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# Foundat Physics II PHYS 351

UO

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This 13 page Class Notes was uploaded by Isom Sawayn on Tuesday September 8, 2015. The Class Notes belongs to PHYS 351 at University of Oregon taught by Parthasarathy in Fall. Since its upload, it has received 73 views. For similar materials see /class/187255/phys-351-university-of-oregon in Physics 2 at University of Oregon.

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Date Created: 09/08/15

Physics 55l Problom Set 2 Solvifmls Hats jluz warren Just Me it hmmml Case 30 t has Wlw nim W Magnify WW7 W4 Xi itth 3 I U LWT5W T131quot 3 A rolling ball a As X gtO h gt 00 and as X gt 00 h gt 00 Also there is onl one value ofX for which Y dh 1 a a m W 7 1s Zero namely x0 From the rst observat10n or by look1ng at the second dx 2 x 217 derivative ofh it should be clear that hX0 is a local minimum These two facts together mean hX must look like h X You could also have deduced this just from the rst statement about the limits ofhX and the shape of the functions VX and X b An equ1l1bnurn po1nt 1s one at vh1ch the net force act1ng on an obect 1s Zero S1nce F E th1s 1s equivalent to the statement that an equilibrium point is one at which the spatial derivative of the potential energy function is Zero For our system the potential energy is simply the gravitational potential energy m g Therefore from a there is only one equilibrium point 23 c From above there is only one equilibrium point x0 The ball rolls back and forth about this point How do we determine the frequency of oscillation There are three reasons to think that a Taylor series expansion might help i We re asked to recall the lectures ofWeek 1 in which we discussed Why any oscillation is harmonic ii We re asked to consider small oscillations about equilibrium which begs for a Taylor series expansion Why iii Most importantly Our ball is oscillating in an unfainiliai potential energy landscape a Ux mg bx We could determine a differential equation describing its motion but I wouldn t have the slightest idea what its solution would be If only we could turn this Ux into a form for which we know the solution perhaps a form that looks like the oscillations ofa mass on a spring What is Ux for a mass on a spring Remembering it or 1230 Jime deriving i from F ikx we know that Ustg 12 k x2 where x is the displacement from equilibrium This is a simple quadratic polynomial A Taylor series is a polynomial expansion Eureka And so we expand In general as I sincerely hope you all know by now the expansion of Ux about xxo is l dZU 2 x x0 3 d 2 x x0 higher order terms Applying this x mm to our Ux expanded about our equilibrium point we note that the xixo term is zero since the derivative of Uis zero there and so W Uxod U dx xxn 1 CPU leU0 T x MJZN 2 dx xxn For small oscillations we can neglect the higher order terms So this looks just like USpnng 1 2 k x2 since it s just a constant times a quadratic displacement You may be worried that there is a constant term Uxo in the above expression that does not show up in Uspnng but this doesn t matter 7 one can arbitrarily dU add or subtract a constant from any potential energy function since F d these constants don t x translate into a physically meaningful force You may be worried that USprmg 12 k x2 while our expression involves xixo But think about what these symbols mean 7 for the spring x is the displacement from equilibrium while for our ball xixo is the displacement from equilibrium Hence our xixo and the springs x map onto the same physical concept Therefore our rolling ball looks just like a massonaspring with an effective spring constant WU a 3W 3 k dxl ExpliCitly k 52 mg 4 6123 0 4 mg Therefore the angular frequency m k m x mm and period T 27103 are given by 46123 215 ais 1 27239 The last two expressions are equivalent either is fine 3g2b53 bss 3g Let s check the dimensions h and x have dimensions oflength so from the statement ofhx it must be 1 th t L32 d 1 Th f eriod L 2 a a an ere ore p LTZ T as it should be Anathcr approachquot LQ 713E1be r 1 F 331 7M2 MHAUUVX X r 5 mm vcm 3 wg h lxv x if g rthm u LLSMFJmw oam maglh vr upms ni quotthhegmlrbnfmm Pam LPoM nggy XA M9mgf ML 7 m t L5 Etghf1zv12lt gummy wmm39 2J1 2 117quot3 O U a 39 T Q a 3L0 m x4 me 1 17 X106 mquotSquot39 WW 303 m Law5 39 i xo Prof Raghuveer Paxthasarathy University of Oregon Fall 2007 Based on notes from Sept 2006 Physics 351 Dimensional Analysis UNITSAND DIMENSIONS All physical quantities have units with which they are measured 7 kilograms acres Newtons etc Some units are somewhat arbitrarily fundamental eg the units ofmass length and time Other derived units can be expressed in terms of these For example the SI units kg m and 5 kilograms meters and seconds are fundaInenml representing the physical quantities mass 2 length L and time The unit of force the Newton is a derived unit 1 N 1 kg 52 The quantities L M and Tare a useful though not unique set of properties on which to base our fundaInental units The dimension ofa physical quantity A often written A is the function of the fundaInental set ofproperties that corresponds to the units ofA Despite the complexity of the preceding sentence the dimension is easy to understand The dimensions of force for example are P ML T2 It is of course important that any physical relation eg an equation have the same dimensions on each side Mass must equal mass for example not mass gtlt length On your problem sets we ll take offmore points for dimensionally incorrect answers than for numerically incorrect answersl More interestingly dimensional relations can be used to mammt correct physical relationships DIMENSIONAL ANALYSIS a skill that is very useful in physics and other sciences and which is rarely mught The preliminary step is to figure out what the dimensions ofa quantity are Any correct expression can guide you Lefs consider energy B What is Recall that kinetic energy Ek 12 771 122 where m is mass and 12 is velocity Therefore MLZ T Z since M and 12 LT Note that numerical factors eg 12 are irrelevant to the dimensions The essence of dimensional analysis is relating the dimensions of the quantity ofinterest to the dimensions of the paraIneters it might depend on I ll explain this with some exaInples 1 What is the period of a pendulum Consider a mass m hanging at the end ofa pendulum string of length Z What is its period of oscillation Of course this is very relevant to this course and of course we all know the answer from firstiyear mechanics but lefs harness dimensional analysis The period 5 has dimensions of time T Without knowing anything we might suppose that E can depend lefs say on Z 771 and the gravitational acceleration g So I fumtz39on ofq m where fumtz39on of is just some algebraic expression 7 products ofpowers of the arguments I VF 774b Lgc where a b and c are exponents we need to determine The dimensions Z L M LT Z and I T So the above expression becomes T L1 Mb LT Z c There is no 2 on the left side of this equation so b must be equal to zero This is our first interesting observation The only a and c that will work think about this are a 12 and c 712 Therefore 1 NZ and 1 I dimensionless constant l Aside from a numerical constant which happens to be 211 by the way dimensional analysis has given us the correct expression for the period ofa pendulum and even told us that the pendulum mass doesn t matter Ofcourse we had to think a little bit at the beginning supplying a physically reasonable guess for the parameters on which 5 depends 2 How energetic is an atomic bomb blast and how do you determine this if the answer is a secret Next a less trivial example Following the development of atomic weapons in the 1940 s the energy E released by an atomic bomb blast was maintained as classi ed secret information sza magazine published photos of the blasts showing the expanding reball of radius r at a series of times following detonation see Figure page 4 From these publicly available photos G 1 Taylor a very clever physicist figured out the value ofE Here s how Taylor reasoned that r the blast radius should depend on t of course E and the density of air It might depend on other parameters such as the atmospheric pressure but for reasons we ll just briefly touch on Taylor concluded that these are irrelevant The blast pressure for example greatly exceeds the ambient atmospheric pressure and so the atmospheric pressure can t play a significant role in the blast s properties The task now is simply to find by dimensional analysis an expression relating r t E and p r El pb Me or 1quot const El pb tc The dimensions r L MLZ T Z p ML 3 mass volume and T Therefore our dimensional relation becomes L MLZ T39Z a ML s b TC Maw L2 lr3b Tara Examining this 0 There are no M s on the left side so ab 0 ie a 7b 0 There are no T s on the left side so 72aC 0 ie c 2a 0 There is 1 power ofL on the left side so 2a73b 1 Therefore combining lines 3 and 1 5a 1 ie a 15 b 715 and c 25 1 Et2 E Therefore rt X dimensionless consmnt p Given the density of air one can read values of rand 1 from the photos and therehj 69611715113 1 The best way to do this is to plot logr vs logl see Figure page 4 The slope should from the above expression be 25 7 measuring it provides verification that the dimensional analysis approach and the choice of dependent paraIneters is reasonable The intercept of the logilog plot gives E Recall that loglB logl logB and logA v XlogQl Note that dimensional climb5239s mmmt reveal anything ahout imemz39onlem mmtzmtx in physical expressions Typically however these constants are things like 211 etc 7 numbers of order 1 FURTHER READING On dimensional analysis and related issues see falling by G l Barenblatt Cambridge Univ Press Cambridge 2003 On G 1 Taylor see eg Modem lamimlphyxm through the ware ofG I Tag01quot M P Brenner amp H A Stone Physim Today May 2000 Figure N ext Page From Barenblatt 2003 see notes p 6 qmdow ave fro Quiescent air Air in motion ground I tvttre l 2 Photograph of the reball nf the atomic explmimt in New Mexico 11 1 15 ms cnnttrmtng in general the spherical symmen ol39tlie gas mminn T iUl39 1950b 1961 1011 quotr Mm 105 s 74 73 72 10ng Figure 03 Lttgurithmic plot ml the rebuil ratlith showing thtttrlr s to the timer t39l uylur WSOh I963 nportim ml Prof Raghuveer Parthasarathy University of Oregon Fall 2007 Physics 351 Nov 19 2007 The Energy of a Vibrating String We have derived in two different ways the general expression describing a string pinned at the ends vibrating in normal mode 71 yx I cosant y or explicitly writing the function n yx t Aquot s1n x cosant n What is the energy ofthe vibrating string We know that in general it has kinetic energy since it s moving and potential energy since the stringis tense and deformed The total energy E KEPE is constant We re ignoring damping With damping we know that 2 E0677 Since E is constant we can make our lives easier by picking a simple con guration in which to evaluate B We realize that when the stringis straight ie y 0 0 the kinetic energy is nonzero since the string is moving past this at configuration and 0 the potential energy is zero since the stringis not deformed Therefore at this particular instant in time E KE and so it suffices to calculate the kinetic energy What is this instance in time It s determined by ant since the cosine factor above is zero then gt 1 The kinetic energy for a segment of string located between x and x dx is just the usual quot mvz 2 1 for this segment which is Eydxyz where IL is the mass density From the above expression 717139 yx t anAn s1n ijs1nant by At our special pOint In time yx t anAn Sin the klnetlc energy of our small plece 1 1 of stringis Eydxy2 EydxltmnAw s1nn7rxL2 The total kinetic energyis simply L L L 1 rm 1 mr KE quotdKEquot 2142 Sin2 x m2 2 sin2 x x g g 2 a n n L 2 a quotA J L which is a standard integral you ve seen recently so mr sin2mr 1 L 1 KE szz mzAZL 2 4 by 1 Therefore the total energy at all times is E z szZL 2 2 rm 139 7239 A71 139 2 Inserting our expression for the normal mode frequenCies aquot T yields E YT u Note that E N n2 For a given amplitude more energy is required to excite higher modes A few notes on Matrix Algebra Raghuveer Parthasarathy Department of Phy cs University of Oregon si for Physics 351 Foundations of Physics Dated November 5 2007 Well very brie y explain a few aspects of linear alge bra This discussion is suf cient for the problems encoun tered in your homework assignments To learn more see any linear algebra textbook eg Howard Anton Elementary Linear Algebra 7th E51 John Wiley and Sons New York 1994 MATRICES A matrix is a rectangular array of numbers The ma trix A below is a 3 X 3 matrix with elements aij where the indices 239 and j run from 1 to 3 an an 03913 a21 a22 a23 A 1 a31 a32 ass A The matrix B below is a 3 X 1 matrix B 1221 2 The product AB is C also a 3 X1 matrix The elements an of C are given by 3 Cil E aijbjly j1 where 2 indicates a sum In other words each element of C is the sum of the products of a row of A with a column of B Here B only has one column alibii M21121 13531 21511 22521 23531 3 agibii M21121 33531 CAB This is shorthand for three linear equations alibii M21121 13531 011 4 21511 22521 23531 C21 5 agibii 32521 33531 631 6 When dealing with N coupled oscillators well have matrix equations that look like AB 0 where 0 is a N X l array of zeros The elements of A will involve the oscillation frequencies w and the elements of B will involve the amplitudes Welll want to nd the condi tions such that the equation has a solution other than the trivial one B 0 A fundamental theorem of lin ear algebra states that a nontrivial solution exists if and only if the determinant of A is zero In the next section well look at how to calculate a determinant Exercises Don t turn these in 1 Let 1 3 6 A 3 72 2 7 2 1 5 2 B 71 8 0 Calculate C AB Answer 71 C 8 9 3 2 Write the following system of equations as a matrix equation 10 11 1 70101 10202 10103 0 2w1C1 2w1C3 w2C1 102639 2w1C3 0 O Answer w1 w2 w1 C1 2101 0 2101 C2 0 13 w2 w2 2101 C3 II DETERMINANTS The determinant of the 2 X 2 matrix A an an 14 a21 a22 detA anagg 7 algagli 15 Note that this is one diagonal minus the other The determinant of the 3 X 3 matrix an a12 dis A a21 a22 a23 16 Mi a32 ass detA a11a22a33 a23a32 17 7ai2a21ass 7 a2sa31 18 a13a21a32 a22a31 19 This looks complicated7 but note that we took each ele ment of the rst row of A we could have used any row and multiplied it by the determinant of the 2 X 2 matrix that is formed by excluding the row and column of the element we7re consideringi We then add all these terms7 ipping the sign ii for each alternate element In other words detA a11det 20 03921 03923 ialgdet an ass 21 a a a13det 21 22 Mi a32 Exercises Don t turn these in 1 Consider a rotation matrix77 R 5213 12 Show that detR ll 2 Evaluate the determinant of the array w1 w2 w1 W 2101 0 2101 i w2 w2 2101 Answer 22 23 24 detW wl 72101102 7 112 4w 7 2101102 25 w12w1w2 74w w2 210110 26 27

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