuity This makes everything in this derivation similar to the shallow water derivation eXchanging the h for a Then W N Um so that and this gives us We should also note that y is an eXponential scale length 42 0 Finally we have an OOM derivation for surface tension waves This regime is only relevant for short wavelengths We ll try to estimate the critical wavelength for which the gravitational restoring force is equal to that of surface tension F9 m ay mass displaced N pbA2 For surface tension a perturbation increases the surface area and so surface tension will try to smooth this perturbation out Using that surface tension is a force per pemendicular length we can say strf ng c39rit N l v P9 Lecture 19 We ll now discuss applications of our previous dispersion relations Recall we have the following And setting these forces comparable regimes o kh ltlt 1 when we have surface waves on deep water and then k2 LUZ gk L p If A gt Am then we have gravity waves and LUZ gk and if A lt Amt then we have capillary waves and k2 wz L p o kh gtgt 1 when we have gravity waves on shallow water and LUZ ghk2 Using careful calculation we nd that Amt 170m 0 In the rst kh ltlt 1 case we can calculate the phase velocity up For A gt Amt we get U h a gravity k If A lt Amt then we get k L cap 10 Jllmnll In mt The mmlmum phase vehmtv s of oxdex k N 15 A caxeful calculatmn actualh gwes 23m us 7 H the wmd blows on the watezs suxiace thh WW then waves axe excxted thh phase velocr mes 1 twp H you dug a stxck at 715 then waves axe exated thh WW wstmk and the stxck pushes m hont of 1t a suxfue of constant phase 1quot ng 01 WW s less than 23 then you cannot duve waves s Gmup velouty lsgwen by n f H Agtgt AW then 1 9 Eu 3 ylc E WWW H A lt Amt then Lgmupm H you dzop a pebble attez the waves navel outwazd thexe s a stm cucle which expands at 17 7 Wave soztmg waves on deep watez axe dispersive mum1 m hug wanS um am we hm mm at equot huglhs At any moment you see waves havmg a vauetv ot peuods The wxdth g N y N Th1 national bandwxdt 12 us some achonal bands T o the size of the stoxm and the duxahon oi the st h is detezmmed by the dstahoe to the stozm oxm Racttohal bandwtdth means that waves as modulated mto sets Th1 s caused by the modulaho waves havmg close but not Identical hequencles The numbex of wa es m a set 395 N speed of boats 9L when L s the walezrlme length of the boat H om was t en t e ship can excite waves h h whose A N L Fox these waves om N ME Wheh ths happens you bow wm end up on the A5 le lol a wave and youl sleln wlll be on a llough and you wlll be chm Thls ls ng ll expellence hlgh ollag anol ll e Rood numb wl lh c uphlllquot lhe mve In fut llyou exc lh you A 1L 2L you wlll expellence low ollag m ls calleol el and comes lnlo effect when ll ls ol ololel 1 blng alleol wavamzki A Lecture 20 Tsunamls Fol shallow walel waves whe lhls case lhe phase and gxoup velocllles ale equa le lch ltlt 1 we dexwed lhe ollspelslon M 9th In l i E K Thls w al happens wllh lsunannls lhey ale genelaleol ln lhe ocean sonnewhele and all ol lhe Foullel connponenls alllve ln phase al a beac Fol lsunannls h N 51cm and A N 100km onnl ls we can c lale l 19 E m hls nneans lhal lol an eallhquake ln lhe mlddle s 7 ol lhe ocean lhele can onh be a couple ol houzs nollce belole a lsunannl hlls Wave Sleepenlng olue lo Sloplng Beuhes ncy 5 Au change e hlch Thele ale sevelal ways lo solve lol how ampllluole wavelenglh anol lleque lol changlng helghl Mao Thele ale a couple ol ways ol dolng lhls On ls WKBJ w you solullons lo equallons ol lhe lolnn y apy 0 when h 3 gtgt A a Howevel we ll do an OOM olellvallon By enelgy consewallon lhe announl ol enelgy closslng b un ary al some glven llme musl b equal lhe ounl leaVlng lh l hus we We ll co pule lhe glavuallonal polenllal enelgy can c mp le an e elgy enslly A leglon n olwalel wllh awave on ll has a hzghw cenlel ol mass l an lhe sanne leglon We can calculale lhe helghl whlch lhe walel ls lalsed Fllsl we calculale lhe unpellulbed water height lty0gt fpdVy frhodV 2 h L ifpdmdydby gm 0 7h 7 fpdV 7 pmbh 7 2 where b is some arbitrary length perpendicular to the direction of the wave Now we compute the perturbed water height ltygtbffydxdy Ah bmyh mtydm dy 0 0 Ah b 101 i j 2 dm b A h 2115 52 EO itTtil h bhz b A b A 2 t J MmJ 5 h b 1 E l O f 2bzh Thus 2 i h 150 ltygt 7 2 4 h So W7 the potential energy of the block of ocean with a wave on top is 1 W my ltygt W0 1314953 and the energy density of the wave itself is 1 2 W 7 W0 11014950 We could have guessed that the energy of a wave goes as its amplitude squared We ll double this energy to account for the kinetic energy of the wave as well we only did the potential energy Now the energy crossing a surface is the energy density times the velocity of the wave times the area of the surface Thus 1 A 2 2 Ex ghxhb Thus by energy conservation 53 constant Thls bleahs olowh lol small h because om fgh and so lal we ve lust thlowh out 5 m effect causes waves steepeh and loll ovel the top lh shallow watel A t u l E i o 30 h a r ow llheal systemsthe equency ol ohe watel paltlcle Jlgglmg lh the wave plopagates the wave to allothel paltlcle ahol lt has to do so at the equency lt ls Jlgglmg lc changes though because we have a xed equency and a chahglhg velocltv 1 X 7 Thls all causes waves to bend to become palallel to the coastllhe o Supelclltlcal vs Subclltlcal Flow Thls ls anothel shallow watel sltuatloh The sholt ahswel to how the watel ls ollsplaceol ls that lt depends on the hood humbel 1 W Ft Uslhg Belhoulll we can say B gm P lt1gt ch 9lth r y 9L y gm 9h 9L whele h ls the helght ol the watel and L ls the helght ol the log Note that the stleamllhes hele ale lholepeholeht ol 3 Thus we can take the avelage ovel the helght ol the ow and leoluce thls to a 1D ploblem Fol ho ollsslpatloh dB in d E9 h ML E HE Lecture 21 Flow ovel a log Hyou y to set up that the Mood numbez s 1 as you go ovex the log then the ow becomes choked and the ow ceaohusts Itself so that ths s not the case H watec s supezcuhcal n is owmg astec than any ohstuzbahce cah navel along the watec v m Thus a txdal boxe would be supexsomc 11 gt th L W ozlsolhezmalh has the chance to cool befoxe the next xaxefachon wave hxts 1t We need to compaxe the me ox the next zaxefachon wave to the thexmal diffusion Mme 1lt 2 G gt TMv N 2 The who ot these times s A2 V m A A A vrMp WMp TMp 4M1 AP 5 WE t g ox an at sTP ox hydzogen ox hehum and watex 1s about 1 1s andy 55 and 1s abou Thus the equation ot motion N105gtgt1 302m A 104m A 5 F mo N poet 7 7AP A Ths m tum gwes us a dispersion elation A 5 5 A77 N ip A PO 2 m2 0 Ths s a simple hazmomc oscillate thh chameleushc Ixequency where we have omitted the factors of because it is unfair to use the full Note that Z 09 independent of so sound waves are not dispersive At STP cs z 340 X The typical uid displacement is of order 5 the typical pressure perturbation is AP N 39yPOEk and the typical velocity is u N wE N 05kg APN YPO u N pocsu Cs where p005 is the Rayleigh acoustic impedance This is Ohm s Law V Z only we use the symbols AP pocsu Sound waves bounce off of water because of an impedance mismatch So to summarize there are 4 quantities associated with sound waves 1AP A NANJNElt1 g Cs vPo p0 The energy density of the wave comes from two components There is the bulk energy density which is the kinetic energy pu2 The total energy comes from this kinetic energy and the thermal energy density of the perturbation internal energy We will assume an equipartition of energy so that TE N KE and thus E N 2KE Another way to estimate the thermal energy is to nd the energy density of the compressed region and rare ed regions and subtract from them the density of a region without the wave Thus 5 l 5 l TEP0lt1Tgt Polt177gt PoJrPo 27r 27r Taking a second order Taylor expansion we ll nd that the rst order terms will cancel and we ll nd that Where we used that P0 poi Yet another way of estimating TE is to use energy conservation to say that the piston does work and this work is what is carried away in the wave During a compression phase lasting time if information about the compression is carried out to a distance about est Then W N F ut N AP Aut N ltyP03gt Aut Thus the energy density is this work divided by the region which knows about the compres sion uz W W wiw pi Acst Y 06 Volume 50 unnan healing At bllth u 20m 4 201cm so you ale sensltlve to a oldels ol nnagnltude ln bandwldth In wavelengths thls collesponds to A 170mm 4 17cm Thls ls why lol because they wlll suffez lntellelence between the lnconnlng and outgolng mve 12m you utslde eal ls used lol helght locallzatlon Youl audllory canal ls ol length 2 7 and connects thlough the ealdlunn to s h T e co as 2 t l s Th t p ane ealdlu ad delolnn wlth applled plessule and coupled thlough the a bones lt excltes the t al wmdowquot w anothel m bl he lea has a ul wlth l el sound p ed whlch ls what allows you to healsounds whlch have wavelengths ln all ol 3 bones ale pxobably lesponslble lol lnnpedance nnatchlng between the tympanlc nnennblane and the owl lndow Faavdvm savdvm Fmt olat Fm 7 4m 1 Fouat 5cm We can also calculate that is P Aw ml 13 x N 20 T e eal bones Wexe deslgne to telmlnate the sound mves entellng but they alen t pellect h cxealesa h d and you can get standlng waves ln the closed tube t at ls the audllory canal ls e at A Ag aoooHc ol you ze at least most sensltlve at thls Ixequency The length ol an umolled cochlea ls a 5m nlllnl ll nllln anllm ulunnlnln ulllu What essentlallv happens ls that waves ale exclted down the basllal nnennblane sheet Thls sheet stalts wlth elastlc modulus M N Wig but ends at M N 105 We can calculate the speed ol these waves on the nnennblane elmN MNW lo s Thls glves us a wavelength ol AM Awasm 5A As the waves go down the sheet the amplltude nnust go up and the mvelength nnust declease Thls nneans that qualllallvehx the 51 distance before non linearity from Problem Set 11 happens depends on wavelength similar to waves on water 2 db reak N F When the waves break they excite hairs connected to nerve bundles on the basilar membrane and this is what goes to your brain To leading order where the waves break depends only on frequency not on amplitude There are about 4000 nerve bers logarithmically placed over the frequency range of human hearing creating actually around 1500 frequency channels Lecture 22 0 Human hearing covers a factor of 1000 in frequency which is 10 octaves o Tonal Quality Timbre is what makes a violin sound differently than a trombone When you play a note you don t get a perfect sine wave outiyou get a waveform which repeats at the fundamental frequency Every instrument has some different mix of the of 2x 3x 4x of the fundamental frequency includegraphicsscale7harmonicspng You cannot hear differences in phase between the harmonics The human ear likes to hear some tones together The theory is that this is because some of the harmonics align Thus we have the following intervals Actually what may determine Octave 2 Fifth 3 2 Fourth 4 3 Major Third 54 Minor Third 65 Major Sixth 53 Minor Sixth 85 the niceness of an interval is the absence of roughness which is when harmonics of the two tones come close to aligning but miss they may beat or just be rough Tritones have 2 harmonics which clash in this way 0 Well tempered vs Just tempered You can t make every interval sound perfect For example 5 6 2 60 01HE1HG1H01Zgtlt3gtlt3 However if we look at some other intervals 5 2 80 ClaAlaDZHG1H0173X3X237g 52 ls means you have lo lxade exaolness ln some lnlemls oz euozs ln olhez lnlewals Bach devlsed lhe welltempered scale whloh dlslubules lhls 21101 We all lnlezwls 01 keys Amphluole Amphluole ls measuxed ln declbels a logaulhnnlo scale P MBA 101 0 7 m lt TheA ln L L A s a bandpass shape oz lhe hequenoy lnlenal ovex whlch you lnleg zale powex Theze axe sevezal bandpass shapes B C ox example Masklng When you ze hslenlng lo a louol sounol u can mask you ablhly lo heal a soll sounol An lnlezesllng delall ls lhal a low hequenoy slgnal can hazdh be masked al all by a hlgh hequenoy slgnal We can xelale Fowex lo Dlsplaoennenl P N whee N 45mins N Po k2cs Thus oz an m ozoheslza 100 dB 5 N hm anol oz a whlspez 10 dB 5 N 0 ail That s a haollon ol an alonn ln ohsluzbanoel o Musloal Inslzumenls Slandlng Waves m Resonanl Cavmes numb man In a llule you gel wavelenglhs oz whlch 1 a 5 L EA EA KA and so A A A 7 AL EL EL The even haxmonlcs axe nnlsslng Lecture 23 chlmes L341 Mnrllu mgquot cnlnnes axe Jusl slxuls wlnlcln axe ngglmg We can calculale llneu zesloung one my 7 iszy l And llns gwes us a fundamental Izequency Fox hlghez hazmomcs we would llke lo Just made Lu lo A Howevez we Wlll be systemahr quot L lneu ends and so llne effecllve lengln ol llne wave on llne slzul ls about llne lengln ol llne slzul Then equallon has llne lxend ngnl but ls off by a ado ol ol llne actual Izequency Tlns Tnls gwes us llne lmnno e 272 12 A Slmllaxly lo a slluls clamped on one end M e 27271Z damned n U 5A Stnngs mush dlnmumv n O 00 cooo We know the Ixequency of these waves and n Howevez these equations always gwes us a note whxch 1s attex than xeahty Th1 us because we have not taken mlo account the stz ness hke a stxut of the we Zaz MHZ Vi 72 7 p W k A W wuesowe can mfez Z Z z A MB TB Ma 3972 8t2 ael A aw Thus both the tension and the shuts of the stnng axe acting to pull astung back to e u hbuum the tension acting on the second denvatwe and the shuts on the ouxth denvatwe T wes u n22 32 1 MTMZi 92 In ths equation MT 1s 1 oval lheslzam Note that you want to decxease the anhazmomcny of you stung undex a lot of sham typically quite close to the yield sham Since yield slums axe axound 5 2 t en e estunat that a 30 cm stung wtth 0 5 mm xadlus has a Ath hazmomc whxch s shaxp by 2 pezcent Acoustic Fowex Multxpole theory Suppose we have a putsmg sphexe whose suuace expands and contmts at velocity U thh penod a and xadxus R Then the neat zone is of xadms es 7mm 21 2 55 In the near zone the pressure and displacement of air have different phase relationships than in the far zone Another way of de ning the near zone is that it is where the spheres perturbing in uence is smeared out conditions are roughly homogeneous inside the near zone Thus the energy ux across the near zone boundary is F N p7chS AP 2 N 76 C p S 5 p6 N 7 0 PO2 9 We can estimate the amplitude of the disturbance as E v lAVl P V0 N 39y47rR2U 3 7r 6 Thus the power emitted from a monopole is 13mm R4z2U2 1447W2P0 N T Dipoles can be constructed out of 2 out of phase pulsing spheres Most instruments are dipoles If these two spheres are separated by distance 1 then they mostly cancel but at the near zone radius mam along the axis of the dipoles you still get a little disturbance because one of the dipoles in closer than the other The pressure difference is AP N APmO r LOl l APmtmoZ neardipole APm 0 Tnear Thus the power from a dipole is d 2 Pdipole N Pmono lt 7 Tnear d 4 Pquad N Pmono Tnear Likewise a quadrupole is Lecture 26 Magnet ism 0 Suppose you have a solenoid which has a magnetic eld induced by current B H If you put a bar inside the solenoid then the magnetic eld inside the solenoid becomes BH47rM 56 where M is the magnetization which is de ned by M nm where n is the number density of magnetic dipoles and m is the dipole moment of those little dipoles 47rM BH 1 V xsusceptibility ymagmztic permeability If X lt 0 the material is diamagnetic if X gt 0 it is paramagnetic and if M gt 0 even when H 0 it is ferromagnetic it has spontaneous magnetism Diamagnetism All materials are diamagnetic The applied eld always modi es circulating electronic current to oppose the eld For most materials it is the electrons moving in their orbits which are important for this effect Consider an electron moving in an orbit of radius r in a eld B not H because the electron responds to all the eld Then in cgs units the magnetic moment is i A i Pemrrz i 5117 c 27rrc 26 Then the magnetic ux is I Bwrz and so the induced electromotive force is cdt 0 dt There is some sleight of hand here The radius of the electron orbit actually changes if you vary B Langevin s trick says that we can ignore this and just account for the change in velocity and it ll come out in the end rdB Then the perturbative force is d1 r medt 5 20 dt d1 7 5 dB dt 7 277150 dt Am aAv 2C and so using the above we have 7522 A AB m 4771562 On average you have electrons spinning in all directions If you vary a magnetic eld along some axis the electrons orbiting such that their induced eld reinforce B will slow down and the ones opposing will speed up and so a net M will be set up opposing B If X ltlt 1 then AB z AH so 75272 A 3 7 4771562 and then 47139 AM XE AH 7 47rnAm 7 AH 5272 747m 4771562 Then 71 N 313 and r N 3A2 so X N 73 10 5 If we were careful we would need to account for 1 a wide variety of frozen pitch angles 2 non circular orbits 3 sum over all electrons but 1 2 tend to cancel ln neutral gas at STP ngas N 15813 so X N 733375 N 73 10 s In general X is independent of T but not in plasmas Paramagnetism In some materials the permanent magnetic dipoles are not frozen in place and rotate to point along the B eld and thus reinforce it The energy of a dipole in an applied eld is U in I ichos0 Thus the total magnetic moment is M mcos0dn where dn is the differential number density of dipoles with angle 9 with the magnetic eld Then U dn 316 W 27139 sin 0d0 W 19 Which gives us M mcle cos0sin we 7r mB 27rclm e Wcosecos0s1n0d0 90 1 771B 27rclm ze Wzdz 71 Then the total 71 is n cle 27rsin0d0 and so 1 77717131 fil 25 W dz M mn 1 7 file kT Zdz Thls ls Langevms Funotlon Golng buk to B H1 10 ll m lt 1 then B N H and so an Mr M oth i s i m a lt o gt m Thls i025 mt bow oz leuomagnetlsm because M Wlll appeae on both sldes ol the equatlon and you ll get a txansoenolental lunotlo m u mm In slunllmltll 0 We can also estlmate X A7rM Awmzn Aamgmn H SkT Sk I leeaby 1 t 2 t u Thls glves us A742 104 Z 71 N N 10 2 X wane ave to be caxeful wlth thls calculatlon Sometlmes t As Thls ls about 100 tlmes m metals you only o m M want to use electzons whlch axe m the conductlon band only those at the top ol the Fezml ea n n 1 X lcT Wham 3M EF etals X olt 5 olt g so theze ls no tempexatuxe olee ln genezal by coollng matenals down Fox a x N 1 we 1K Thls leaols to the pzeohotlon that m neeol somethlng amund m pendence ou n e h e pazamagnellsm household magnet Wlth 1 Gauss In ozdez lo Feuomagnetlsm Fox some matenals the plus ol the eleotzons nd lt enexgellcalh Iavoxable 1 change lntezaotlon whlch ls caused by the lntezaotlon the spatlal n s to allg n du t an quot wavelunotlons ol the eleotxons Spm ahgnment tends to move the eleotxons Ianhez apaxt o average7 and this decreases the Coulomb interaction between them Then 2 104G 2Tesla 1 B N 47rM N m3 h 0 surface 0 T Note that this is insensitive to size This is because7 the total number of dipoles in an object goes as the volume 737 and dipole elds fall off as r3

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "I made $350 in just two days after posting my first study guide."

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.