Second Order Equations

Transcription

1 Chapte 2 Second Ode Equations 2 Second Deivatives in Science and Engineeing Second ode equations involve the second deivative d 2 y=dt 2 Often this is shotened to y, and then the fist deivative is y In physical poblems, y can epesent velocity v and the second deivative y D a is acceleation : the ate dy =dt that velocity is changing The most impotant equation in dynamics is Newton s Second Law F D ma Compae a second ode equation to a fist ode equation, and allow them to be nonlinea : Fist ode y D f t; y/ Second ode y D Ft; y; y / () The second ode equation needs two initial conditions, nomally y/ and y / the initial velocity as well as the initial position Then the equation tells us y / and the movement begins When you pess the gas pedal, that poduces acceleation The bae pedal also bings acceleation but it is negative (the velocity deceases) The steeing wheel poduces acceleation too Steeing changes the diection of velocity, not the speed Right now we stay with staight line motion and one dimensional poblems : d 2 y dt > (speeding up) d 2 y 2 dt < 2 (slowing down): The gaph of yt/ bends upwads fo y > (the ight wod is convex) Then the velocity y (slope of the gaph) is inceasing The gaph bends downwads fo y < (concave) Figue 2 shows the gaph of y D sin t, when the acceleation is a D d 2 y=dt 2 D sin t The impotant equation y D y leads to sin t and cos t Notice how the velocity dy=dt (slope of the gaph) changes sign in between zeos of y

2 74 Chapte 2 Second Ode Equations y y D sin t a D y > v D 2 3 t v D y < y D sin t y is going down and bending up y D cos t Figue 2: y > means that velocity y (o slope) inceases The cuve bends upwad The best examples of F D ma come when the foce F is y, a constant times the position o displacement yt/ This poduces the oscillation equation Fundamental equation of mechanics m d2 y C y D (2) dt2 Thin of a mass hanging at the bottom of a sping (Figue 22) The top of the sping is fixed, and the sping will stetch Now stetch it a little moe (move the mass downwad by y/) and let go The sping pulls bac on the mass Hooe s Law says that the foce is F D y, popotional to the stetching distance y Hooe s constant is The mass will oscillate up and down The oscillation goes on foeve, because equation (2) does not include any fiction (damping tem b dy=dt) The oscillation is a pefect cosine, with y D cos t and D p =m, because the second deivative has to poduce =m to match y D =m/y Oscillation at fequency D m y D y/ cos m t : (3) At time t D, this shows the exta stetching y/ The deivative of cos t has a facto D p =m The second deivative y has the equied 2 D =m, so my D y The movement of one sping and one mass is especially simple Thee is only one fequency When we connect N masses by a line of spings thee will be N fequencies then Chapte 6 has to study the eigenvalues of N by N matices m d2 y dt 2 D y y m m y < y > sping pushes down y > y < sping pulls up Figue 22: Lage D stiffe sping D faste Lage m D heavie mass D slowe

3 2 Second Deivatives in Science and Engineeing 75 Initial Velocity y / Second ode equations have two initial conditions The motion stats in an initial position y/, and its initial velocity is y / We need both y/ and y / to detemine the two constants c and c 2 in the complete solution to my C y D : Simple hamonic motion y D c cos m t C c 2 sin m t : (4) Up to now the motion has stated fom est (y / D, no initial velocity) Then c is y/ and c 2 is zeo : only cosines As soon as we allow an initial velocity, the sine solution y D c 2 sin t must be included But its coefficient c 2 is not just y / dy At t D ; dt D c 2 cos t matches y / when c 2 D y / : (5) The oiginal solution y D y/ cos t matched y/, with zeo velocity at t D The new solution y D y /=/ sin t has the ight initial velocity and it stats fom zeo When we combine those two solutions, yt/ matches both conditions y/ and y / : Unfoced oscillation yt/ D y/ cos t C y / sin t with D m : (6) With a tigonometic identity, I can combine those two tems (cosine and sine) into one Cosine with Phase Shift We want to ewite the solution (6) as yt/ D R cos t / The amplitude of yt/ will be the positive numbe R The phase shift o lag in this solution will be the angle By using the ight identity fo the cosine of t, we match both cos t and sin t : R cost / D R cos t cos C R sin t sin : (7) This combination of cos t and sin t agees with the solution (6) if R cos D y/ and R sin D y / : (8) Squaing those equations and adding will poduce R 2 : y Amplitude R R 2 D R 2 cos 2 C sin 2 / D y// 2 / 2 C : (9) The atio of the equations (8) will poduce the tangent of : Phase lag tan D R sin R cos D y / y/ : () Poblem 4 will discuss the angle we should choose, since diffeent angles can have the same tangent The tangent is the same if is inceased by o any multiple of The pue cosine solution that stated fom y / D has no phase shift : D Then the new fom yt/ D R cos t / is the same as the old fom y/ cos t

4 76 Chapte 2 Second Ode Equations Fequency o f If the time t is measued in seconds, the fequency is in adians pe second Then t is in adians it is an angle and cos t is its cosine But not eveyone thins natually about adians Complete cycles ae easie to visualize So fequency is also measued in cycles pe second A typical fequency in you home is f D 6 cycles pe second One cycle pe second is usually shotened to f D Hetz A complete cycle is 2 adians, so f D 6 Hetz is the same fequency as D 2 adians pe second The peiod is the time T fo one complete cycle Thus T D =f This is the only page whee f is a fequency on all othe pages f t/ is the diving function Fequency D 2f Peiod T D f D 2 : y D A cos t D A cos m t A D T D f D 2 m f D 2 t D t D T time A Figue 23: Simple hamonic motion y D A cos t : amplitude A and fequency Hamonic Motion and Cicula Motion Hamonic motion is up and down (o side to side) When a point is in cicula motion, its pojections on the x and y axes ae in hamonic motion Those motions ae closely elated, which is why a piston going up and down can poduce cicula motion of a flywheel The hamonic motion speeds up in the middle and slows down at the ends while the point moves with constant speed aound the cicle sin t e it x D cos t y D sin t cos t 2 t t 2 Figue 24: Steady motion aound a cicle poduces cosine and sine motion along the axes

5 2 Second Deivatives in Science and Engineeing 77 Response Functions I want to intoduce some impotant wods The esponse is the output yt/ Up to now the only inputs wee the initial values y/ and y / In this case yt/ would be the initial value esponse (but I have neve seen those wods) When we only see a few cycles of the motion, initial values mae a big diffeence In the long un, what counts is the esponse to a focing function lie f D cos t Now is the diving fequency on the ight hand side, whee the natual fequency n D p =m is decided by the left hand side : comes fom y p, n comes fom y n When the motion is diven by cos t, a paticula solution is y p D Y cos t : Foced motion y p t/ at fequency my C y D cos t y p t/ D cos t: () m 2 To find y p t/, I put Y cos t into my C y and the esult was m 2 /Y cos t This matches the diving function cos t when Y D = m 2 / The initial conditions ae nowhee in equation () Those conditions contibute the null solution y n, which oscillates at the natual fequency n D p =m Then D m n 2 If I eplace by m 2 n in the esponse y pt/, I see 2 n 2 in the denominato : Response to cos t y p t/ D cos t (2) m 2 n 2 Ou equation my C y D cos t has no damping tem That will come in Section 23 It will poduce a phase shift Damping will also educe the amplitude jy/j The amplitude is all we ae seeing hee in Y/ cos t : Fequency esponse Y/ D m D (3) 2 m 2 n 2: The mass and sping, o the inductance and capacitance, decide the natual fequency n The esponse to a diving tem cos t (o e it ) is multiplication by the fequency esponse Y/ The fomula changes when D n we will study esonance With damping in Section 23, the fequency esponse Y/ will be a complex numbe We can t escape complex aithmetic and we don t want to The magnitude jy/j will give the magnitude esponse (o amplitude esponse) The angle in the complex plane will decide the phase esponse (then D because we measue the phase lag) The esponse is Y/e it to f t/ D e it and the esponse is gt/ to f t/ D ıt/ These show the fequency esponse Y fom equation (3) and the impulse esponse g fom equation (5) Ye it and gt/ ae the two ey solutions to my C y D f t/

6 78 Chapte 2 Second Ode Equations Impulse Response = Fundamental Solution The most impotant solution to a linea diffeential equation will be called gt/ In mathematics g is the fundamental solution In engineeing g is the impulse esponse It is a paticula solution when the ight side f t/ D ıt/ is an impulse (a delta function) The same gt/ solves mg C g D when the initial velocity is g / D =m Fundamental solution mg C g D ıt/ with zeo initial conditions (4) Null solution also gt/ D sin nt m n has g/d and g /D m : (5) To find that null solution, I just put its initial values and =m into equation (6) The cosine tem disappeaed because g/ D I will show that those two poblems give the same answe Then this whole chapte will show why gt/ is so impotant Fo fist ode equations y D ay C q in Chapte, the fundamental solution (impulse esponse, gowth facto) was gt/ D e at The fist two names wee not used, but you saw how e at dominated that whole chapte I will fist explain the esponse gt/ in physical language We stie the mass and it stats to move All ou foce is acting at one instant of time : an impulse A finite foce within one moment is impossible fo an odinay function, only possible fo a delta function Remembe that the integal of ıt/ jumps to when we pass the point t D If we integate mg D ıt/, nothing happens befoe t D In that instant, the integal jumps to The integal of the left side mg is mg Then mg D instantly at t D This gives g / D =m You see that computing with an impulse ıt/ needs some faith The point of gt/ is that it solves the equation fo any focing function f t/ : Rt my C y D f t/ has the paticula solution yt/ D gt s/f s/ ds (6) That was the ey fomula of Chapte, when gt s/ was e at s/ and the equation was fist ode Section 23 will find gt/ when the diffeential equation includes damping The coefficients in the equation will stay constant, to allow a neat fomula fo gt/ You may feel uncetain about woing with delta functions a means to an end We will veify this final solution yt/ in thee diffeent ways : Substitute yt/ fom (6) diectly into the diffeential equation (Poblem 2) 2 Solve fo yt/ by vaiation of paametes (Section 26) 3 Solve again by using the Laplace tansfom Ys/ (Section 27)

7 2 Second Deivatives in Science and Engineeing 79 REVIEW OF THE KEY IDEAS my Cy D : A mass on a sping oscillates at the natual fequency n D p =m 2 my C y D cos t : This diving foce poduces y p D cos t/=m n Thee is esonance when n D The solution y p Dt sin t includes a new facto t 4 mg Cg D ıt/ gives gt/ D sin n t/=mn D null solution with g / D =m 5 Fundamental solution g : Evey diving function f gives yt/ D R t gt s/f s/ ds 6 Fequency : adians pe second o f cycles pe second (f Hetz) Peiod T D=f Poblem Set 2 Find a cosine and a sine that solve d 2 y=dt 2 D 9y This is a second ode equation so we expect two constants C and D (fom integating twice) : Simple hamonic motion yt/ D C cos t C D sin t: What is If the system stats fom est (this means dy=dt D at t D ), which constant C o D will be zeo? 2 In Poblem, which C and D will give the stating values y/ D and y / D? 3 Daw Figue 23 to show simple hamonic motion y D A cos t / with phases D =3 and D =2 4 Suppose the cicle in Figue 24 has adius 3 and cicula fequency f D 6 Hetz If the moving point stats at the angle 45 ı, find its x-coodinate A cos t / The phase lag is D 45 ı When does the point fist hit the x axis? 5 If you dive at 6 miles pe hou on a cicula tac with adius R D 3 miles, what is the time T fo one complete cicuit? You cicula fequency is f D and you angula fequency is D (with what units?) The peiod is T 6 The total enegy E in the oscillating sping-mass system is E D inetic enegy in mass C potential enegy in sping D m 2 dy 2 C dt 2 y2 : Compute E when y D C cos t C D sin t The enegy is constant 7 Anothe way to show that the total enegy E is constant : Multiply my C y D by y : Then integate my y and yy :

8 8 Chapte 2 Second Ode Equations 8 A foced oscillation has anothe tem in the equation and in the solution : d 2 y C 4y D F cos t has y D C cos 2t C D sin 2t C A cos t: dt2 (a) Substitute y into the equation to see how C and D disappea (they give y n ) Find the foced amplitude A in the paticula solution y p D A cos t (b) In case D 2 (focing fequency D natual fequency), what answe does you fomula give fo A? The solution fomula fo y beas down in this case 9 Following Poblem 8, wite down the complete solution y n C y p to the equation m d 2 y dt 2 C y D F cos t with n D p =m (no esonance): The answe y has fee constants C and D to match y/ and y / (A is fixed by F ) Suppose Newton s Law F D ma has the foce F in the same diection as a : my D C y including y D 4y: Find two possible choices of s in the exponential solutions y D e st The solution is not sinusoidal and s is eal and the oscillations ae gone Now y is unstable Hee is a fouth ode equation : d 4 y=dt 4 D 6y Find fou values of s that give exponential solutions y D e st You could expect fou initial conditions on y : y/ is given along with what thee othe conditions? 2 To find a paticula solution to y C 9y D e ct, I would loo fo a multiple y p t/ D Ye ct of the focing function What is that numbe Y? When does you fomula give Y D? (Resonance needs a new fomula fo Y ) 3 In a paticula solution y D Ae it to y C 9y D e it, what is the amplitude A? The fomula blows up when the focing fequency D what natual fequency? 4 Equation () says that the tangent of the phase angle is tan D y /=y/ Fist, chec that tan is dimensionless when y is in metes and time is in seconds Next, if that atio is tan D, should you choose D =4 o D 5=4? Answe : Sepaately you want R cos D y/ and R sin D y /=: If those ight hand sides ae positive, choose the angle between and =2 If those ight hand sides ae negative, add and choose D 5=4 Question : If y/ > and y / <, does fall between =2 and o between 3=2 and 2? If you plot the vecto fom ; / to y/; y /=/, its angle is

9 2 Second Deivatives in Science and Engineeing 8 5 Find a point on the sine cuve in Figue 2 whee y > but v D y < and also a D y < The cuve is sloping down and bending down Find a point whee y < but y > and y > The point is below the x-axis but the cuve is sloping and bending 6 (a) Solve y C y D stating fom y/ D and y / D (This is y n ) (b) Solve y C y D cos t with y/ D and y / D (This can be y p ) 7 Find a paticula solution y p D R cost / to y C y D cos t sin t 8 Simple hamonic motion also comes fom a linea pendulum (lie a gandfathe cloc) At time t, the height is A cos t What is the fequency if the pendulum comes bac to the stat afte second? The peiod does not depend on the amplitude (a lage cloc o a small metonome o the movement in a watch can all have T D ) 9 If the phase lag is, what is the time lag in gaphing cost /? 2 What is the esponse yt/ to a delayed impulse if my C y D ıt T /? 2 (Good challenge) Show that y D Why is y D R t R t gt s/f s/ ds has my C y D f t/ g t s/f s/ ds C g/f t/? Notice the two t s in y 2 Using g/ D, explain why y D R t g t s/f s/ ds C g /f t/ 3 Now use g / D =m and mg C g D to confim my C y D f t/ 22 With f D (diect cuent has D ) veify that my C y D fo this y : Step esponse yt/ D Z t sin n t s/ m n ds D y p C y n equals cos nt: 23 (Recommended) Fo the equation d 2 y=dt 2 D find the null solution Then fo d 2 g=dt 2 D ıt/ find the fundamental solution (stat the null solution with g/ D and g / D ) Fo y D f t/ find the paticula solution using fomula (6) 24 Fo the equation d 2 y=dt 2 D e it find a paticula solution y D Y/e it Then Y/ is the fequency esponse Note the esonance when D with the null solution y n D 25 Find a paticula solution Ye it to my y D e it The equation has y instead of y What is the fequency esponse Y/? Fo which is Y infinite?