Class Note for MATH 254 at UA
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Date Created: 02/06/15
MATH 254 Lectures 162 1 0 2nd order7 linear7 constant coef cient equations With time dependent forcing 0 Method of undetermined coef cients 0 Use of SYSTEMS o Forced Vibrations 0 Applications to massspring systems and to RLC circuits 0 Notion of resonance 0 Notion of beating 0 Method of variation of parameters Consider ode s of the form My magi qlttgty glttgt 6 1 2 If there is a coefficient in front of divide it through the equation Let ot7 Lt7 gt7 be continuous over some interval 1 The general solution of 61 in I is the sum of o the general solution of My magi My o 62gt 0 any specific or particular solution of 61 The former is often called the complementary solution the latter is the particular solution Let Y1t7 Y2t be any two solutions of 61 Then easy to prove Y1t 7 Y2t satisfies 6 Therefore7 we may write the difference between any two solutions of 611 as Ayl t Byg t7 where yl t and yg t are two linearly independent solutions of 62 Now let Yt be the general solution of 61 This means that Yt satisfies 611 for all t E I and satisfies two arbitrarily assigned initial conditions Yt0 Y0 Y t0 YO at some point to E 1 Let Y2t be any particular solution of 61 It does not have to satisfy the initial conditions Yt0 Y0 Y t0 YOi Then we may write YO Y2 t Amt By2 t and we choose A7 B to satisfy Y0 32050 Ay1t0 By2t0 Y6 YWO Ayl o ByWo Since y1t and 92t are linearly independent on 1 these two linear algebraic equations can be solved to give unique values of A and Bi Recall yhyg liii Wy1y2 My 7 ygy l 0 In what is to come7 we write ypt for Y2t7 the particular solution of 61 We often denote the complementary solution7 6 27 by yo i Example 1 1 1 7 7 7 7 y to 729 t7 ya o m 10 Check that t are lii solutions of y y7 ivy 0 on both the intervals 700 0 and 0 00 excluding t 01 Try the particular solution yp A1t A2 A1 A2 constanti 0 A1 7 A1t 142 Equate powers oft this doesnlt work Try the particular solution yp Altloge t A275 A3 9 AllogetA1A2 y A110get A1 A2 A1t10ge t A2t As A1 5142 arbitrary A3 01 yp tlogetJr St is a particular solution the general solution is y tloge t A 3t So y10A3B 719 7 19 yl10logelA3 x AWFELB and 1 19 19 t itl t itiii y2 oge 2 2 Note that the solution becomes discontinuous at t 0 where pt qt and gt are all discontinuousi Exercise y 5y 4y 6 7 90 07 W0 110 Step 1 Find the general solution of y By 4y 01 Step 2 Try yp Ale substitute into y 5y7 4yp e and determine A11 Answer yt 7amp6quot e 4 eti Exercise 9 y 17 90 07 40 1 Answer yt 1 sin ti How to nd yCt and ypt In general this is dif cult and in most cases approximations for example expand ing the solution in series form numerical approximations of solutions representa tions of solutions in integral form must be made But for a special and important class of cases we can calculate yC t and ypt explicitlyi These are the cases where pt and qt are constants and gt E C where C is the class of functions which are sums of terms which are of the form taem cos wt or taem sin 7t where a is a nonnegative integer and 7 are real The Method of Undetermined Coef cients y py qy 9 63 Step 1 Determine two lli solutions y1t yg t of y py 4y 0 64 We know that yl and yg are in the class C Step 2 Derive the auxiliary set of lli functions gt obtained from gt by repeated differentiation This set is nite because gt is in the class C Example gt t2 gtt2t1l gt tsint gt t sint7tcos t7sint7cos it because gt mintyt sinttcost g t 2 costitsint g t 73 sint 7 tcos t and so on All derivatives can be written as a linear combination of members of 96 Step 3 Find out if any member of the auxiliary set solves the homogeneous equation 64 if not7 do not modify the auxiliary set if one does7 multiply the auxiliary set by t where m is the lowest integer such that no member of the new auxiliary set solves 64 This is the modi ed auxiliary set Step 4 Seek yp t in the form of a linear combination of members of the auxiliary or7 if it has been modi ed the modi ed auxiliary setl Substitute this expression into 63 and determine the unknown coefficients by using the linear independence of the members of the auxiliary set Remark lf gt 91t 92 it eg gt t2 tsint7 determine ypt for 91t and 920 separately and add if 706 satis es y 109 19 05 for i 17 27 1 2 then yp 917 gt 917 satls es y py W 91 92 Suppose that 9W 1t7 2t A A WW Then let yp A1 1t l l l AMMU Substitute into 63 and equate coefficients of all the linearly independent functions We will discuss why these will be enough information to determine the 7 constants A1 i Ari You might give some thought to this and try to construct a proof that this must be so But7 rst7 you should do lots of examples Worked Examples see HW on NSS pg 2117 s 17779713729 The best way to learn the method is to do lots of examples 1 y 7 2y 7 3y 7 362 ND 7 07y 0 7 1 ii y 7 2y 7 3y 7 73wquot ND 7 MO 7 0 iii y 9y Fe 6 y0 y 0 0 iv y may coswt7 w we y0 yO 0 V y may cosw0t7 y0 yO 0 vi y W W 7 iFCOSWL 02 lt 4qy0 7 ayy 0 7 5 vii y 7 t3 ND 7 MO 7 1 viii y y 7 t6quot ND 7 MO 7 0 ix y y 7 tsint y0 7 y 0 7 0 x y Zy By 4equot cos2t7 y01y 0 0 Example y 7 2y 7 3y 3am y0 0y 0 1 First nd yC i H Step 1 y any 7 27 7 3 0 7 7 3717 so yC AegtBe4i Now nd ypi H Step 2 Repeated differentiation of gt 362 simply gives multiples of 62 so 9W 7 62 Step 3 The auxiliary set does not need modi cationi Step 4 yp Ale yg 7 2y7 7 3 Semi So7 4A1 7 4A1 7 3A162 362 7 A1 71 Remark You could have added any solution of y 7 2y 7 3y 0 to yp and this term would disappear because y l py l qyl 0 The general solution is yt A6 Bequot 7 62 Using the initial conditions y00AB7landyO13A7B72giveA1B0sowehave the particular solution y 63 7 62 Example ii y 7 2y 7 3y 73mquot y0 y 0 0 Step 1 yC Aeg Be h Step 2 gt 7376quot gt te e t Step 3 The modi ed auxiliary set is gt t2equottequoti Remark If you had sought solutions in the form yp Altaquot Agequot7 A2 would disappear because 671 7 26quot 7 36quot 0 Step 4 Let yp A1t2equot Agtequot7 y 7A1t2equot 2A1 7 A2tequot Agequot7 y A1t2equot 74A1t A2736quot 2A1 7 2A2equoti Then y 7 2y 7 3yp 7376quot 7t2equotA1 2A1 7 3A1 tequot74A1 A2 7 4A1 2A2 7 3A2 equot2A1 7 2A2 7 2A2 7376quot Vti Now t2equottequotequot are liii Therefore7 the coef cients of each of these expressions Inust vanish7 so A1 2A1 7 3A1 07 this is true automatically 78A1 7 73 7 A1 7 38 2A1 7 4A2 7 0 7 A2 7 316 yt 1463th Bequot 3726quot t t y00AB 7 3 y07073A7B A B There is a unique solution 3 7t 3 2 7t 3 7t 7 7t 7t i 646 8 e 16 e 3 t if 3 y 646 Example iii Do yourself gm 7 at 7 am 72gt em yt Acos3t Bsin3t 7 7 it W12 63 g y00 0A A7i y070707337 7B7 i Example iv yC Acos wot B sinoJot7 yp A1 coswt A2 sinwti y wgyp coswt A1 320 A2 0i yA coswot B sinwot cos wt 0 yltogtyltogt707A76i7WBo coswt 7 cos wot tsinwot A so yt W wlewo yt WUse LlHopitalls Rule Example V y wgy coswot y0 y 0 0 Step 1 Solve y wgy 0 y eTE r2 41 0 r iiwo yC Acosw0tBsinw0ti Step 2 gt coswot gt cosw0t7sinw0t Step 3 The modi ed auxiliary set is gt t cosw0t7tsinw0t Step 4 Let yp Altcoswot Agtsinwot7 so that y A1 cos wot 7 Alwot sinwot A2 sinwot Agwot cos wot7 y 72A1w0 sinwot 7 AWE cos wot 2A2w0 cos wot 7 AWE sinwoti y wgyp cos wot 72A1w0 sinwot 2A2w0 coswot coswoti But sinwot and cos wot are linearly independent7 so A1 07 A2 yt Acos wot B sinwot it sinwot 2mg y00 A0 y 00 B0i Therefore7 yt t sinwoti Compare the previous exercise Examine Vi This is a very important example Work through it carefully y W W iFCOSWtL 02 lt 447 y0 a y 0 BA Step 1 The solution of the homogeneous equation y py 4y 0 for p2 lt 4g is 2 2 yC e g Acosx q7 pZtBsinx q7 pjtgti Step 2 gt tiosth7 so gt coswt7sinwti 7 Step 3 The auxiliary set does not need to be changed Step 4 Let yp A1 coswt A2 sinwt7 y 7A1w sinwt Agw coswt7 y 7Alw2 cos wt 7 Agw2 sinwti Then y 1091er qyp cos wt7 so 7Alw2 Alq cos wt Agpw cos wt 7 Alpw sinwt A24 7 A2w2 sinwt A10 402 PwA2 E Aipw A2 WQ 0 7 F ti 7 F p A vaArmWM p yt e pt2 ltAcos q72tBsin q72tgt E g7mwcoswt mpsinwt 10 t F 7 cos wt7 m A and B are determined from the two equations F M 7 mw A im a w lt 7mwgt2 mp 2 Li EL 5772A q 4Bw77nw2mp For long times7 gt 7 00 so 675 lt17 m 7 wt 7 com 7 as 615 El n 1 m Where 1 1 7 mw mp WW7c0sq i7 slnqt ZW ZM 6 6 Interpretation of Vi and its solutions 65 and 66 The response of the darnped oscillator system to periodic here sinusoidal forcing is described by the equation W 7 cosw 7 ab Where is the ampli cation factor and qt is the phase lag Draw the graph of against w for m q 1 and p 1021 We will study this example closelyi Example y t3 y0 y 0 11 By straightforward integration7 y At Bi y0 1 B 1 yO 1 A 17 so we have yt 1 t by the method of undetermined coef cients Step 1 The homogeneous equation is y 0 This gives y 6quot with 7 2 0 which has repeated root 07 so yC AtBeot AtBl Step 2 gt t3 so gt t3t2t1i Step 3 The modi ed auxiliary set is gt t57t47t37t2 We cannot just multiply by t because the element t would satisfy y 0 Step 4 Let yp A1t5 AM Agts A4t2 y 5A1t4 4A2t3 3Agt2 2A4t y 20A1t3 12A2t2 6A3t 2A4 t3 1 141 7A2A3A40 Hence we have yt At B 271079 as we found above by direct integration Exercise y y t6quot y0 y 0 0 Answer yt 7 cost t Dequot Exercise y y tsint y0 y 0 0 Answer yt 7t2 cos it it sin 7 Exercise y 2y 5y 4equotcos 2t y0 17 yO 0 Answer yt 6quot cos 2t mquot sin 27 Exercise Solve 327 Ay sin7rz7 y0 07 yl 0 For what values of A is there a unique solution For what values of A is there no solution For what values of A is there an in nite number of solutions Message BVP s are fundamentally different from lVP sl 10 Forced Vibrations The response ofa weakly damped massspring system to periodic forcing Models 1 Massspring system under the in uences of gravity7 friction and external forcing d2z 1 dz k 7 t dt2 m dt m1 7 mcosw Mass In kgsl friction force 712 tension or spring force 7161 external force Fcos wt Note Gravity disappears from equation 2 Massspring With damping and external forcing dQI dz my HIE k1 Fcoswt dim id 1amp5 t dt2 m dt m 7 mcosw 11 3i Pendulumi Newton7s second law in angular direction d d9 d9 mglg 75 7 mg s1nt9 Fti g Where 1 is the velocity of the bob in angular direction 7 dz is friction 7mg sin 9 is gravity and Ft is the applied external force in angular direction In the direction along the pendulum the tension in the rod which is absorbed at the hinge O is balanced by the gravity and centrifugal forces That is T mg cost9 mld 9dt2i d2 9 5 d6 9 F t 42 EEYSIDL W 67 is second order nonlinear and nonautonomous if F depends on t For small oscillations however sint9 is well approximated by 9 and then 67 is linear 6 7 4 Electrical Circuitsi See Lectures 789 1 have already told you how to derive these equations You may also consult NSS 5i5i For now we will just deal with the equation Consider the RLC circuit pictured to which the ernf voltage Vt is applied Then the charge satis es d2Q dQ 1 L RE 5Q 7 Wt 68 and the current satis es in Rdi 1 1dV faEFZW 69 L inductance R resistance C capacitance 12 lf Vt V0 coswt then the RHS of 68 is 0 7 t L cosw Remark Because we have already used 4 as a coef cient in 63 we will use cap 4 or Q to denote the charge on the capacitor of the RLC circuiti Each of these models is described by a linear second order constant coef cient nonhomogeneous odei dQI dz F0 Wpgqz gcoswt 610 where the external force is periodic with frequency am In the massspring system p 12721 4 km F0 is the magnitude of the force and m is the mass In the electiic circuit p RL q lLC VBL Recall from exercise vi that for p2 lt 4g the general solution of 610 is a zt 10t 11 t where b 10 t e ptQ ltA cos Wt sin Mt cl zpt coswt 7 611 d M er cosqb sm Today we are going to look at this solution study it investigate how it changes as we change the parameters p q and w and learn from it Observations l The general solution consists of two parts the complementary function 10t in which the unknown constants A and B are determined by the initial conditions and the particular solution zpti 2 As t A 00 10t decays to zero on a time scale proportional to pZ 1i It is called the transient part of the solution because while very important for small t and for matching the initial conditions it eventually dies away 3 As t A 00 zt A zpti The long term response of the massspring pendulum electric circuit to periodic forcing is given by 1pc coswt 7 aa 612 13 4 5 6 7 For the electric circuit7 V V0 cos wt and Qpt coswt 7 7 Vo 7 Q17 MW 0050 1 iwL R 1 iwL2Rzlsln 1 iwL2R2 ln observation 9 below7 we discuss the long term current of the RLC circuiti We are interested in those cases when the damping is small that is not only is p2 lt 4g but p lt Z iiei 102 ltlt or that the decay rate 102 of the transient is a lot less than its natural frequency q 7 1024 which is almost equal to In these cases7 the natural oscillations of the spring 10t has frequency and is weakly damped7 whereas the forced oscillation zpt has the same frequency w as the external force It is natural to ask When is the response to the forcing maximum Clearly this occurs for the maximum value of m 1 Zw 7 quotquot1 7 613 cosqb W mp 2 which occurs when w i namely when the forcing frequency is the same as the natural frequency of the oscillator with zero dampingi The strength of the response when if q is lZw 1mp7 which is very large when mp is very small Recall from exercises iv as we A w and v7 the response of the un damped oscillator qr 0 to the external forcing Fem coswt when q w was dQI F0 t z i cosw dt2 q m giving F0 t 1t Acoswt Bsinwt m w sinwt namely zt grows like t The effect of damping7 no matter how weak is to arrest this unbounded behavior and give rise to a bounded response 117 t with an amplitude which is proportional to the inverse of the damping We call this strong response of an oscillator which is brought into contact with another oscillator of the same frequency by the name resonance Observe that Zw has the dimensions or units of friction or resistance To see this look at 1 1 ZE 7 mp For mass spring systems lmp 1 where the friction force was 12 electric circuits7 Z 1max For 8 9 Where the resistance of the circuit is R and the voltage drop due to resistance is Ri Ohm7s LaWi Let us examine response When the circuit or massspring system is tuned ie When 4 is chosen to be w Then cosqb 0 sinqb 1 Z mp b so that 45 7r2 and zpt sinwt or E dt So in a forced massspring system Which is tuned so that the natural fre quency7 of the undamped unforced system is chosen to be the forcing b F0 cos wti frequency w the massspring system behaves as if the 3117 and 41 terms are ignored they exactly cancel each other out and the remaining force is the instantaneous friction and the external force 721227 12 k1 F0 coswti When Is is tuned so that km if then the forced response F0 zp I E slnwt balancesl 2 mddtip kzp ie it makes it zero leaving friction to balance the external eld d 12 F0 coswt Note that if the long term charge Qpt is given by 613 then the long term current ip is given by 7V ip Two sinwt 7 so 614 At resonance namely When w g0 g and since sinwt 7 7 sin 7 wt 7 cos wt 0 z 7 coswt 6 15 17 R lt gt This is very interesting because it is just like Ohm s law Rip Vt for a time dependent voltagei Look at equation 68 at the resonance Where J JLT 2 d Q L RE On resonance the rst and third terms on the LHS balance iiei cancel leaving dQ RE Ri V0 coswt 617 So that at each instant of time the ohmic resistance of the circuit balances the external voltagei Since Zw Which has the units of the resistance R is like a resistance With terms added to account for the lack of tuning Zw 15 dQ Vocoswt 6 16 is called the impedance of the RLC circuit R the resistance and i 7 wL the reactancei 10 Recall again the graph of lZw Which is the amplitude of the response of the weakly damped oscillator to external forcing For p very small 1 N 0 for q f w2 M N for L w2 The response of the oscillator as a function of w or if you x w and vary 4 as a function of q is very sensitive to w the smaller p is For p very very small if w Up a 123 the response amplitude decays very quickly as a is increased The Width of the response is inversely proportional to the height The latter is called the Q7 of the circuit or oscillator 11 Analogy between mechanical and electrical systems Mechanical massspring damping and forcing Electrical RLC circuit in series b In Ft Displacement zt Velocity Vt Mass m frictiondamping 1 spring constant k External force Ft 2 LR c2vlttgt charge current Q it inductance L resistance R 71 1 capacitance 5 External voltage Vt 16 12 Beating between oscillators The notion of amplitude modulation Let us return to exercise iv d2 ft wgz coswt 618 10 dim 0 Solution t t cosw icoswe I t ltgt mg 7012 Let us imagine that w is close to we and write7 w woiw 7 woiw 2 2 7 we 7 and then rewrite zt as 2sinw07wt wew t 2 39 t 619 z Mg 7 if sin 2 Now if we m w so that we 7 w we w 2 2 the factor sin 0 in z it changes much faster than the factor sin 07 We see 2 2 this when we use SYSTEMS to plot zt with we l and w 0in ln fact7 the response of the oscillator to the forcing frequency which is almost in resonance is an amplitude modulation of the average frequency 2 m0 zt 72 gt X sin Log 75 7 Slowly varying X We call the fast frequency amplitude the carrier frequency For tsmalL weith weiw 2 N 2 t wow sin we w 2 sin zt m There is rapid resonant growth and the amplitude grows like ti However by the time L327 m assume we gt w7 the fact that the frequencies of the natural oscillation 17 and the forcing are slightly different has registered The amplitude reaches its maximum O For g lt t lt 7r7 the phase of the forcing is such that it is opposing the motion zt and so zt decreases back to zero at we 7 w 2 Then the cycle repeatsi This phenomenon is know as beating We will talk about the beating of oscillators during the lecturer t7ri I will also tell you about the phenomenon of synchronization in which a nonlinear oscillator imagine the restoring force to be proportional to kz 7 713 can adjust its amplitude so as to tune its frequency which now depends on its amplitude to match that of the forcing frequency This phenomenon was rst observed by Swiss clockmakers who noticed that clocks on the same wall which would act as a coupling between the clocks would synchronize iiei keep exactly the same time Method of Variation of Parameters Recall the method of how to nd particular solutions yp t of d2y Llyl W where pt and qt were constant and gt was a linear combination of terms each of which had the form of a product of a polynomial7 an exponential and a sinusoidal factori Namely7 gt was a linear combination of terms of the form taem cos wt or taem sin yti 6 21 magi qlttgty W 620 The method of undetermined coef cients was very useful because it was reasonably easy to implement But what happens when pt7 qt are not constants of gt is not of the required form Here I am going to show you a very general method which may be used if you happen to know two liii solutions y1t7 yg t of the homogeneous equation d2y L 7 y M In fact7 all you need to know is one solution yl t of 623 because by reduction of order lectures 1116 you can nd the second magi qlttgty or 622 I will rst apply the method to the rst order equation dy 7 a WM i W Let y1t satisfy pty 0 Then set ypt y1tvt and nd d1 dyl 7 91 11E POer 7 90 18 OI d1 1 E 905 Thus 1 7 9 5 S v 7 913d mt M f M ds 915 is a particular solution Note that yl t exp lt7 ft ptdtgt so that the particular solution found here agrees exactly with what you would have found by using an integrating factori Let y1t7 yg t be two liii solutions of 622 Set 91205 yi 7311105 y2tv2 t 623 and differentiating nd y y1v1 y2v yivl y v2 624 Now since we introduced two unknown functions 111t7 v2t in place of one un known ypt7 we are at liberty to choose one relation between v1 and v2 for our convenience We will choose yivi 9211 0 625 so that now y y lvl y vgi 6 26 Differentiate to nd 7 yp 7 91111 92112 9101 2112 6 27 Substitute 6237 6 267 6 277 in 623 to nd 9101 112 yivi y v 1075 91111 112 qty1v1 y2v2 975 But y py qyi 07 1727 so that 9174 y v W 628 Now7 6 25 and 628 are two equations for the derivatives 11 and v of the un known functions 111t7 v2 We find7 v t 79ty2t 9091 t 1 Wlt91792gt 7 Wy17y27 where Wy1y2 yly 7 y lyg constant exp lt7 ft pt dtgti Thus 1 10 7 ft 95 w ffgzd57 3902 t ft 95 Wyff2d37 mt ma 178 WEf2ds y2lttgt ms 2st 6 30 v t 6 29 Example Find the general solution of t2y 7 tt mg t 2y 23 tgt 0 19 Writing this in the form 620 t 2 t 2 7 y t y t2 y2ttgt0i From an earlier lecture7 we know that 9105 757 9205 tet Set yp t tvl te v2 ypt v1 t le v2 tv l te v i Choose tv l te v 07 631 so that now y v1 t Delvgi Then y 11 t 2e v2 t Delv i Substitute yp y77 yz in equation t 2 t v1 t l t 02 Check that the coef cients of v1t 22 tt 2t2 and v2t 2e 7 t lt 2e t tt 2e t2 are zero We nd t2 t2 11 t 2etv2 t Deli 7 tvl te vg 2t 11 t Deli 2t 63 Solving 6317 6327 to nd 0 te t 2t t1e 1 2t 7 7 7t v1 7 t te 2 v2 t te 26 1 tle 1 tle v1 72tcl7 v2 72 t02i ypt t72t cl te 72 quot 02 7272 cl 7 2t te The last two terms are solutions of the homogeneous equation and Will automati cally be included in the complementary functions yC t7 so we can take ypa 72 Check H4 7 tt 274t t 272t2 2t General solution W Mt yp At Ate 72 Exercises Some of these may be done using the method of undetermined coef cients i y y coswt7 if f 1 ii y y cos t 20 iii y y tant7 0 lt t lt 7r2 iv to ty 7 y ea tgt 0 21 TRIAL EXAM 2 To help you prepare for Exam 2 Work in groups Should take 45 hours Qili Linear independence 1i and linear dependence lid of functions on inter vals In What follows7 I Will give you a set of functions and an interval ll You have to say Whether they are lii or lid on that interval Function set lnterval lii or lid 1 1 t R ml 711 ii 1 t R ml 01 iii sin t7 tsint 07 7r iv 62 tam t262 7007 00 v 17 6 7 5 26 te 70000 vi 5 26 6 7 te 7007 00 vii tatwm 711 viii t3 t2 Rm 01 sin t7 cos t7 sin 2t 70000 On i7 set k1 k2 1cth 1mm 0 for all t 6 711 22 Choose t0 k10 a k ff0 Choose Add t 70 k3 0 k4 7 2k 72k t 0 k20i lt 4 2 k30 k1k2k3k40 WU WWW z positive t t2 ml t2 1 2t Qltl2t 0 2 2 0 0 0 det 0i OOOH So W0fortgt0 W0fortlt0 Wnot de ned att0i We can draw no conclusion from W test in this case Q2 The ode7 unit a1 a0 7 00 lt t lt 00 has associated With it the characteristic eqni an a1quot a0 0i Below7 I give the roots of You give the general solution of 23 Roots of General solni of n 5li7ii7i 1t Ale A2 cos t A3 sint A4t cost A5t sint n 7001712i712i7 717 2i 71 7 2239 n 40000 n10000112 awful 71i 1721 Q3 Solve the following initial value problems i ii iii 1V V vi Vii viii ix xi d2z dz dz W2E5IO 10 1E0 Oi 22 z 0 ma L d2 aw 2 0 54z0 1010gt0A T 44z0 10 1o o 24z0 10 1o o fo 77 071017di 05 1234i 127 02 9011 0101d3 gt 0 ill 02 9011 cosgt 10 d3 0 317 71 te 10 170 0 e 2 z ea zlt0gt 1 12 1ng oi es257 zlt0gt9oi 24 Qi4ai A very light spring its extension under its own right is negligible hangs from a hinge Fig A A weight of 1 kg is added to the pan Fig B and the spring stretches by 20cmsi The pan is stretched a further 4 cmsi Fig C and then is released With zero velocityi If g gravity is taken to be lOmsec2 and if the friction force experienced by the spring is numerically equal to 2 in Newtons nd the equation Which zt satis esi What are the initial conditions 10 0 Express the solution zt Ce quot cosbt 7 go and determine C a b and gal A B C 20 cm i 4cm 940 mg Q4bi If in addition the spring experiences an external force numerically equal to 1 cos wt in Newtons nd zt for 0J2 50 if 49 if 25 ie Aux50 7 5 In particular Write zt Ce quot cosbt 7 gal D coswt 7 02 and determine C a 12 gal D gogi You Will nd a lt 0 Therefore as t 7gt 00 zt 7gt Dcoswt 7 02 What is the value of 02 When w m Q5 An LRC circuit With inductance L ll henry R 2 ohms and C 0 2 farads is set in motion by a voltage of 1 cos wt applied at time t 0 Given the initial qt and current are zero determine qt and for t gt 0 When w Q6 Write down the general solution to dQI dz F0 mg 12 km 7 s1nwt 7 go Where Z E 7 mm2 b2 What happens When k is chosen km w Show then that 12 fo F0 coswti QT Consider the ode d2z dz 4 7 0 M p dt I 7 for p 5 4 and 2 25 a Write the ode as a system of 2 rst order ode7s dz a 9 dj 7 dt b For p 57 Write the solutions as A a equot B a 4 Fillin squares Given A a 07 draW the trajectory 1t7 yt in the Xy plane and in partic ular note the direction along Which it approaches 07 0 c For p 4 Write the general solution A a D B a Dr Fillin squares d For p 2 Write the solution in the form zt Ce quot cosbt 7 go W and draw the trajectory zt7 yt in the Xy planer Use SLUPES or SYSTEMS if you wish 26 MATH 254 Mock exam 2 m 1 hours Name 1 Given y1z 1 satis es d2 1d 1 y i7y0forzgt0 z M w m nd a 2nd liii solni 921 Solve With initial condition yl 07 l ll 91 Hint you should nd ye2 262i 2 Solve the following initial value problems 2 a g 7 2 y ylt0gt 0 lt0gt 0 91 2 b 27 y coszy 90 07 0 0 91 2 c 754y0 y01700 91 d 1 2ij 21 cos2t 10 7 0 g Write zt Acos2t 7 go Find A7 cos g0 sin gal A 27 cosgo singo zt 3 An LRC Circuit7 shown below7 is acted upon by an emf of Vt VI cos wt for t gt 0 W B L C R E D VAivEVt tgt0 0tlt0i For L ll heniry7 R i2 ohIns7 C 2 famcls7 V0 i volts and w 2 radsec7 Write down the ode for Lt7 the Charge in Coulombs on the capac itori Do not solve the equation Given the initial Charge is zero and the initial current is zero7 What are q0 and 0 ode 40 ltogt 28
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