vibration monochromatique - Free
TB3 / TD Physique Vibratoire : vibrations monochromatiques. Séance n°2 .... de
la vibration quasi-sinusoïdale étudiée dans l'exercice corrigé numéro 4, page 1.
part of the document
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\pvpara\posy0\dxfrtext141\dfrmtxtx141\dfrmtxty0\adjustright \shading2000 \fs20\lang1036\cgrid {\b\fs32 Rappels de Cours
\par }\pard \nowidctlpar\widctlpar\adjustright {\page
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\par }\pard \nowidctlpar\widctlpar\box\brdrsh\brdrs\brdrw15\brsp20 \pvpara\posy0\dxfrtext141\dfrmtxtx141\dfrmtxty0\adjustright \shading2000 {\b\fs32 Exercices avec \'e9l\'e9ments de correction
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\par
\par
\par }\pard\plain \s1\sb240\nowidctlpar\widctlpar\outlinelevel0\adjustright \b\f1\ul\lang1036\cgrid {Exercice 1 : vibration monochromatique
\par }\pard\plain \qj\nowidctlpar\widctlpar\adjustright \fs20\lang1036\cgrid {\fs24
\par Une vibration monochromatique a pour expression r\'e9elle : v}{\dn6 1}{\fs24 (t) = - 5 cos (6}{\f3\fs24 \'70}{\fs24 t-}{\f3\fs24 \'70}{\fs24 /4)
\par
\par 1.1) Mettre v}{\dn6 1}{\fs24 (t) sous la forme : v}{\dn6 1}{\fs24 (t)= A}{\dn6 1}{\fs24 cos (}{\f3\fs24 \'77}{\fs24 t-}{\f3\fs24 \'6a}{\dn6 1}{\fs24 ) en pr\'e9cisant les valeurs de A}{\dn6 1}{\fs24
\par (avec A}{\dn6 1}{\fs24 >0), }{\f3\fs24 \'77}{\fs24 et }{\f3\fs24 \'6a}{\dn6 1}{\fs24 .
\par 1.2) Ecrire le signal analytique }{\fs24\ul v}{\dn6 1}{\fs24 (t) associ\'e9 \'e0 v}{\dn6 1}{\fs24 (t), ainsi que l'\'e9quation qui lie }{\fs24\ul v}{\dn6 1}{\fs24 (t) \'e0 d}{\fs24\ul v}{\dn6 1}{\fs24 (t)/dt
\par
\par 1.3) M\'eames questions pour v}{\dn6 2}{\fs24 (t) = 3 sin (}{\f3\fs24 \'70}{\fs24 t-}{\f3\fs24 \'70}{\fs24 /6).
\par }\pard\plain \s1\sb240\nowidctlpar\widctlpar\outlinelevel0\adjustright \b\f1\ul\lang1036\cgrid {Exercice 2 :
\par }\pard\plain \nowidctlpar\widctlpar\adjustright \fs20\lang1036\cgrid {\fs24
\par Quelle est l'amplitude A}{\fs24\dn6 r}{\fs24 et le retard de phase }{\f3\fs24 \'6a}{\fs24\dn6 r}{\fs24 de la vibration r\'e9sultant de la superposition de
\par v}{\fs24\dn6 1}{\fs24 ( t ) = A}{\fs24\dn6 1}{\fs24 cos (}{\f3\fs24 \'77}{\fs24 t-}{\f3\fs24 \'6a}{\fs24\dn6 1}{\fs24 ) avec v}{\fs24\dn6 2 }{\fs24 ( t ) = A}{\fs24\dn6 2}{\fs24 cos (}{\f3\fs24 \'77}{\fs24 t-}{\f3\fs24 \'6a}{\fs24\dn6 2}{\fs24 ) ?
\par }\pard \qj\nowidctlpar\widctlpar\adjustright {\fs24 On \'e9tablira le r\'e9sultat par la m\'e9thode trigonom\'e9trique, puis par la m\'e9thode complexe.
\par }\pard\plain \s1\sb240\nowidctlpar\widctlpar\outlinelevel0\adjustright \b\f1\ul\lang1036\cgrid {Exercice 3 :
\par }\pard\plain \qj\nowidctlpar\widctlpar\adjustright \fs20\lang1036\cgrid {\fs24
\par Montrer que l'amplitude de la vibration r\'e9sultant de la superposition de
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t }{\b\fs24 +}{\fs24 }{\f3\fs24 \'6a}{\fs24 ) peut s'\'e9crire : }{\pard\plain \nowidctlpar\widctlpar\adjustright \lang1036\cgrid {\object\objemb\objw1219\objh800{\*\objclass Equation.3}{\*\objdata 01050000020000000b0000004571756174696f6e2e330000000000
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\par }\pard\plain \s1\sb240\nowidctlpar\widctlpar\outlinelevel0\adjustright \b\f1\ul\lang1036\cgrid {Exercice 4 :
\par }\pard\plain \nowidctlpar\widctlpar\adjustright \fs20\lang1036\cgrid {\fs24
\par Une vibration quasi-monochromatique }{\fs24\ul v}{\fs24 (t) peut \'eatre consid\'e9r\'e9e comme la somme de trois vibrations monochromatiques }{\fs24\ul v}{\dn6 o}{\fs24 (t), }{\fs24\ul v}{\dn6 1}{\fs24 (t) et }{\fs24\ul v}{\dn6 2 }{\fs24
(t), d'amplitudes respectives A}{\dn6 o }{\fs24 , A}{\dn6 1 }{\fs24 = A}{\dn6 o}{\fs24 /2 = A}{\dn6 2}{\fs24 et de fr\'e9quences respectives }{\f3\fs24 \'6e}{\dn6 0}{\fs24 , }{\f3\fs24 \'6e}{\dn6 1}{\fs24 =}{\f3\fs24 \'6e}{\dn6 0}{\fs24 - }{\f3\fs24
\'44\'6e\'2f\'32}{\fs24 et }{\f3\fs24 \'6e}{\dn6 2}{\fs24 =}{\f3\fs24 \'6e}{\dn6 0}{\fs24 + }{\f3\fs24 \'44\'6e\'2f\'32}{\fs24 avec }{\b\f3\fs24 \'44\'6e}{\b\fs24 0.
\par
\par 2.2) Quelle est l\rquote amplitude A et la phase \'e0 l\rquote origine du temps de s}{\fs24\sub 1}{\fs24 (t) ?
\par }\pard\plain \s1\sb240\nowidctlpar\widctlpar\outlinelevel0\adjustright \b\f1\ul\lang1036\cgrid {Exercice 3 : modulation d\rquote amplitude
\par }\pard\plain \nowidctlpar\widctlpar\adjustright \fs20\lang1036\cgrid {
\par }{\fs24 Dessiner le chronogramme de la vibration quasi-sinuso\'efdale \'e9tudi\'e9e dans l\rquote exercice corrig\'e9 num\'e9ro 4, page 1. On supposera l\rquote hypoth\'e8se }{\pard\plain \nowidctlpar\widctlpar\adjustright \lang1036\cgrid {\object\objemb
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\par
\par }\pard\plain \s1\sb240\nowidctlpar\widctlpar\outlinelevel0\adjustright \b\f1\ul\lang1036\cgrid {Exercice 4 : vibration quasi-monochromatique amortie exponentiellement.
\par }\pard\plain \nowidctlpar\widctlpar\adjustright \fs20\lang1036\cgrid {\fs24
\par Soit : v (t) = A}{\dn6 o}{\fs24 [ cos 2}{\f3\fs24 \'70\'6e}{\f3\dn6 \'6f}{\fs24 t ] exp(-t / }{\f3\fs24 \'74}{\fs24 ) pour t >0 (voir }{\b\fs24 figure 2, page 2}{\fs24 ).
\par On appelle }{\f3\fs24 \'74}{\fs24 la dur\'e9e de relaxation en amplitude de la vibration, qui correspond \'e0 la tangente \'e0 l'origine de la repr\'e9sentation temporelle.
\par
\par 4.1) Quel est le spectre }{\fs24\ul V}{\fs24 (n) de cette vibration ?
\par
\par 4.2) En d\'e9duire le carr\'e9 du module du spectre et trouver la relation entre t et la largeur totale \'e0 mi-hauteur }{\f3\fs24 \'44\'6e}{\dn6 1/2 }{\fs24 de | }{\fs24\ul V}{\fs24 (n)|}{\up6 2}{\fs24 ?
\par
\par }\pard\plain \s1\sb240\nowidctlpar\widctlpar\outlinelevel0\adjustright \b\f1\ul\lang1036\cgrid {Exercice 5 : }{\f0 Force \'e9lectromotrice d'un alternateur
\par }\pard\plain \qj\nowidctlpar\widctlpar\adjustright \fs20\lang1036\cgrid {\fs24
\par On ajoute en s\'e9rie les f.e.m de 30 conducteurs r\'e9guli\'e8rement r\'e9partis sur un arc de 60\'b0 du stator d'un alternateur monophas\'e9.
\par Le rotor comporte un p\'f4le nord et un p\'f4le sud et tourne \'e0 la vitesse angulaire }{\f3\fs24 \'77}{\fs24 .
\par La f.e.m. dans le premier conducteur est e}{\dn6 1}{\fs24 = E cos }{\f3\fs24 \'77}{\fs24 t, avec E = 10V.
\par
\par 5.1) D\'e9terminer le retard de phase de la f.e.m. dans le n}{\up6 i\'e8me}{\fs24 conducteur, par rapport au premier.
\par
\par 5.2) En d\'e9duire l'amplitude et la phase de la f.e.m. r\'e9sultante e de l'alternateur.
\par
\par Indications : le d\'e9phasage entre les diff\'e9rentes f.e.m. vient du fait que pour une position du rotor
\par (aimant), les enroulements ne voient pas le m\'eame flux en m\'eame temps, mais avec un d\'e9calage temporel qui est le m\'eame entre deux enroulements successifs. D'autre part il est conseill\'e9 d'utiliser la notation complexe.
\par }\pard\plain \s1\sb240\nowidctlpar\widctlpar\outlinelevel0\adjustright \b\f1\ul\lang1036\cgrid {
\par }}
