DKiPI_3_2018_s_izradom -- Sage

Zadavanje vektora i matrica, osnovne naredbe i operacije

- MPZI_predavanje_02 i MPZI_vj02 - vektori i matrice

Vektor se zadaje nizom brojeva (listom brojeva)

a = vector ([1., 2., 3., 4.])
a

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(1.00000000000000,\,2.00000000000000,\,3.00000000000000,\,4.00000000000000\right)$

Broj komponenata vektora (nazvan i njegovom dimenzijom ili dimenzijom vektorskog prostora kojem pripada) daju funkcije

len (a)

$\newcommand{\Bold}[1]{\mathbf{#1}}4$

a.degree()

$\newcommand{\Bold}[1]{\mathbf{#1}}4$

Pristupanje komponentama vektora vrši se indeksiranjem (prvi element liste ima indeks 0!)

a[0]

$\newcommand{\Bold}[1]{\mathbf{#1}}1.00000000000000$

a[2]

$\newcommand{\Bold}[1]{\mathbf{#1}}3.00000000000000$

Za vektore sa $n$ komponenata ( $n$ -dimenzionalne vektore) skalarni je umnožak $s = \mathbf{a}\cdot\mathbf{b} = \displaystyle\sum_{i=0}^{n-1}\, a_i\, b_i$ .

Skalarni produkt vektora u matričnoj notaciji:

$\mathbf{a}\cdot\mathbf{b} = \mathbf{a}^T \mathbf{b}$

b = vector([5., 6., 7., 8.])

a*b

$\newcommand{\Bold}[1]{\mathbf{#1}}70.0000000000000$

b*a

$\newcommand{\Bold}[1]{\mathbf{#1}}70.0000000000000$

a.dot_product (b)

$\newcommand{\Bold}[1]{\mathbf{#1}}70.0000000000000$

Matrice

Osnovni način zadavanja matrica je nizom čije su komponenete nizovi brojeva jednakih duljina

A = matrix( [[1., 2.5, 3., 5.], [2., 3., 4., 5.], [2., 3., 8., 2.], [4., 3., 1., 3.]])
A

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr} 1.00000000000000 & 2.50000000000000 & 3.00000000000000 & 5.00000000000000 \\ 2.00000000000000 & 3.00000000000000 & 4.00000000000000 & 5.00000000000000 \\ 2.00000000000000 & 3.00000000000000 & 8.00000000000000 & 2.00000000000000 \\ 4.00000000000000 & 3.00000000000000 & 1.00000000000000 & 3.00000000000000 \end{array}\right)$

Pristupanje komponentama matrice:

A[1, 3]

$\newcommand{\Bold}[1]{\mathbf{#1}}5.00000000000000$

Pristupanje cijelom retku ili cijelom stupcu matrice:

A[:, 0]

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{r} 1.00000000000000 \\ 2.00000000000000 \\ 2.00000000000000 \\ 4.00000000000000 \end{array}\right)$

A[1, :]

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr} 2.00000000000000 & 3.00000000000000 & 4.00000000000000 & 5.00000000000000 \end{array}\right)$

Pristupanje 'podmatricama':

A[0:2,0:3]

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 1.00000000000000 & 2.50000000000000 & 3.00000000000000 \\ 2.00000000000000 & 3.00000000000000 & 4.00000000000000 \end{array}\right)$

Konstruiranje jedinične matrice B reda 4x4

B = identity_matrix(ZZ, 4)
B

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)$

Vektor možemo "pretvoriti" u jednorednu ili u jednostupčanu matricu:

a.column()

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{r} 1.00000000000000 \\ 2.00000000000000 \\ 3.00000000000000 \\ 4.00000000000000 \end{array}\right)$

a.row()

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr} 1.00000000000000 & 2.00000000000000 & 3.00000000000000 & 4.00000000000000 \end{array}\right)$

a

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(1.00000000000000,\,2.00000000000000,\,3.00000000000000,\,4.00000000000000\right)$

Neke martrične operacije

Matrice možemo množiti brojem i, ako su istog tipa (jednakih brojeva redaka i stupaca), zbrajati ili oduzimati:

Kao i kod vektora, zbrajanje i oduzimanje matrica izvode se po parovima komponenata, dok se pri množenju matrice brojem sve njezine komponente množe tim brojem.

C = 2*B - A
show(C)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr} 1.00000000000000 & -2.50000000000000 & -3.00000000000000 & -5.00000000000000 \\ -2.00000000000000 & -1.00000000000000 & -4.00000000000000 & -5.00000000000000 \\ -2.00000000000000 & -3.00000000000000 & -6.00000000000000 & -2.00000000000000 \\ -4.00000000000000 & -3.00000000000000 & -1.00000000000000 & -1.00000000000000 \end{array}\right)$

Umnožak $\mathbf{C} = \mathbf{A\,B}$ dviju matrica $\mathbf{A}$ i $\mathbf{B}$ definiran je ako je broj stupaca prve jednak broju redaka druge, jer je opći izraz za komponente umnoška $c_{i,j} = \displaystyle\sum_{k = 0}^{n - 1} a_{i,k}\,b_{k,j}$ , gdje je $n$ broj stupaca prve i broj redaka druge matrice.

A*C

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr} -30.0000000000000 & -29.0000000000000 & -36.0000000000000 & -28.5000000000000 \\ -32.0000000000000 & -35.0000000000000 & -47.0000000000000 & -38.0000000000000 \\ -28.0000000000000 & -38.0000000000000 & -68.0000000000000 & -43.0000000000000 \\ -16.0000000000000 & -25.0000000000000 & -33.0000000000000 & -40.0000000000000 \end{array}\right)$

Produkt vektora $\mathbf{a} \mathbf{b}^T$ daje matricu.

Da bi primjenili operacije s matricama na vektore, prvo ih je potrebno transformirati u jednoredne ili jednostupčane matrice

a.column()*b.row()

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr} 5.00000000000000 & 6.00000000000000 & 7.00000000000000 & 8.00000000000000 \\ 10.0000000000000 & 12.0000000000000 & 14.0000000000000 & 16.0000000000000 \\ 15.0000000000000 & 18.0000000000000 & 21.0000000000000 & 24.0000000000000 \\ 20.0000000000000 & 24.0000000000000 & 28.0000000000000 & 32.0000000000000 \end{array}\right)$

a.row()*b.column()

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{r} 70.0000000000000 \end{array}\right)$

Vektor se, u stvari, u određenim kontekstima može smatrati jednostupčanom matricom i bez eksplicitne prevorbe, tako da je definirano i množenje matrice i vektora; rezultat je vektor (a ne jednostupčana matrica):

A*a

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(35.0000000000000,\,40.0000000000000,\,40.0000000000000,\,25.0000000000000\right)$

a*A

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(27.0000000000000,\,29.5000000000000,\,39.0000000000000,\,33.0000000000000\right)$

Transponirana matrica: $\mathbf{A}^T$

A.T

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr} 1.00000000000000 & 2.00000000000000 & 2.00000000000000 & 4.00000000000000 \\ 2.50000000000000 & 3.00000000000000 & 3.00000000000000 & 3.00000000000000 \\ 3.00000000000000 & 4.00000000000000 & 8.00000000000000 & 1.00000000000000 \\ 5.00000000000000 & 5.00000000000000 & 2.00000000000000 & 3.00000000000000 \end{array}\right)$

Inverzna matrica: $\mathbf{A}^{-1}$

A^-1

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr} -2.55000000000000 & 2.95000000000000 & -0.475000000000000 & -0.350000000000000 \\ 4.90000000000000 & -6.10000000000000 & 1.05000000000000 & 1.30000000000000 \\ -0.900000000000000 & 1.10000000000000 & -0.0500000000000000 & -0.300000000000000 \\ -1.20000000000000 & 1.80000000000000 & -0.400000000000000 & -0.400000000000000 \end{array}\right)$

A.inverse()

DKPI - ZADATAK

Za zadani okvir potrebno je odabrati dinamičke stupnjeve slobode te za odabrani model odrediti oblike titranja metodom Stodola.

Zatim treba odrediti zakon prisilnog titranja sustava koje će nastati od naglog ubrzanja podloge $u_g(t) = 0,25 g$ trajanja 4 s.

EI = 8138.02;
EIg = 33333.33;
L = 4.;
h = 3.;

Matrica krutosti

k = matrix( [
[ 24*EI/h^3, -24*EI/h^3, 0., 6*EI/h^2, 6*EI/h^2, 6*EI/h^2, 6*EI/h^2, 0., 0., 0., 0., 0.],
[ -24*EI/h^3, 60*EI/h^3, -36*EI/h^3, -6*EI/h^2, -6*EI/h^2, 0., 0., 6*EI/h^2, 0., 6*EI/h^2, 6*EI/h^2, 6*EI/h^2],
[ 0., -36*EI/h^3, 66*EI/h^3, 0., 0., -6*EI/h^2, -6*EI/h^2, -6*EI/h^2, 6*EI/h^2, -3*EI/h^2, -3*EI/h^2, 0.],
[6*EI/h^2, -6*EI/h^2, 0., 4*EI/h + 4*EIg/L, 2*EIg/L, 2*EI/h, 0., 0., 0., 0., 0., 0.],
[6*EI/h^2, -6*EI/h^2, 0., 2*EIg/L, 4*EI/h + 4*EIg/L, 0., 2*EI/h, 0., 0., 0., 0., 0.],
[6*EI/h^2, 0., -6*EI/h^2, 2*EI/h, 0., 8*EI/h + 4*EIg/L, 2*EIg/L, 0., 0., 2*EI/h, 0., 0.],
[6*EI/h^2, 0., -6*EI/h^2, 0., 2*EI/h, 2*EIg/L, 8*EI/h + 8*EIg/L, 2*EIg/L, 0., 0., 2*EI/h, 0.],
[0., 6*EI/h^2, -6*EI/h^2, 0., 0., 0., 2*EIg/L, 4*EI/h + 4*EIg/L, 0., 0., 0., 2*EI/h],
[0., 0., 6*EI/h^2, 0., 0., 0., 0., 0., 4*EI/h + 4*EIg/L, 2*EIg/L, 0., 0.],
[0., 6*EI/h^2, -3*EI/h^2, 0., 0., 2*EI/h, 0., 0., 2*EIg/L, 7*EI/h + 8*EIg/L, 2*EIg/L, 0.],
[0., 6*EI/h^2, -3*EI/h^2, 0., 0., 0., 2*EI/h, 0., 0., 2*EIg/L, 7*EI/h + 8*EIg/L, 2*EIg/L],
[0., 6*EI/h^2, 0., 0., 0., 0., 0., 2*EI/h, 0., 0., 2*EIg/L, 8*EI/h + 4*EIg/L]
])

Kondenzacija matrice krutosti:

$\mathbf{K} = \mathbf{k}_{tt} - \mathbf{k}_{t0} \mathbf{k}_{00}^{-1} \mathbf{k}_{0t}$

ktt = k[0:3,0:3];
kt0 = k[0:3, 3:13];
k0t  =k[3:13, 0:3];
k00 = k[3:13, 3:13];
show(ktt)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 7233.79555555556 & -7233.79555555556 & 0.000000000000000 \\ -7233.79555555556 & 18084.4888888889 & -10850.6933333333 \\ 0.000000000000000 & -10850.6933333333 & 19892.9377777778 \end{array}\right)$

k = ktt - kt0*k00^(-1)*k0t
k.n(digits = 6)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 5661.66 & -6206.39 & 541.260 \\ -6206.39 & 15522.9 & -9867.12 \\ 541.260 & -9867.12 & 17698.0 \end{array}\right)$

Matrica fleksibilnosti:

$\mathbf{F} = \mathbf{k}^{-1}$

F = k^-1
F.n(digits = 6)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 0.000461635 & 0.000271987 & 0.000137522 \\ 0.000271987 & 0.000260034 & 0.000136658 \\ 0.000137522 & 0.000136658 & 0.000128488 \end{array}\right)$

Matrica masa

m = matrix( [[12.28, 0, 0], [0, 15., 0], [0, 0, 22.39]] )
m.n(digits = 4)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 12.28 & 0.0000 & 0.0000 \\ 0.0000 & 15.00 & 0.0000 \\ 0.0000 & 0.0000 & 22.39 \end{array}\right)$

Odredivanje vlastitih frekvencija i vektora metodom Stodola

$\mathbf{D} v_m^{(n)} = \lambda_m v_m^{(n+1)}$

gdje je : $\mathbf{D} = \mathbf{F} \;\mathbf{M}$ i $\lambda_m = \frac{1}{\omega_m^2}$

D = F*m
D

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 0.00566888111477983 & 0.00407981095483138 & 0.00307912567201004 \\ 0.00334000523502195 & 0.00390050307325327 & 0.00305976775063485 \\ 0.00168877459813682 & 0.00204986673780808 & 0.00287685349445953 \end{array}\right)$

Da bi proveli iterativni postupak metode Stodola, programirati će se jednostavna funkcija koja izvršava iterativni proces do zadovoljenja zadanog uvjeta konvergencije

Više o osnovama programiranja:

- MPZI_predavanje_09 i MPZI_vj09

- MPZI_predavanje_10 i MPZI_vj10

- MPZI_predavanje_11 i MPZI_vj11

- MPZI_predavanje_12 i MPZI_vj12

- MPZI_predavanje_13 i MPZI_vj13

"Programska se funkcija definira na sljedeći način:

def naziv (parametri) :

tijelo funkcije

Definicija funkcije započinje ključnom riječju def iza koje slijedi naziv funkcije. Dobro odabrani naziv trebao bi biti ključem značenja i svrhe funkcije. Nakon naziva se između okruglih zagrada navode parametri funkcije. Parametri su varijable koje pri upotrebi funkcije prihvaćaju vrijednosti koje se prenose u funkciju—koje ulaze u nju. (Par zagrada treba napisati i ako funkcija nema parametara.) Prvi redak definicije funkcije završava dvotočkom.

Tijelo funkcije niz je naredbi koje propisuju kako funkcija radi ono čemu je namijenjena.

Vrijednost koju funkcija izračunava i 'vraća' naredbom return je vrijednost funkcije.

Upotreba funkcije naziva se pozivom funkcije:

naziv (argumenti)

Nakon naziva se između okruglih zagrada navode funkcijski argumenti—vrijednosti koje se prenose u funkciju."

Iteracija:

$\mathbf{D} v_m^{(n)} = \lambda_m v_m^{(n+1)}$

uvjet konvergencije $\Delta = \vert \lambda_m^{(n+1)} - \lambda_m^{(n)} \vert < 10^{-6}$

def Stodola (A):
    v = vector(RR, [1, 1, 1])   #početni vektor za iteraciju
    delta = oo
    lam = 0                     #početna vrijednost svojstvene vrijednosti lambda
    br_it = 0                   #brojač iteracija
    while abs(delta) > 10^(-6) :
        br_it = br_it + 1
        rez = max(A*v)
        delta = lam - rez
        lam = rez
        v = A*v/lam
    return sqrt(1/lam), v, br_it

1. oblik

Stodola (D)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(9.93680252049857, \left(1.00000000000000,\,0.755811420415918,\,0.446590923439053\right), 7\right)$

w1 = Stodola(D)[0]
w1.n(digits = 4)

$\newcommand{\Bold}[1]{\mathbf{#1}}9.937$

fi1 = Stodola(D)[1]
fi1.n(digits = 3)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(1.00,\,0.756,\,0.447\right)$

2. oblik

Matrica čišćenja:

$\mathbf{S}_1 = \mathbf{I} - \frac{\mathbf{\phi}_1 \mathbf{\phi}_1^T \mathbf{M}}{\mathbf{\phi}_1^T \mathbf{M} \mathbf{\phi}_1}$

S1 = identity_matrix (RR, 3) - (fi1.column()*fi1.row()*M)/(fi1*M*fi1)
S1

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 0.514898740743150 & -0.447856358067349 & -0.395000841601390 \\ -0.366645071804470 & 0.661505049866817 & -0.298546147156229 \\ -0.216641819332964 & -0.200008584517349 & 0.823596209390032 \end{array}\right)$

Nova dinamička matrica ('očišćena od prvog oblika'):

$\mathbf{D}_1 = \mathbf{D} \mathbf{S}_1$

D1 = D*S1
D1

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 0.000755989779400860 & -0.000455880468468910 & -0.000921268421263936 \\ -0.000373209392012029 & 0.000472380082738953 & 0.0000362280777592202 \\ -0.000505272398287678 & 0.0000242733622052170 & 0.00109031842871194 \end{array}\right)$

Stodola(D1)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(24.3095125276492, \left(-1.17279713669530,\,0.388332263575262,\,1.00000000000000\right), 8\right)$

w2 = Stodola(D1)[0]
w2.n(digits=5)

$\newcommand{\Bold}[1]{\mathbf{#1}}24.310$

fi2 = Stodola(D1)[1]
fi2.n(digits = 3)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(-1.17,\,0.388,\,1.00\right)$

Normiranje vektora $\mathbf{v}_2$ na jedinicu

fi2 = fi2/fi2[0]
fi2

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(1.00000000000000,\,-0.331116312808797,\,-0.852662381848744\right)$

2. oblik

Matrica čišćenja:

$\mathbf{S}_2 = \mathbf{S}_1 - \frac{\mathbf{\phi}_2 \mathbf{\phi}_2^T \mathbf{M}}{\mathbf{\phi}_2^T \mathbf{M} \mathbf{\phi}_2}$

S2 = S1 - (fi2.column()*fi2.row()*M)/(fi2*M*fi2)
S2

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 0.108314488018755 & -0.283410089409097 & 0.237095655928775 \\ -0.232018393196248 & 0.607054207733532 & -0.507843608757772 \\ 0.130037278017211 & -0.340225731637632 & 0.284631304247713 \end{array}\right)$

D2 = D*S2
D2

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 0.0000678318943632407 & -0.000177549481265380 & 0.000148576723906746 \\ -0.000145333629044484 & 0.000380213899010568 & -0.000318039139130355 \\ 0.0000814101668774230 & -0.000213015116337138 & 0.000178231961717735 \end{array}\right)$

Stodola(D2)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(39.9591747793073, \left(0.833505236615000,\,-1.78480288115190,\,1.00000000000000\right), 3\right)$

w3 = Stodola(D2)[0]
w3.n(digits=5)

$\newcommand{\Bold}[1]{\mathbf{#1}}39.959$

fi3 = Stodola(D2)[1]
fi3.n(digits = 3)

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(0.833,\,-1.78,\,1.00\right)$

Normiranje vektora $\mathbf{v}_3$ na jedinicu

fi3 = fi3/fi3[1]
fi3

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(-0.467001283680729,\,1.00000000000000,\,-0.560285962422140\right)$

Frekvencije sustava:

frek = [w1, w2, w3]
frek

$\newcommand{\Bold}[1]{\mathbf{#1}}\left[9.93680252049857, 24.3095125276492, 39.9591747793073\right]$

Periodi:

periodi = [n(2*pi/w1), n(2*pi/w2), n(2*pi/w3)]
periodi

$\newcommand{\Bold}[1]{\mathbf{#1}}\left[0.632314599612706, 0.258466116917532, 0.157240116741182\right]$

fi = column_matrix( [fi1, fi2, fi3] )
show( fi.n(digits=3) )

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 1.00 & 1.00 & -0.467 \\ 0.756 & -0.331 & 1.00 \\ 0.447 & -0.853 & -0.560 \end{array}\right)$

Modalna analiza

$\mathbf{u} = \mathbf{\Phi} \mathbf{q}$

$M_n\ddot{q}_n + K_n q_n = P_n(t)$ n=1,2,3

gdje je: $M_n = \mathbf{\phi}_n^T \mathbf{m} \mathbf{\phi}_n$ ; $K_n = \mathbf{\phi}_n^T \mathbf{k}\mathbf{\phi}_n$ i $P_n = \mathbf{\phi}_n^T \mathbf{p}_{eff}$

Efektivno opterećenje za horizontalno gibanje podloge:

$\mathbf{p}_{eff}(t) = - \mathbf{M}\mathbf{l} u_g(t)$

gdje je $\mathbf{l}$ vektor utjecaja

l = vector( [1, 1, 1] )
ug = 0.25*10
P = -M*l*ug
P

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(-30.7000000000000,\,-37.5000000000000,\,-55.9750000000000\right)$

P1 = fi1*P
P2 = fi2*P
P3 = fi3*P
M1 = fi1*M*fi1
M2 = fi2*M*fi2
M3 = fi3*M*fi3
K1 = fi1*K*fi1
K2 = fi2*K*fi2
K3 = fi3*K*fi3

var('t') 
q1 = function('q1')(t)
q2 = function('q2')(t)
q3 = function('q3')(t)

de1 = M1*diff(q1, t, 2) + K1*q1 == P1 
de2 = M2*diff(q2, t, 2) + K2*q2 == P2 
de3 = M3*diff(q3, t, 2) + K3*q3 == P3

sol1 = desolve( de1, q1, ics = [0, 0, 0] )
sol2 = desolve( de2, q2, ics = [0, 0, 0] )
sol3 = desolve( de3, q3, ics = [0, 0, 0] )

sol1

$\newcommand{\Bold}[1]{\mathbf{#1}}\frac{3364024942368}{100053051567865} \, \cos\left(\frac{1}{21804527988} \, \sqrt{1782676980975055} \sqrt{26333971} t\right) - \frac{3364024942368}{100053051567865}$

sol2

$\newcommand{\Bold}[1]{\mathbf{#1}}-\frac{5903974404321}{3578329562859854} \, \cos\left(\frac{1}{1117600889499} \, \sqrt{5943243628927378} \sqrt{124177876611} t\right) + \frac{5903974404321}{3578329562859854}$

sol3

$\newcommand{\Bold}[1]{\mathbf{#1}}-\frac{14260664380421}{68616988528874850} \, \cos\left(\frac{1121}{5404698185373} \, \sqrt{1801566061791} \sqrt{20602241561} t\right) + \frac{14260664380421}{68616988528874850}$

q = vector([sol1, sol2, sol3])

u = fi*q

s1 = plot( u[0], [0,4], color='blue', plot_points=500)
s2 = plot( u[1], [0,4], color='red', plot_points=500)
s3 = plot( u[2], [0,4], color='green', plot_points=500)
show(s1 + s2 + s3)

DKiPI_3_2018_s_izradom

2251 days ago by mdemsic@grad.hr