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M3P2
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0c848c41
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0c848c41
authored
Apr 12, 2023
by
Claude Meny
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Update cheatsheet.fr.md
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0c848c41
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@@ -845,6 +845,8 @@ $`\quad\boldsymbol{\mathbf{=\color{brown}{2\,A\cdot cos\Big(\dfrac{\varphi_1-\va
**$`\boldsymbol{\mathbf{\underline{U_1}(x,t) = A\cdot e^{\,i\;(\omega t - kx + \varphi_1)}}}`$**
**$`\boldsymbol{\mathbf{\underline{U_2}(x,t) = A\cdot e^{\,i\;(\omega t - kx + \varphi_2)}}}`$**
<br>
<!---------------------
soit encore :
<br>
$
`\begin{align}\underline{U_1}(x,t) &= \underline{A_1}\cdot e^{\,i\;(\omega t - kx)}\\
...
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@@ -854,6 +856,7 @@ $`\quad\boldsymbol{\mathbf{=\color{brown}{2\,A\cdot cos\Big(\dfrac{\varphi_1-\va
&\quad\quad\text{avec }\underline{A_2} = A_2\,e^{\,i\;\varphi_2}\end{align}`
$.
<br>
ou $
`\underline{A_2}`
$ et $
`\underline{A_2}`
$ sont les amplitudes complexes des deux ondes.
---------------------------->
*
Calcul de l'onde complexe résultante :
<br>
...
...
@@ -863,66 +866,32 @@ $`\quad\boldsymbol{\mathbf{=\color{brown}{2\,A\cdot cos\Big(\dfrac{\varphi_1-\va
<br>
$
`\color{blue}{\scriptsize{\quad\text{mettons en commun les termes qui peuvent l'être}}}`
$
<br>
$
`\quad =A\;e^{\,i\;(\omega t\,-\, kx)}`
$ $
`\cdot\big(\,e^{\,i\varphi_1}\,+\,e^{\,i\varphi_1}\big)`
$
$
`\quad =A\;e^{\,i\;(\omega t\,-\, kx)}`
$
$
`\cdot\big(\,e^{\,i\varphi_1}\,+\,e^{\,i\varphi_1}\big)`
$
<br>
$
`\quad =A\;e^{\,i\;(\omega t\,-\, kx)}
\cdot\left(\,e^{\,i\left(\frac{\varphi1 + \varphi2}{2}\;+\;\frac{\varphi1-\varphi2}{2}\right)
}
\,+\,
$
`\quad =A\;e^{\,i\;(\omega t\,-\, kx)}
`
$
$
`
\cdot\left(\,e^{\,i\left(\frac{\varphi1 + \varphi2}{2}\;+\;\frac{\varphi1-\varphi2}{2}\right)
}
`
$
$
`
\,+\,
e^{\,i\;\left(\frac{\varphi1 + \varphi2}{2}\;-\;\frac{\varphi1-\varphi2}{2}\right)}\right)`
$
<br>
$
`\color{blue}{\scriptsize{\quad \text{utilisons } exp\,(a+b)\;=\; exp\,(a)\;+\; exp\,(b)}}`
$
$
`\color{blue}{\scriptsize{\quad\text{et regroupons encore les termes communs.}}}`
$
<br>
$
`
A\;e^{\,i\;(\omega t\,-\, kx)
\,e^{\,i\left(\frac{\varphi1 + \varphi2}{2}\right)}`
$
$
`
\quad = A\;e^{\,i\;(\omega t\,-\, kx)}
\,e^{\,i\left(\frac{\varphi1 + \varphi2}{2}\right)}`
$
$
`\cdot\big(\,e^{\,i\left(\frac{\varphi1-\varphi2}{2}\right)}\,+\,e^{\,-\,,i\;\left(\frac{\varphi1-\varphi2}{2}\right)}\big)`
$
$
`\quad =A\;\big[\,e^{\,i\;(\alpha\, +\, \varphi_1)} + e^{\,i\;(\alpha\, + \,\varphi_2)} \,\big]`
$
<br>
$
`\color{blue}{\scriptsize{\quad\text{Rappel : } e^{\,i\;(a+b)}\;=\;e^{\,i\,a}\times e^{\,i\,b}}}`
$
$
`\color{blue}{\scriptsize{\quad \text{utilisons } exp\,(i\,a)\,+\,exp\,(-\,i\,a)\;=\;2\,cos\,a}`
$
<br>
$
`\quad =A\;\big[ \,e^{\,i\,\alpha}\;e^{\,i\,\varphi_1}\; + \; e^{\,i\,\alpha}\;e^{\,i\,\varphi_2}\,\big]`
$
$
`\quad =A\;e^{\,i\,\alpha}\;\big[\,e^{\,i\,\varphi_1}\; + \; e^{\,i\,\varphi_2}\,\big]`
$
*
Pour exprimer l'onde en notation réelle, il faut
**décomposer l'onde complexe**
en ses
*parties réelle et imaginaire*
:
$
`\quad = A\;e^{\,i\;(\omega t\,-\, kx)}\,e^{\,i\left(\frac{\varphi1 + \varphi2}{2}\right)}`
$
$
`\cdot cos\left(\dfrac{\varphi1-\varphi2}{2}\right)`
$
<br>
$
`\color{blue}{\scriptsize{\quad\text{Rappel : }e^{\,i\,a} = cos (a) + i\,sin (a)}}`
$
$
`\color{blue}{\scriptsize{\quad\text{et pour simplifier l'écriture, posons la notation :}}}`
$
$
`\color{blue}{\scriptsize{\quad\ cos\,(a) = "ca" \text{ , et } sin\,(a) = "sa"}}`
$
$
`\quad = A\;cos\left(\dfrac{\varphi1-\varphi2}{2}\right)\;e^{\,i\;(\omega t\,-\, kx\,+\,\left(\frac{\varphi1 + \varphi2}{2}\right)}`
$
<br>
$
`\quad =A\cdot(c\alpha\,+\,i\,s\alpha) \cdot (c\varphi_1\,+\,i\,s\varphi_1\,+\,c\varphi_2\,+\,i\,s\varphi_2)`
$
L'onde réelle est donc :
<br>
$
`\quad
=A\cdot(c\,\alpha\;+\;i\,s\,\alpha) \cdot \big[\,(c\varphi_1+c\varphi_2)`
$$
`\,+\, i\,(s\varphi_1+s\varphi_2)\,\big]`
$
$
`\quad
\boldsymbol{\mathbf{=\color{brown}{2\,A\cdot cos\Big(\dfrac{\varphi_1-\varphi_2}{2}\Big) \cdot cos\Big(}\color{blue}{\underbrace{\color{brown}{\omega t - kx + \dfrac{\varphi_1+\varphi_2}{2}}}_{\text{pulsation }\omega\text{ inchangée}}}\color{brown}{\Big)}}}`
$
*
L'onde réelle est la partie réelle de $
`\underline{U}(x,t)`
$ :
<br>
$
`\mathbf{U(x,t)} = \mathscr{Re}[\,\underline{U}(x,t)\,]`
$
<br>
$
`\quad =A\cdot\big[\,c\alpha\,(c\varphi_1+ c\varphi_2) - s\alpha\,(s\varphi_1+ s\varphi_2)\,\big]`
$
<br>
$
`\color{blue}{\scriptsize{\quad\left| \begin{align} &cos(a+b)=cos(a)\,cos(b)-sin(a)\,sin(b)\\
&cos(a-b)=cos(a)\,cos(b)+-sin(a)\,sin(b)\end{align}\right.}}`
$
$
`\color{blue}{\scriptsize{ \quad\Longrightarrow cos(a+b)+cos(a-b)=2\,cos(a)\,cos(b)}}`
$
$
`\color{blue}{\scriptsize{ \quad\text{En posant } p=a+b \text{ et } q=a-b\;,}}`
$
$
`\color{blue}{\scriptsize{\quad \text{nous obtenons } a = (p+q)\,/\,2 \text{ et } b = (p-q)\,/\,2.}}`
$
$
`\color{blue}{\scriptsize{ \quad\text{Nous retrouvons ainsi }}}`
$
$
`\color{blue}{\scriptsize{ \quad\quad cos(p) + cos(q) = 2\,cos\Big(\dfrac{p+q}{2}\Big)\,cos\Big(\dfrac{p-q}{2}\Big)}}`
$
<br>
$
`\color{blue}{\scriptsize{ \quad\text{De même nous retrouverions }}}`
$
$
`\color{blue}{\scriptsize{ \quad\quad sin(p) + sin(q) = 2\,sin\Big(\dfrac{p+q}{2}\Big)\,cos\Big(\dfrac{p-q}{2}\Big)}}`
$
<br>
$
`\quad \begin{align}=A\;\Big[\,&c\alpha\cdot 2\,c\Big(\dfrac{\varphi_1+\varphi_2}{2}\Big)\,c\Big(\dfrac{\varphi_1-\varphi_2}{2}\Big)\\
&- s\alpha\cdot 2\,s\Big(\dfrac{\varphi_1+\varphi_2}{2}\Big)\,c\Big(\dfrac{\varphi_1-\varphi_2}{2}\Big)\end{align}`
$
<br>
$
`\quad \begin{align}=2\,A\;&c\Big(\dfrac{\varphi_1-\varphi_2}{2}\Big)\\
&\times \Big[ \,c\alpha\,c\Big(\dfrac{\varphi_1+\varphi_2}{2}\Big)-s\alpha\,s\Big(\dfrac{\varphi_1+\varphi_2}{2}\Big)\,\Big]
\end{align}`
$
<br>
$
`\quad =2\,A\;c\Big(\dfrac{\varphi_1-\varphi_2}{2}\Big)\,c\Big(\alpha + \dfrac{\varphi_1+\varphi_2}{2}\Big)`
$
<br>
**$`\boldsymbol{\mathbf{\quad =2\,A\;cos\Big(\dfrac{\varphi_1-\varphi_2}{2}\Big)\,cos\Big(\omega t - kx + \dfrac{\varphi_1+\varphi_2}{2}\Big)}}`$**
<br>
Bien sûr nous obtenons le même résultat qu'avec le calcul en notation réelle.
...
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