POPREČNE OSCILACIJE ŽICE

Poprečne oscilacije žice je gibanje  napete žice  duljine l koje nastaje uslijed nekog vanjskog djelovanja.

Pretpostavka:
 1.  poprečne oscilacije-gibanje u xy ravnini - progib žice u smjeru x osi jednak nuli
 2.  nema kidanja-mali progib  i male deformacije

Primjer 1.

\(Žica\ duljine\ l = 
          0.5\ m\ konstantne\ napetosti\ 0.01\ koja\ je\ učvršćena\ na\ oba\ kraja\ \ u\ početnom\ trenutku\ u\ obliku\ sinusoide\ 0.7\ Sin[
                x  π\/0.25]\  . \ 
\((a)\) 
            Koji\ oblik\ će\ poprimiti\ nakon\ \ 5.8 
             sec . \ \ 
\((b)\) 
            Koji\ oblik\ poprima\ u\ prizvoljnom\ trenutku\ \(\(t\)\(.\)\)\)

\(Ilustracija\ rješenja - \ 
            Početno\ rubni\ problem\  - oscilacija\ žice\) 

 \(\((a)\)\ \ Interesira\ nas\ progib\ žice\ na\ \ mjestu\ x\ \ \ \ u\ trenutku\ t = 5.8\ sec\)

\(u \((x, 5.8)\) = \(\(0.7\)\(\ \)\(Cos[\ 
              5.8 \( 2\ π\)\/5]\)\(\ \)\(Sin[
              4\ π\ x]\)\(
\)\(
\)\(
\)\)\)

(b) Promjena progiba u vremenu t = 0 do t = 
              7 sec


\(u \((x, t)\) = \(\(0.7\)\(\ \)\(Cos[\ 
              t \( 2\ π\)\/5]\)\(\ \)\(Sin[
              4\ π\ x]\)\(
\)\)\)

RJEŠENJE RUBNOG PROBLEMA-slobodne oscilacije žice

\(RUBNI\ PROBLEM\ za\ slobodne\ oscilacije\ žice\ l = 
          0.5\) , p (x) = p = 0.01, ρ (x) = ρ = 1, (c^2) = (p/ρ),α (x) = 0.7 Sin[x  (π/0.25)],β (x) =0:
(((∂^2) u (x, t))/∂) (t^2)=\(\(c\^2\) \(∂\^2\ u \((x, t)\)\)\/∂\ x\^2\ \((valna\ jednadžba)\)\)
\(\(\ \ \ \ \ \ \ \)\(u \((0, t)\) = 0, \ 
      u \((l, t)\) = 
        0\ \((\ rubovi\ žice\ učvršćeni)\)\)\)
\(\(\ \ \ \ \ \ \ \)\(u \((x, 
            0)\) = α\ \((x)\), \ \(\(∂\ u \((x, 0)\)\)\/\(\(∂\ t\)\(\ \)\) = β\ \((x)\)\ \((početni\ položaj\ i\ početna\ brzina)\);\)\)\)
                                                                                                     α (o) = 0,  α (l) = 0

Fourierova metoda (separacija varijabli )
\(\(u \((x, 
            t)\) = ∑\+\(n = 1\)\%∞\((A\_n\ cos \( \(n\)\(\ \)\(π\)\(\ \)\(c\)\(\ \)\)\/l\ t + 
                B\_n\ sin\ \(\(n\)\(\ \)\(π\)\(\ \)\(c\)\(\ \)\)\/l)\)\ sin\ \(n\ π\)\/l\ x;\)\)
\(\(\(A\_n = \(2\/l\) \(∫\_0\%l α \((x)\) 
              sin\ \(\(n\ π\ x\)\/l\) dx\);\)\(
\)
    \)\)
(Subscript[B, n]) = (2/(c n π)) (∫_0^l) β (x) 
              sin ((n π x)/l) dx;

Rješavanje  u Mathematica

l := 0.5;
p = 0.01;
c = Sqrt[p];
alfa = 0.7 Sin[x  (π/0.25)]

A[
          n_] := (2/l) (∫_0^l) 
              alfa*  Sin [((n π x)/l)] \[DifferentialD]x;
B[n_] := 0;

Table[{A[n], B[n]}, {n, 10}]

u[n_, x_, 
        t_] := (A[
                n] Cos[((n π c )/l) t] + 
            B[n] Sin [((n π c )/l) t]) *
        Sin [((n π)/l) x]

u[2, x, t]

u[x_, t_] := (∑_(n = 1)^10) u[n, x, t]

u[x, t]

u[x, 5.8]

Animacija za Primjer 1. (primjer1žica.nb)

u[x_, 0] :=  0.7 Sin[x  (π/0.25)]
u[x_, t_] := 
      0.7 Cos[ t (( 2 π)/5)] Sin[
          4 π x]
l = 0.5

\(\(\(
\)\(Table[
      Show[Plot[u[x, 0] // Evaluate, {x, 0, 0.5}, 
          PlotRange -> {{0, 1}, {\(-3\), 7}}, Frame -> True, 
          FrameTicks -> {{0, l}, {\(-1\), 
                1}}, \ \ PlotStyle\  -> {AbsoluteThickness[1], Hue[1]}, 
          DisplayFunction -> Identity], 
Plot[
          u[x, t] // Evaluate, {x, 0, 0.5}, Frame -> True, 
          FrameTicks -> {{0, l}, {\(-1\), 0, 
                1}}, 
\ \ PlotStyle\  -> {AbsoluteThickness[3], Hue[t/2]}, 
          DisplayFunction -> 
            Identity], 
Graphics[{AbsolutePointSize[5], 
            Hue[1], Point[{l/3, 
                u[l/3, t]}]}], 
Graphics[{AbsolutePointSize[
              5], Hue[1], 
            Point[{0, 0}]}], 
Graphics[{AbsolutePointSize[
              5], Hue[t/2], 
            Point[{0.8, t}]}], 
Graphics[{AbsolutePointSize[
              5], Hue[1], 
            Point[{l, 0}]}], 
Graphics[{AbsolutePointSize[
              5], Point[{l/3, 0}]}], 
Graphics[
          Text[StyleForm["u(x,0)", FontFamily -> Times, 
              FontSize -> 18], {l/3, 2}]], 
Graphics[
          Text[StyleForm["l", FontFamily -> Times, 
              FontSlant -> "Italic", 
              FontSize -> 16], {l, \(-0.4\)}]], 
Graphics[
          Text[StyleForm["0", FontFamily -> Times, 
              FontSlant -> "Italic", 
              FontSize -> 16], {0, \(-0.4\)}]], 
Graphics[
          Text[StyleForm["x", FontFamily -> Times, 
              FontSize -> 16], {l/3, \(-0.3\)}]], 
Graphics[
          Text[StyleForm[t\ \ "s", FontFamily -> Times, 
              FontSize -> 16], {0.9, t}]], 
Graphics[
          Text[StyleForm[\ "u(x,t)"\ , FontFamily -> Times, 
              FontSize -> 18], {l/
                3, \(-2\)}]], 
DisplayFunction -> $DisplayFunction, AspectRatio -> 1]
, \ {t, 0, 7, 
        0.2}]\)\)\)

Mathematica rješenje

Clear[alfa];
l = 1/2;
c = 0.1;
alfa[x_] := 0.7 Sin[4  x π] ;
u = y[x, t];
rjes = 
      NDSolve[{(∂_(t, t))u == c* (∂_(x, x))u,  
          y[x, 0] == 
            alfa[x], ((∂_t) u /. t -> 0) == 0, 
          y[0, t] == 0, y[l, t] == 0}, 
        y, {x, 0, l}, {t, 0, 8}]

Plot3D[Evaluate[y[x, t] /. First[rjes]],
   {x, 0, l}, {t, 0, 8}, PlotPoints->30]

<<Graphics`Animation`

Animate[
        Plot[Evaluate[y[x, t] /. First[rjes]], {x, 0, l}, 
          PlotRange -> {-1, 1}], {t, 0, 8, 0.2}];

Izvod VALNE JEDNADŽBE za poprečne oscilacije žice

Vektorska polja 
(Overscript[u,  ->])  : [0, l]×
              R -> (Subscript[X, 0]) (E)
 (Overscript[φ,  ->])  : [0, l]×
              R -> (Subscript[X, 0]) (E)
 (Overscript[ψ,  ->])  : [0, l]×
              R -> (Subscript[X, 0]) (E)
 (Overscript[f,  ->])  : [0, l]×
              R -> (Subscript[X, 0]) (E)
1.  kinematičko       
\(progib\ žice\ \ \(u\& -> \) \((x, 
                t)\) = \(u\_x\) \((x, 
                  t)\) \(i\& -> \)\  + \(u\_y\) \((x, 
                  t)\) \(j\& -> \)\) 
Pretpostavka 1. => (Subscript[u, x]) (x, t) ≈ 0   
       
 (Overscript[u,  ->])  (x, 
              t) =  (Subscript[u, y]) (x, t) (Overscript[j,  ->])   = 
            u (x, t) (Overscript[j,  ->])  

 Pretpostavka 2.  => \(mali\ progib\ \(∂\ u \((x, t)\)\)\/∂\ x ≪ 1\)
\(male\ deformacije - 
          apsolutni\ prirast\ progiba\ u\ odnosu\ na\ progib\ u\ ishodištu\ malen\ u\ odnosu\ na\ duljinu\ žice\)
\(\(\(|\)\(u \((x, t)\) - 
              u \((0, t)\)\)\(|\)\) = \(\(\(|\)\(∫\_0\%x\(\(∂\ u \((s, t)\)\)\/∂\ 
                        s\) ds\)\(|\)\(\(<\)\(∫\_0\%x\)\)\(|\)\(\(∂\ u \((s, t)\)\)\/∂\ 
                    s\)\(|\)\(ds ≪ \ ∫\_0\%x 1 ds\)\) = x ≪ l\)\) 
   2.  dinamička
\(\(-\ 
              gustoća\)\ količine\ gibanja\ \ \(φ\& -> \) \((x, 
              t)\) = \(φ\_x\) \((x, 
                t)\) \(i\& -> \)\  + \(φ\_y\) \((x, 
                t)\) \(j\& -> \)\) 
\((količina\ gibanja\ \ po\ jedinici\ duljine\ \ žice\ u\ točki\ x\ u\ trenutku\ t)\) 
- 
              kontaktna sila   (Overscript[ψ,  ->])  (x, 
              t) = (Subscript[ψ, x]) (x, 
                t) (Overscript[i,  ->])   + (Subscript[ψ, y]) (x, 
                t) (Overscript[j,  ->])  
\((količina\ gibanja\ koja\ se\ u\ točki\ x\ prenese\ s\ desna\ na\ lijevo\ u\ jedinici\ vremena\ uslijed\ \ neprekidnosti\ žice\ i\ napetosti)\)
- 
              gustoća vanjske sile  (Overscript[f,  ->])  (x, 
              t) = (Subscript[f, x]) (x, t) (Overscript[i,  ->])   + (Subscript[f, y]) (x, 
                t) (Overscript[j,  ->])  
\((količina\ gibanja\ po\ jedinici\ duljine\ žice\ \ koja\ se\ izvana\ prenese\ na\ žicu\ u\ jedinici\ vremena\ u\ točki\ x\ u\ trenutku\ t\ uslijed\ djelovanja\ vanjske\ sile\ , \ \ gravitacijske\ sile\ i\ \(\(sl\)\(.\)\))\)

ZAKONI PONAŠANJA

1.  Zakon 
 \(\(φ\& -> \) \((x, 
              t)\) = ρ \((x)\)\ \(∂\ \(u\& -> \) \((x, t)\)\)\/∂\ t\ \ \ , 
\ ρ \((x)\)\  - \ 
          linearna\ gustoća\ mase\ žice\ u\ točki\ x\ u\ tenutku\ t\)
 (Overscript[φ,  ->])  (x, 
            t) = ρ (x) ((∂ u (x, 
                  t))/∂) 
                t (Overscript[j,  ->]) 
2.  Zakon 

 (Overscript[ψ,  ->])  (x, t) =  
            p (x, t) (Overscript[T,  ->])  (x, t)  ,
 \(p \((x, t)\)\  - \ 
            napetost\ žice\ u\ točki\ x\ u\ trenutku\ t\),
 \(\(T\& -> \)\ \((x, t)\)\  - 
            jedinični\ vektor\ tangente\ \ na\ žicu\ u\ točki\ x\ u\ trenutku\ t\)  , (Overscript[T,  ->])  (x, 
              t)  = (Overscript[i,  ->])   + ((∂ u (x, 
                      t))/∂) 
                    x (Overscript[j,  ->]) 
 ((∂ u (x, t))/∂) 
            t
 \(\(ψ\& -> \) \((x, t)\) = 
        p \((x)\) \(i\& -> \)\  + 
          p \((x)\)\ \(\(∂\ u \((x, t)\)\)\/∂\ 
                  x\) \(j\& -> \)\ \ \ \ \ , 
      p \((x)\)\  - \ 
        napetost\ žice\ koja\ ne\ ovisi\ o\ vremenu\)
\(Zakon\ održanja\ količine\ gibanja\ \ u\ diferencijalnom\ obliku\)
\(\(Ukupna\ količina\ gibanja\ je\ jednaka\ \(nulvektoru : 
\(∂\ \(φ\& -> \) \((x, t)\)\)\/∂\ t\  - \(∂\ \(ψ\& -> \) \((x, t)\)\)\/∂\ x - \(f\& -> \) \((x, t)\)\)\)\(=\)\(0\& -> \)\(\ \)\)

\(Primjenjujući\ 1. \ i\ 2. \ zakon\ ponašanja\ u\ zakonu\ održanja\ količine\ gibanja\ \(dobivamo : 

\(\(*\)\(\ \)\(običnu\)\(\ \)\(diferencijalnu\)\(\ \)\(jednadžbu\)\(\ \)\(za\)\(\ \)\(napetost\)\(\ \)\(žice\)\)\)\) 
 p' (x)  +  (Subscript[f, x]) (x)  =  0;
\(\(\(*\)\(\ \)\(parcijalnu\)\(\ \)\(diferencijalnu\)\(\ \)\(jednadžbu\)\(\ \)\(za\)\(\ \)\(progib\)\(\ \)\(žice\)\)\  - 
        VALNU\ JEDNADŽBA\)

\(VALNA\ JEDNADŽBA\)

\(ρ \((x)\)\ \(∂\^2\ u \((x, t)\)\)\/∂\ t\^2 = \(∂\/∂\ 
                    x\) \((p \((x)\) \(∂\ u \((x, t)\)\)\/∂\ x)\) + \(f\_y\) \((x, t)\)\)

\(VALNA\ JEDNADŽBA\ U\ sredstvu\ s\ otporom\) 
Sila otpora sredstva  (Overscript[((Subscript[F, ot])),  ->])  (x, 
            t) = - q (x) u (x, t) (Overscript[j,  ->])  ,   q (x) ≥  0, x∈[0, l].

ρ(x) (((∂^2) u (x, t))/∂) (t^2)=\(\(∂\/∂\ 
                    x\) \((p \((x)\) \(∂\ u \((x, t)\)\)\/∂\ x)\) + \(f\_y\) \((x, t)\) - \ q \((x)\) u \((x, t)\)\)

RAVNOTEŽA ŽICE

\(\(-vanjsko\)\ djelovanje\ ne\ ovisi\ o\ vremenu\ \ \(f\&→\) \((x, 
            t)\) = \(\(f\&→\) \((x)\)\(\ \)\(ili\)\(\ \)\(se\)\(\ \)\(mijenja\)\(\ \)\(tako\)\(\ \)\(sporo\)\(\ \)\(da\)\(\ \)\(žica\)\(\ \)\(poprimi\)\(\ \)\(konačni\)\(\ \)\(položaj\)\(\ \)\(prije\)\(\ \)\(no\)\(\ \)\(što\)\(\ \)\(se\)\(\ \)\(djelovanje\)\(\ \)\(izmjeni\)\(\ \)\)\)
u (x, t) = u (x)
\(JEDNADŽBA\ RAVNOTEŽE\ ŽICE\) 
(p (x) u' (x))' + (Subscript[f, y]) (x) =  0;

\(JEDNADŽBA\ RAVNOTEŽE\ ŽICE\ \ u\ sredstvu\ s\ otporom\)
(p (x) u' (x))' -  
            q (x) u (x) + (Subscript[f, y]) (x) =  0;

KONCETRIRANO DJELOVANJE

\(KONCETRIRANO\ DJELOVANJE\ \ vanjske\ sile\ \ \ \(F\&→\) \((x\_\(\(0\)\(\ \)\), t)\) = F \((t)\) j\&→\)

za x<(Subscript[x, 0 ]) i x>(Subscript[x, 0 ]) vrijedi valna jednadžba:\(ρ \((x)\)\ \(∂\^2\ u \((x, t)\)\)\/∂\ t\^2\)=\(\(∂\/∂\ 
                      x\) \((p \((x)\) \(∂\ u \((x, t)\)\)\/∂\ x)\) + \(f\_y\) \((x, t)\)\)
\(\(-u\)\ točki\ \(x\_\(\(0\)\(\ \)\)\) 
              napete\ žice\ objesimo\ uteg\ mase\ m\ \ uz\ pretpostavku\ \  - \ nema\ kidanja\ žice\ \((u \((x, t)\)\ neprekinuta\ funkcija)\)\)
    u ((Subscript[x, 0 ]) + 0, t)  =  
              u ((Subscript[x, 0 ]) - 0, t);

\((u\ točki\ x\_\(\(0\)\(\ \)\)\ ne\ vrijedi\ izvod\ zakona\ o\ održanju\ količine\ gibanja\ ali\ 
promjena\ količine\ gibanja\ komada\ žice\ u\ ε\ okolini\ oko\ \(x\_\(\(0\)\(\ \)\)\) 
              u\ jedinici\ vremena\ jenaka\ je\ sili\ koja\ djeluje\ na\ taj\ mali\ dio\ žice)\)

	(Subscript[ψ, y]) ((Subscript[x, 0 ]) + 0, 
                  t)  -  (Subscript[ψ, y]) ((Subscript[x, 0 ]) - 0, t) + 
              F (t) = 0 
ili prema 2.  zakonu ponašanja

	\(\(∂\ u \((x\_0 + 0, t)\)\)\/∂\ x - \(∂\ u \((x\_0 - 0, t)\)\)\/∂\ x = \(-\(\(F \((t)\)\)\/\(p \((x\_0)\)\)\)\)\)

POČETNI I RUBNI UVJETI

U1. Kinematički (DIRICHLETOVI, geometrijski) rubni uvjeti određuju vrijednost kinematičkog polja (Overscript[u, →])(x,t) na rubovima.
	Dirichletov rubni uvjet za poprečne oscilacije žice: u(0,t)= a, u(l,t)= b (rubovi učvršćeni).

U2. Dinamički (NEUMANNOV, prirodni) rubni uvjeti određuju vrijednost dinamičkog polja  polja (Overscript[ψ, →])(x,t) na rubovima.
	Neumannov  rubni uvjet za poprečne oscilacije žice: (Subscript[ψ, y])(0,t)= a, (Subscript[ψ, y])(l,t)= b (na rubu zadana kontaktna sila):
prema 2. zakonu ponašanja p (0) ((∂ u (0, t))/∂) x= a, p (l) ((∂ u (l, t))/∂) x= b;
ako je homogeni Neumannov uvjet = rub slobodan!

P. Početni uvjeti  za kinematičko polje u trenutku t=0.
Početni uvjeti moraju biti kompatibilni s rubnim uvjetima \(\((na\ rubovima\ moraju\ zadovoljavati\ rubne\ uvjete\ \(∀ \ t\))\) . 
Početni\) uvjeti za poprečne oscilacije žice : u (x, 0)=α (x), \(\(∂\ u \((x, 0)\)\)\/\(\(∂\ t\)\(\ \)\) = β\ \((x)\)\ \((početni\ položaj\ i\ početna\ brzina)\)\);
                                                                                                     α (o) = a,  α (l) = b

PRIMJERI RUBNIH PROBLEMA za  ravnotežu žice

RAVNOTEŽA ŽICE

Primjer A

Homogena žica napeta je horizontalnoutegom mase M>0 na kraju x=0. Odredite ravnotežni progib ako je drugi kraj (a)učvršćen; (b) slobodan

\(\((a)\)\ Rubni\ \(problem\  : 
\((p \((x)\)\ u' \((x)\))\)' + \ \(f\_y\) \((x)\)\) = 0\)
\(	\(u \((0)\) = \(u \((l)\) = 0\);\)\)
\(\((b)\)\ Rubni\ \(problem\  : 
\((p \((x)\)\ u' \((x)\))\)' + \ \(f\_y\) \((x)\)\) = 0\)
\(	u \((0)\) = 0, \ \(u' \((l)\) = 0;\)\)
\(Napetost\ p \((x)\) = g\ M, \ 
    linijska\ gustoća\ homogene\ žice\ ρ \((x)\) = ρ, \ gustoć\ vanjske\ sile\ je\ \ \(f\_y\) \((x)\) = \(ρ\ \(g : 
\((a)\)\ Rubni\ \(problem : 
g\ M\ u'' \((x)\) + \ ρ\ g\)\) = 0\)\)
\(		u \((0)\) = \(u \((l)\) = 0. \)\)
\(\((b)\)\ Rubni\ \(problem : 
g\ M\ u'' \((x)\) + \ ρ\ g\) = 0\)
\(		u \((0)\) = 0, \ u' \((l)\) = 0. \)

\(Rješenje\ rubnog\ \(problema : 
Rješenje\ obične\ diferencijalne\ jednadžbe\ u \((x)\)\) = \(-\(ρ\/\(2\ M\)\)\) x\^2 + \(c\_1\) x\  + 
        c\_2\)
(a) u (x) = (ρ/(2 M)) 
          x (l - x);
(b) u (x) = (ρ/(2 M)) 
        x (2  l - x).

(a)

(b)

Rad za primjer

Clear[ur];
ur[l_, ro_, M_, x_] := (ro/(2 M)) 
          x (l - x);
l = 10;
ro = 0.1;
\(Table[
      Show[Plot[\(-ur[l, ro, ro, x]\) // Evaluate, {x, 0, l}, 
          PlotRange -> {{\(-1\), l}, {\(-20\), 20}}, Frame -> True, 
          FrameTicks -> {{0, 
                l}, {}}, \ \ \ PlotStyle\  -> {AbsoluteThickness[1], Hue[1]}, 
          DisplayFunction -> 
            Identity], 
Plot[\(-ur[l, ro, M, x]\) // 
            Evaluate, {x, 0, 
            l}, 
\ \ PlotStyle\  -> {AbsoluteThickness[3], Hue[M/2]}, 
          DisplayFunction -> 
            Identity], 
Graphics[{AbsolutePointSize[5], 
            Hue[1], Point[{l/3, 
                ur[l, ro, M, 
                  0]}]}], 
Graphics[{AbsolutePointSize[5], 
            Hue[1], Point[{0, 
                0}]}], 
Graphics[{AbsolutePointSize[5], 
            Hue[M/2], 
            Point[{0, \(-M\)}]}], 
Graphics[{AbsolutePointSize[5], Hue[1], 
            Point[{l, 0}]}], 
Graphics[{AbsolutePointSize[
              5], Point[{l/3, 0}]}], 
Graphics[
          Text[StyleForm["u(x)", FontFamily -> Times, 
              FontSize -> 18], {l/3, 5}]], 
Graphics[
          Text[StyleForm["l", FontFamily -> Times, 
              FontSlant -> "Italic", 
              FontSize -> 16], {l + 
                0.5, \(-0.4\)}]], 
Graphics[
          Text[StyleForm["0", FontFamily -> Times, 
              FontSlant -> "Italic", 
              FontSize -> 
                16], {\(-0.5\), \(-0.4\)}]], 
Graphics[
          Text[StyleForm["x", FontFamily -> Times, 
              FontSize -> 16], {l/3, \(-1\)}]], 
Graphics[
          Text[StyleForm[M\ \ "kg", FontFamily -> Times, 
              FontSize -> 
                16], {1, \(-M\)}]], 
DisplayFunction -> $DisplayFunction, AspectRatio -> 1]
, \ {M, 
        0.1, 7, 0.5}]\)

Clear[ur];
ur[l_, ro_, M_, x_] := (ro/(2 M)) 
          x (2*l - x);
l = 10;
ro = 0.1;
\(Table[
      Show[Plot[\(-ur[l, ro, ro, x]\) // Evaluate, {x, 0, l}, 
          PlotRange -> {{\(-1\), l}, {\(-60\), 10}}, Frame -> True, 
          FrameTicks -> {{0, 
                l}, {}}, \ \ \ PlotStyle\  -> {AbsoluteThickness[1], Hue[1]}, 
          DisplayFunction -> 
            Identity], 
Plot[\(-ur[l, ro, M, x]\) // 
            Evaluate, {x, 0, 
            l}, 
\ \ PlotStyle\  -> {AbsoluteThickness[3], Hue[M/2]}, 
          DisplayFunction -> 
            Identity], 
Graphics[{AbsolutePointSize[5], 
            Hue[1], Point[{l/3, 
                ur[l, ro, M, 
                  0]}]}], 
Graphics[{AbsolutePointSize[5], 
            Hue[1], Point[{0, 
                0}]}], 
Graphics[{AbsolutePointSize[5], 
            Hue[M/2], 
            Point[{0, \(-M\)}]}], 
Graphics[{AbsolutePointSize[5], Hue[1], 
            Point[{l, 0}]}], 
Graphics[{AbsolutePointSize[
              5], Point[{l/3, 0}]}], 
Graphics[
          Text[StyleForm["u(x)", FontFamily -> Times, 
              FontSize -> 18], {l/3, 5}]], 
Graphics[
          Text[StyleForm["l", FontFamily -> Times, 
              FontSlant -> "Italic", 
              FontSize -> 16], {l + 
                0.5, \(-0.4\)}]], 
Graphics[
          Text[StyleForm["0", FontFamily -> Times, 
              FontSlant -> "Italic", 
              FontSize -> 
                16], {\(-0.5\), \(-0.4\)}]], 
Graphics[
          Text[StyleForm["x", FontFamily -> Times, 
              FontSize -> 16], {l/3, \(-3\)}]], 
Graphics[
          Text[StyleForm[M\ \ "kg", FontFamily -> Times, 
              FontSize -> 
                16], {1, \(-M\)}]], 
DisplayFunction -> $DisplayFunction, AspectRatio -> 1]
, \ {M, 
        0.1, 7, 0.5}]\)

Rješavanje u Mathematica

DSolve[ {M y''[x] == - ρ ,y[0] == 0, y[d]==0}, y[x], x ]

DSolve[ {M y''[x] == - ρ ,y[0] == 0, y'[d]==0}, y[x], x ]

RAVNOTEŽA ŽICE s koncentriranim djelovanjem

Primjer B

\(Odrediti\ ravnotežno\ stanje\ žice\ ako\ je\ vanjska\ sila\ F\ koncentrirana\ u\ točki\ x\_0 ∈ \(\(\([0, 
              l]\) . \ 
            Teška\)\(\ \)\(žica\)\(\ \)\(sastavljena\)\(\ \)\(od\)\(\ \)\(dva\)\(\ \)\(homogena\)\(\ \)\(komada\)\(\ \)\(jednake\)\(\ \)\(duljine\)\(\ \)\(s\)\(\ \)\(linijskim\)\(\ \)\(gustoćama\)\(\ \)\(mase\)\(\ \)\(ρ\_1\) \(i\)\(\ \)\(ρ\_2\)\(\ \)\(i\)\(\ \)\(napeta\)\(\ \)\(horizontalno\)\(\ \)\(utegom\)\(\ \)\)\)
\(mase\ M > 
        0. \ Odrediti\ ravnotežni\ progib\ ako\ je\ žica\ u\ sredini\ opterećena\ utegom\ mase\ M\_1 > \ 
        0\ i\ ako\ je\ drugi\ kraj\ \(\(učvršćen\)\(.\)\)\)

\(Napetost\ \ p \((x)\) = g\ M, \ 
    linijske\ gustoće\ homogene\ žice\ \(ρ\_1\) \((x)\) = ρ\_1, \(ρ\_2\) \((x)\) = ρ\_2, \ 
    gustoće\ vanjskih\ sila\ su\ \ \(f\_1\) \((x)\) = \(\(\(ρ\)\(\ \)\)\_1\) 
        g, \(f\_2\) \((x)\) = \(ρ\_2\ \(g : 
Rubni\ \(problem : 
g\ M\ u'' \((x)\) + \ ρ\_1\ g\)\) = 
        0\)\ , \ x < l\/2\)
\(		g\ M\ u'' \((x)\) + \ ρ\_2\ g = 0, \ \ x > 
      l\/2\)
\(		u \((0)\) = \(u \((l)\) = 0. \)\)
\(uvjeti\ koncentriranog\ djelovanja\ u\ x = \(\(l\/2\) : 
u \((l\/2 + 0)\) = 
        u \((l\/2 - 0)\)\), 
u' \((l\/2 + 0)\) - 
        u' \((l\/2 - 
              0)\) = \(-\(M\_1\/\(g\ M\)\)\)\)

Rješenje rubnog problema 
u (x)={({{-((Subscript[ρ, 1])/(2 M)) (x^2) +  
                      C x,,  x < (l/2)}, {-((Subscript[ρ, 2])/(2 M)) ((x^2) - (l^2)) + 
                      D (x - l),,  x > (l/2)}})
C=\(M\_1\/\(2\ M\) + \(l\/\(8\ M\)\) \((3  ρ\_1 + ρ\_2\ )\)\),
\(D = \(-\(M\_1\/\(2\ M\)\)\) + \(l\/\(8\ M\)\) \((\(-ρ\_1\) + \ 5  ρ\_2\ )\)\)

Rad za primjer

Clear[urk1, urk2, Cu, Du, M1];
l = 10;
ro1 = 0.9;
ro2 = 0.3;
M = ro1 + ro2;
Cu[M1_] := 
        (M1/(2 M)) + (l/(8 M)) (3  ro1 + 
                ro2);
Du[
          M1_] := -(M1/(2 M)) + (l/(8 M)) (-ro1 +  
                5  ro2 );
urk1[l_, ro1_, ro2_, M_, M1_, x_] := -(ro1/(2 M)) (x^2) +  
          Cu[M1] x;
urk2[l_, ro1_, ro2_, M_, M1_, 
          x_] := -(ro2/(2 M)) ((x^2) - (l^2)) + 
          Du[M1] (x - l);
\(Table[
      Show[Plot[
          Which[x < l\/2, \(-urk1[l, ro1, ro2, M, 0, x]\), 
              x > l\/2, \(-urk2[l, ro1, ro2, M, 0, x]\)] // Evaluate, {x, 0, 
            l}, PlotRange -> {{\(-2\), l}, {\(-30\), 10}}, 
          Frame -> True, 
          FrameTicks -> {{0, 
                l}, {}}, \ \ \ PlotStyle\  -> {AbsoluteThickness[1], Hue[1]}, 
          DisplayFunction -> Identity], 
Plot[
          Which[x < l\/2, \(-urk1[l, ro1, ro2, M, M1, x]\), 
              x > l\/2, \(-urk2[l, ro1, ro2, M, M1, x]\)] // Evaluate, {x, 0, 
            l}, 
\ \ PlotStyle\  -> {AbsoluteThickness[3], Hue[M1/2]}, 
          DisplayFunction -> 
            Identity], 
Graphics[{AbsolutePointSize[5], 
            Hue[1], Point[{l/
                  2, \(-urk2[l, ro1, ro2, M, M1, 
                    l/2]\)}]}], 
Graphics[{AbsolutePointSize[5], Hue[1], 
            Point[{0, 0}]}], 
Graphics[{AbsolutePointSize[
              5], Hue[1], 
            Point[{l/2, 0}]}], 
Graphics[{AbsolutePointSize[
              5], Hue[M1/2], 
            Point[{l/
                  2, \(-M1\)}]}], 
Graphics[{AbsolutePointSize[5], Hue[M/2], 
            Point[{0, \(-M\)}]}], 
Graphics[{AbsolutePointSize[5], Hue[1], 
            Point[{l, 0}]}], 
Graphics[
          Text[StyleForm["u(x)", FontFamily -> Times, 
              FontSize -> 18], {l/3, 5}]], 
Graphics[
          Text[StyleForm["l", FontFamily -> Times, 
              FontSlant -> "Italic", FontSize -> 16], {l + 0.5, 
              0.4}]], 
Graphics[
          Text[StyleForm["0", FontFamily -> Times, 
              FontSlant -> "Italic", FontSize -> 16], {\(-0.5\), 
              0.4}]], 
Graphics[
          Text[StyleForm["M", FontFamily -> Times, 
              FontSlant -> "Italic", 
              FontSize -> 
                14], {\(-2\), \(-M\)}]], 
Graphics[
          Text[StyleForm[\*""(Subscript[x, 0])"", FontFamily -> Times, 
              FontSize -> 16], {l/2, 3}]], 
Graphics[
          Text[StyleForm[M1\ "kg", FontFamily -> Times, 
              FontSize -> 16], {l/2 + 
                1, \(-M1\)}]], 
Graphics[
          Text[StyleForm[M\ \ "kg", FontFamily -> Times, 
              FontSize -> 
                14], {\(-1\), \(-M\)}]], 
DisplayFunction -> $DisplayFunction, AspectRatio -> 1]
, \ {M1, 
        0.1, 7, 0.5}]\)

SLOBODNE OSCILACIJE ŽICE

\(Slobodne\ oscilacije\ žice\ \ nastaju\ uslijed\ početnih\ uvjeta\ bez\ utjecaja\ vanjske\ sile\ \((\(f\_y\) \((x, t)\) = 
              0)\) . 

RUBNI\ PROBLEM\ za\ slobodne\ oscilacije\ žice\ p \((x)\) = 
      p, \ ρ \((x)\) = ρ, \ 
    c\^2 = \(\(p\/ρ\) : 
\(∂\^2\ u \((x, t)\)\)\/∂\ t\^2 = \(c\^2\) \(∂\^2\ u \((x, t)\)\)\/∂\ x\^2\ \((valna\ jednadžba)\)\)\)
\(\ \ \ \ \ \ \ u \((0, t)\) = 0, \ 
    u \((l, t)\) = 
      0\ \((\ rubovi\ žice\ učvršćeni)\)\)
\(\ \ \ \ \ \ \ u \((x, 
          0)\) = α\ \((x)\), \ \(\(∂\ u \((x, 0)\)\)\/\(\(∂\ t\)\(\ \)\) = β\ \((x)\)\ \((početni\ položaj\ i\ početna\ brzina)\);\)

    \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ α\ \((o)\) = 
      0, \ α\ \((l)\) = 0\)

Fourierova metoda (separacija varijabli )
\(\(u \((x, 
            t)\) = ∑\+\(n = 1\)\%∞\((A\_n\ cos \( \(n\)\(\ \)\(π\)\(\ \)\(c\)\(\ \)\)\/l\ t + 
                B\_n\ sin\ \(\(n\)\(\ \)\(π\)\(\ \)\(c\)\(\ \)\)\/l)\)\ sin\ \(n\ π\)\/l\ x;\)\)
\(\(\(A\_n = \(2\/l\) \(∫\_0\%l α \((x)\) 
              sin\ \(\(n\ π\ x\)\/l\) dx\);\)\(
\)
    \)\)
(Subscript[B, n]) = (2/(c n π)) (∫_0^l) β (x) 
              sin ((n π x)/l) dx;

Primjer 2

\(Riješite\ rubni\ problem\ oscilacije\ homogene\ žice\ duljine\ π\ \ koja\ je\ na\ rubovima\ učvršćena\ , \((\ c = 1)\), \ 
      a\ početna\ brzina\ je\ β \((x)\) = \(1\/4\) sin\ 2\ x\)

Rješavanje  u Mathematica

l := Pi;

c = 1;
alfa = 0;
beta = 1/4 Sin[2 x]

A[n_] := 0;
B[
            n_] := (2/(c n π)) (∫_0^l) beta* 
                Sin [((n π x)/l)] \[DifferentialD]x;;

Table[{A[n], B[n]}, {n, 10}]

u[n_, x_, 
        t_] := (A[
                n] Cos[((n π c )/l) t] + 
            B[n] Sin [((n π c )/l) t]) *
        Sin [((n π)/l) x]

u[2, x, t]

u[x_, t_] := (∑_(n = 1)^10) u[n, x, t]

u[x, t]

Animacija za Primjer 2.

Clear[u];
u[x_, 0] :=  0;
u[x_, t_] := (1/8) Sin[ 2 t] Sin[2 x]
l = π;

\(\(\(
\)\(Table[
      Show[Plot[u[x, t] // Evaluate, {x, 0, l}, 
          PlotRange -> {{\(-1\), l + 1}, {\(-1\), 3}}, 
          Frame -> True, 
          FrameTicks -> {{0, l}, {\(-1\), 
                1}}, \ \ 
\ \ PlotStyle\  -> {AbsoluteThickness[3], 
              Hue[t/2]}, 
          DisplayFunction -> 
            Identity], 
Graphics[{AbsolutePointSize[5], 
            Hue[1], Point[{l/3, 
                u[l/3, t]}]}], 
Graphics[{AbsolutePointSize[
              5], Hue[1], 
            Point[{0, 0}]}], 
Graphics[{AbsolutePointSize[
              5], Hue[t/2], 
            Point[{t, 2}]}], 
Graphics[{AbsolutePointSize[
              5], Hue[1], 
            Point[{l, 0}]}], 
Graphics[{AbsolutePointSize[
              5], Point[{l/3, 0}]}], 
Graphics[
          Text[StyleForm["l", FontFamily -> Times, 
              FontSlant -> "Italic", 
              FontSize -> 16], {l + 
                0.2, \(-0.3\)}]], 
Graphics[
          Text[StyleForm["0", FontFamily -> Times, 
              FontSlant -> "Italic", 
              FontSize -> 
                16], {\(-0.2\), \(-0.3\)}]], 
Graphics[
          Text[StyleForm["x", FontFamily -> Times, 
              FontSize -> 16], {l/3, \(-0.3\)}]], 
Graphics[
          Text[StyleForm[t\ \ "s", FontFamily -> Times, 
              FontSize -> 16], {t, 2.2}]], 
Graphics[
          Text[StyleForm[\ "u(x,t)"\ , FontFamily -> Times, 
              FontSize -> 18], {l/3, 
              1}]], 
DisplayFunction -> $DisplayFunction, AspectRatio -> 1]
, \ {t, 0, 5, 
        0.2}]\)\)\)

Mathematica rješenje

Clear[alfa, beta];
l = π;
c = 1;
alfa[x_] := 0 ;
beta[x_] := 1/4 Sin[2 x] 
u = y[x, t];
rjes = 
      NDSolve[{(∂_(t, t))u == c* (∂_(x, x))u,  
          y[x, 0] == 
            alfa[x], ((∂_t) u /. t -> 0) == beta[x], 
          y[0, t] == 0, y[l, t] == 0}, 
        y, {x, 0, l}, {t, 0, 8}]

Plot3D[Evaluate[y[x, t] /. First[rjes]],
   {x, 0, l}, {t, 0, 8}, PlotPoints->30]

<<Graphics`Animation`

Animate[
        Plot[Evaluate[y[x, t] /. First[rjes]], {x, 0, l}, 
          PlotRange -> {-1, 1}], {t, 0, 8, 0.2}];

PRISILNE  OSCILACIJE ŽICE

\(Prisilne\ oscilacije\ žice\ \ nastaju\ uslijed\ zadane\ \ vanjske\ sile\ \((gustoće\ vanjske\ sile\ , \ \(f\_y\) \((x, t)\) ≠ 
              0)\) . 

RUBNI\ PROBLEM\ za\ prisilne\ oscilacije\ žice\ p \((x)\) = 
      p, \ ρ \((x)\) = ρ, \ c\^2 = p\/ρ, \ 
    h \((x, t)\) = \(\(\(\(f\_y\) \((x, 
                  t)\)\)\/ρ\) : 
\(∂\^2\ u \((x, t)\)\)\/∂\ t\^2 = \(c\^2\) \(∂\^2\ u \((x, t)\)\)\/∂\ x\^2\  + \ \ h \((x, 
              t)\) \((valna\ jednadžba)\)\)\)
\(\ \ \ \ \ \ \ u \((0, t)\) = 0, \ 
    u \((l, t)\) = 
      0\ \((\ rubovi\ žice\ učvršćeni)\)\)
\(\ \ \ \ \ \ \ u \((x, 
          0)\) = α\ \((x)\), \ \(\(∂\ u \((x, 0)\)\)\/\(\(∂\ t\)\(\ \)\) = β\ \((x)\)\ \((početni\ položaj\ i\ početna\ brzina)\);\)\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ α\ \((o)\) = 
      0, \ α\ \((l)\) = 0\)

Fourierova metoda (separacija varijabli )
u (x, 
            t) = (∑_(n = 1)^∞) 
                (Subscript[D, n ]) (t) sin ((n π)/l) x;
(Subscript[B, n]) (t) = (2/l) (∫_0^l) h (x, t) 
              sin ((n π x)/l) dx;
\(\(D\_n'' \((t)\) + \((\(c\ n\ π\)\/l)\)\^2 + \ \(D\_n\) \((t)\) = \(B\_n\) \((t)\);\)\)
\(\(\(D\_n\) \((0)\) = \(2\/l\) \(∫\_0\%l α \((x)\) 
              sin\ \(\(n\ π\ x\)\/l\) dx\);\)
\)
\(\(D\_n' \((0)\) = \(2\/\(c\ n\ π\)\) \(∫\_0\%l β \((x)\) sin\ \(\(n\ π\ x\)\/l\) dx\);\)\)

Primjer 3

\(Riješite\ rubni\ problem\ \ prisilnih\ oscilacija\ \ homogene\ žice\ duljine\ π\ \ uslijed\ djelovanja\ vanjske\ sile\ h \((x, t)\) = 
        x\ i\ koja\ je\ na\ rubovima\ učvršćena\ , \((\ c = 1)\), \ 
      a\ početna\ brzina\ je\ β \((x)\) = \(1\/4\) 
          sin\ 2\ \(\(x\)\(.\)\)\)

Mathematica rješenje

Clear[alfa, beta];
l = π;
c = 1;
alfa[x_] := 0 ;
beta[x_] := 1/4 Sin[2 x] 
u = y[x, t];
rjes = 
      NDSolve[{(∂_(t, t))u == 
            c* (∂_(x, x))u  + 2 x,  
          y[x, 0] == 
            alfa[x], ((∂_t) u /. t -> 0) == beta[x], 
          y[0, t] == 0, y[l, t] == 0}, 
        y, {x, 0, l}, {t, 0, 8}]

Plot3D[Evaluate[y[x, t] /. First[rjes]],
   {x, 0, l}, {t, 0, 8}, PlotPoints->30]

<<Graphics`Animation`

Animate[
        Plot[Evaluate[y[x, t] /. First[rjes]], {x, 0, l}, 
          PlotRange -> {-1, 10}], {t, 0, 8, 0.2}];