Mah 50-004 Week 5: Secion 7.4, mass-spring sysems. These are noes for Monday. There will also be course review noes for Tuesday, posed laer. Mon Apr 3 7.4 mass-spring sysems. Announcemens: Warm up exercise: Here are wo sysems of differenial equaions, and he eigendaa is as shown. The firs order sysem could arise from an inpu-oupu model, and he second one could arise from an undamped wo mass, hree spring model. Wrie down he general soluion o each sysem. a) = 3 4 3 b) = 3 4 3 eigendaa: For he marix 3 4 3 for he eigenvalue = 5, v =, T is an eigenvecor; for he eigenvalue =, v =, T is an eigenvecor
Exercise ) Consider a rain wih wo cars conneced by a spring: a) Verify ha he linear sysem of DEs ha governs he dynamics of his configuraion (i's acually a special case of wha we did before, wih wo of he spring consans equal o zero) is = k m = k m b). Use he eigenvalues and eigenvecors compued below o find he general soluion. For = 0 and is corresponding eigenvecor v remember ha you ge wo soluions x = v and x = v, raher han he expeced cos v, sin v. Inerpre hese soluions in erms of rain moions. You will use hese ideas in some of your homework problems and in your lab exercise abou molecular vibraions.
In case we need o refer o i, on Friday we showed: Soluion space algorihm: Consider a very special case of a homogeneous sysem of linear differenial equaions, x = A x. If A n n is a diagonalizable marix and if all of is eigenvalues are non-posiive hen for each eigenpair j, v wih 0 here are wo linearly independen sinusoidal soluions o x = A x given by j j x j = cos j v y j j = sin j v j wih And for an eigenpair funcions j, v j wih j j = j. = 0 here are wo indpenden soluions given by consan and linear x j = v j y j = v j This procedure consrucs n independen soluions o he sysem x soluion space. = A x, i.e. a basis for he Remark: Wha's amazing is ha he fac ha if he sysem is conservaive, he acceleraion marix will always be diagonalizable, and all of is eigenvalues will be non-posiive. In fac, if he sysem is ehered o a leas one wall, all of he eigenvalues will be sricly negaive, and he algorihm above will always yield a basis for he soluion space. (If he sysem is no ehered and is free o move as a rain, like on he previous page, hen = 0 will be one of he eigenvalues, and will yield he consan velociy and displacemen conribuion soluions c c v, where v is he corresponding eigenvecor. Togeher wih he soluions from sricly negaive eigenvalues his will sill lead o he general homogeneous soluion.)
Forced oscillaions (sill undamped): If he forcing is sinusoidal, M x = K x F x = A x M F. M x = K x cos G 0 wih F 0 = M G 0. x = A x cos F 0 From vecor space heory we know ha he general soluion o his inhomogeneous linear problem is of he form x = x P x H, Forced oscillaion paricular soluion algorihm: x = A x cos F 0 As long as he driving frequency is NOT one of he naural frequencies, we don' expec resonance; he mehod of undeermined coefficiens predics here should be a paricular soluion of he form x P = cos c where he consan vecor c is wha we need o find. (I's value will depend on he angular frequency he forcing funcion.) Exercise ) Subsiue he guess x P = cos c ino he DE sysem x = A x cos F 0 o find a marix algebra formula for c = c. Noice ha his formula makes sense precisely when is NOT one of he naural frequencies of he sysem. of Soluion: c = A I F0. Noe, marix inverse exiss precisely if eigenvalue, i.e. is no one of he naural frequencies. is no an
Friday model: m = k k m = k k 3 0 = a, 0 = a 0 = b, 0 = b Exercise ) Coninuing wih he configuraion from Friday shown above, bu now for an inhomogeneous forced problem, le k = m, and force he second mass sinusoidally: x 0 = cos x 3 We know from previous work ha he naural frequencies are =, = 3 and ha x H = C cos C cos 3. Find he formula for x P, as on he preceding page. Noice ha his seady periodic soluion blows up as or 3. (If we don' have ime o work his by hand, we may skip direcly o he echnology check on he nex page. Bu since we have quick formulas for inverses of by marices, his is definiely a compuaion we could do by hand.)
Soluion: As long as, 3, he general soluion x = x P x H is given by 3 = cos 6 3 3 3 C cos C cos 3. Inerpreaion as far as inferred pracical resonance for slighly damped problems: If here was even a small amoun of damping, he homogeneous soluion would acually be ransien (i would be exponenially decaying and oscillaing - underdamped). There would sill be a sinusoidal paricular soluion, which would have a formula close o our paricular soluion, he firs erm above, as long as, 3. (There would also be a relaively smaller sin d erm as well.) So we can infer he pracical resonance behavior for differen values wih sligh damping, by looking a he size of he c erm for he undamped problem...see nex page for visualizaions.
resar : wih LinearAlgebra : A Mari,,,,, : F0 Vecor 0, 3 : Iden IdeniyMari : c c ; A wih plos : wih LinearAlgebra : Iden. F0 : # he formula we worked ou by hand plo Norm c,, = 0..4, magniude = 0..0, color = black, ile = `undamped paricular soluion ampliudes as proxy for pracical resonance` ; # Norm(c( ),) is he magniude of he c( ) vecor 4 3 4 4 3 4 undamped paricular soluion ampliudes as proxy for pracical resonance 0 magniude 6 3 3 () 0 3 4
There are srong connecions beween our discussion here and he modeling of how earhquakes can shake buildings: As i urns ou, for our physics lab springs, he modes and frequencies are almos idenical:
An ineresing shake-able demonsraion: hp://www.youube.com/wach?v=m_xjokahzm Below is a discussion of how o model he unforced "hree-sory" building shown shaking in he video above, from which we can see which modes will be excied. There is also a "wo-sory" building model in he video, and is marix and eigendaa follow. Here's a schemaic of he hree-sory building: For he unforced (homogeneous) problem, he acceleraions of he hree massive floors (he op one is he roof) above ground and of mass m, are given by x 3 = k m 0 0 Noe he value in he las diagonal enry of he marix. This is because x 3 is measuring displacemens for he op floor (roof), which has nohing above i. The "k" is jus he linearizaion proporionaliy facor, and depends on he ension in he walls, and he heigh beween floors, ec, as discussed on he previous page. x 3.
k Exercise 3 Here is eigendaa for he unscaled marix m =. For he scaled marix you'd have he same k eigenvecors, bu he eigenvalues would all be muliplied by he scaling facor m and he naural k frequencies would all be scaled by m bu he eigenvecors describing he modes would say he same. Use his informaion describe he fundamenal modes, and he order in which hey will appear. Remark) All of he ideas we've discussed in secion 7.4 also apply o molecular vibraions. The eigendaa in hese cases is relaed o he "specrum" of ligh frequencies ha correspond o he naural fundamenal modes for molecular vibraions. See your lab quesion abou he fundamenal modes of carbon dioxide.
So far we've no considered he possibiliy of complex eigenvalues and eigenvecors. Linear algebra heory works he same wih complex number scalars and vecors - one can alk abou complex vecor spaces, linear combinaions, span, linear independence, reduced row echelon form, deerminan, dimension, basis, ec. Then he model space is n raher han n. Definiion: v n (v 0) is a complex eigenvecor of he marix A, wih eigenvalue if A v = v. Jus as before, you find he possibly complex eigenvalues by finding he roos of he characerisic polynomial A I. Then find he eigenspace bases by reducing he corresponding marix (using complex scalars in he elemenary row operaions). The bes way o see how o proceed in he case of complex eigenvalues/eigenvecors is o work an example. There is a general discussion on he page afer his example ha we will refer o along he way: Glucose-insulin model (adaped from a discussion on page 339 of he ex "Linear Algebra wih Applicaions," by Oo Brescher) Le G be he excess glucose concenraion (mg of G per 00 ml of blood, say) in someone's blood, a ime hours. Excess means we are keeping rack of he difference beween curren and equilibrium ("fasing") concenraions. Similarly, Le H be he excess insulin concenraion a ime hours. When blood levels of glucose rise, say as food is digesed, he pancreas reacs by secreing insulin in order o uilize he glucose. Researchers have developed mahemaical models for he glucose regulaory sysem. Here is a simplified (linearized) version of one such model, wih paricular represenaive marix coefficiens. I would be mean o apply beween meals, when no addiional glucose is being added o he sysem: G 0. 0.4 G = H 0. 0. H Exercise a) Undersand why he signs of he marix enries are reasonable. Now le's solve he iniial value problem, say righ afer a big meal, when G 0 H 0 = 00 0
G H = 0. 0.4 0. 0. G H G 0 H 0 = 00 0 b) The firs sep is o ge he eigendaa of he marix. Do his, and compare wih he Wolfram check on he nex page.
c) Exrac a basis for he soluion space o his homogeneous sysem of differenial equaions from he eigenvecor informaion above:
d) Solve he iniial value problem.
Here are some picures o help undersand wha he model is predicing... you could also consruc hese graphs using pplane. () Plos of glucose vs. insulin, a ime hours laer: wih plos : G 00 exp. cos. : H 50 exp. sin. : plo plo G, = 0..30, color = green : plo plo H, = 0..30, color = brown : display plo, plo, ile = `underdamped glucose-insulin ineracions` ; 00 60 underdamped glucose-insulin ineracions 0 0 0 30 ) A phase porrai of he glucose-insulin sysem: pic fieldplo. G.4 H,. G. H, G = 40..00, H = 5..40 : soln plo G, H, = 0..30, color = black : display pic, soln, ile = `Glucose vs Insulin phase porrai` ; H Glucose vs Insulin phase porrai 40 30 0 0 40 0 0 40 60 80 00 G The example we jus worked is linear, and is vasly simplified. Bu mahemaicians, docors, bioengineers, pharmaciss, are very ineresed in (especially more realisic) problems like hese.
Soluions o homogeneous linear sysems of DE's when marix has complex eigenvalues: x = A x Le A be a real number marix. Le = a b i C v = i n saisfy A v = v, wih a, b,, n. Then z = e v is a complex soluion o x = A x because z = e v and his is equal o A z = A e v = e A v = e v. Bu if we wrie z in erms of is real and imaginary pars, z = x i y hen he equaliy z = A z x i y = A x i y = A x i A y. Equaing he real and imaginary pars on each side yields x = A x y = A y i.e. he real and imaginary pars of he complex soluion are each real soluions. If A i = a b i i hen i is sraighforward o check ha A i = a b i i. Thus he complex conjugae eigenvalue yields he complex conjugae eigenvecor. The corresponding complex soluion o he sysem of DEs e a i b i = x i y so yields he same wo real soluions (excep wih a sign change on he second one). Anoher way o undersand how we ge he wo real soluions is o ake he wo complex soluions z = x i y w = x i y and recover x, y as linear combinaions of hese homogeneous soluions: x = z w y = i z w.