Guadagno e Fase in un sistema massa-molla sottoposto ad una forzante
26 Aug 2001

Scriviamo l'equazione che regge il moto
                                    m y''[t] + c y'[t] +k y[t]==[Graphics:../Images/index_gr_83.gif] Sin[wf t]
                                    ed operiamo le solite sostituzioni
Attenzione: se si usa il simbolo [Graphics:../Images/index_gr_84.gif] in luogo di  ωf la sol2 assume una forma diversa che richede
Transient e Steady diversi; inoltre, dà problemi nel Plot (da indagare meglio).
Nell'eqauzione che segue [Graphics:../Images/index_gr_85.gif] = [Graphics:../Images/index_gr_86.gif]/m.
  

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Anche se raccoglie in un modo un po' strano la parte con gli Exp ...

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i primi due termini vanno a zero per t ->∞; costituiscono la SOLUZIONE TRANSITORIA.
Il terzo termine non si annulla e costituisce la  SOLUZIONE STAZIONARIA.

Estraiamo la  SOLUZIONE TRANSITORIA

Transient  =sol2[[1]]
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Estraiamo la  SOLUZIONE  STAZIONARIA e manipoliamola algebricamente

Steady=Apart[sol2[[2]]]
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Ora calcoliamo l'ampiezza ( ⩵ guadagno)

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Sarà utile in seguito per osservare l'andamento del massimo dell'ampiezza

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La posizione del massimo dell'ampiezza dipende da ζ

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Usiamo la prima delle due  possibilità offerete da Mathematica  per calcolare l'Arctan

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Seconda possibilità

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Sotto forma compatta

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Per studiarne l'andamento dell'ampiezza, in funzione di ωf, per diversi valori di ζ,
dobbiamo  assegnare alcune costanti; osserviamo che, in condizioni statiche, la deflessione
sarebbe  [Graphics:../Images/index_gr_122.gif]/k = 0.2

m=2   (* kg   *);  k=5   (* N/m *)  ;  [Graphics:../Images/index_gr_123.gif]=1 (* N  *);  [Graphics:../Images/index_gr_124.gif] = [Graphics:../Images/index_gr_125.gif]/m (* metro/[Graphics:../Images/index_gr_126.gif]*);

[Graphics:../Images/index_gr_127.gif] //N    
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Plot[
Evaluate[Table[amp,{ζ,0.1,0.5,0.1}]],
{[Graphics:../Images/index_gr_129.gif],0.1,5.0},PlotRange->All,AxesOrigin->{0.1,0},
AxesLabel->{"[Graphics:../Images/index_gr_130.gif]","amp"},
PlotLabel->"Guadagno vs. [Graphics:../Images/index_gr_131.gif]",
PlotStyle->Table[RGBColor[2g,1-2g,0],{g,0.1,0.5,0.1}]];

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Vediamo come si sposta il max dell'ampiezza in funzione di ζ

Table[{ζ,terzozero},{ζ,0.1,0.5,0.1}]//TableForm
0.1` 1.565247584249853`
0.2` 1.51657508881031`
0.30000000000000004` 1.4317821063276353`
0.4` 1.3038404810405297`
0.5` 1.118033988749895`

Esaminiamo l'andamento della fase in funzione di ωf per diversi valori di ζ

Plot[
Evaluate[Table[ang0,{ζ,0.1,0.5,0.1}]],
{[Graphics:../Images/index_gr_133.gif],0.1,5.0},PlotRange->All,AxesOrigin->{0.1,0},
AxesLabel->{"[Graphics:../Images/index_gr_134.gif]","Fase"},
PlotLabel->"Fase (rad) vs. [Graphics:../Images/index_gr_135.gif]",
PlotStyle->Table[RGBColor[2g,1-2g,0],{g,0.1,0.5,0.1}]];

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Un' altra rappresentazione  della fase ...

Plot[
Evaluate[Table[-ang+Pi/2,{ζ,0.1,0.5,0.1}]],
{[Graphics:../Images/index_gr_137.gif],0.1,5.0},PlotRange->All,AxesOrigin->{0.1,0},
AxesLabel->{"[Graphics:../Images/index_gr_138.gif]","Fase"},
PlotLabel->"Fase (rad) vs. [Graphics:../Images/index_gr_139.gif]",
PlotStyle->Table[RGBColor[2g,1-2g,0],{g,0.1,0.5,0.1}]];

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Per osservare l'andamento dell'ampiezza  della soluzione stazionaria in funzione del tempo,
per una data frequeanza ωf della forzante ed uno dato ζ,

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tr=Plot[Evaluate[simpsolstaz0],{t,0,6},
PlotPoints->150,PlotStyle->{{Thickness[0.004],RGBColor[0,0,1]}},
PlotRange->All];

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Un semplice zoom nei primi istanti

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tr=Plot[Evaluate[simpsolstaz0],{t,0,0.4},
PlotPoints->150,PlotStyle->{{Thickness[0.004],RGBColor[0,0,1]}},
PlotRange->All];

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Avviciniamoci alla zona della risonanza (da dx)

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tr=Plot[Evaluate[simpsolstaz0],{t,0,6},
PlotPoints->150,PlotStyle->{{Thickness[0.004],RGBColor[0,0,1]}},
PlotRange->All];

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Riduciamo ancora la frequenza della forzante

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tr=Plot[Evaluate[simpsolstaz0],{t,0,6},
PlotPoints->150,PlotStyle->{{Thickness[0.004],RGBColor[0,0,1]}},
PlotRange->All];

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Utilizziamo NDSolve per studiare
     la risposta (totale=transitoria +stazionaria)  e
     lo sfasamento tra la forzante e la risposta (stazionaria)

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Converted by Mathematica      May 26, 2003