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M3P2
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1efc4282
Commit
1efc4282
authored
Mar 25, 2023
by
Claude Meny
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Update cheatsheet.fr.md
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12.temporary_ins/69.waves/30.n3/20.overview/cheatsheet.fr.md
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1efc4282
...
@@ -51,36 +51,6 @@ $`\newcommand{\ddpt}[1]{\overset{\large\bullet\bullet}{#1}}`$
...
@@ -51,36 +51,6 @@ $`\newcommand{\ddpt}[1]{\overset{\large\bullet\bullet}{#1}}`$
<br>
<br>
*
Soient deux ondes harmoniques synchrones, d'amplitudes égales, se propageant vers les $
`x`
$ positifs :
$
`U_1(x,t) = A\cdot cos(kx - \omega t + \varphi_1)`
$.
$
`U_2(x,t) = A\cdot cos(kx - \omega t + \varphi_2)`
$.
*
Calcul de l'onde résultante :
<br>
$
`\color{brown}{\mathbf{U(x,t)}}\; = U_1(x,t) + U_2(x,t)`
$
<br>
$
`\begin{align} \quad &=A\;\big[\,cos(\underbrace{kx - \omega t}_{\color{blue}{\text{ posons }\\ kx - \omega t \,=\, \alpha}} + \varphi_1) + cos(\underbrace{kx - \omega t}_{\color{blue}{=\; \alpha}} + \varphi_1)\,\big]
&\\
&=A\;\big[\,cos\Big(\alpha + \dfrac{\varphi_1+\varphi_1}{2} + \dfrac{\varphi_2-\varphi_2}{2}\Big) \\
&\quad\quad\quad\quad + \,cos\Big(\alpha + \dfrac{\varphi_2+\varphi_2}{2} + \dfrac{\varphi_1-\varphi_1}{2}\Big)\,\Big]\\
&\\
&=A\;\big[\,cos\Big(\underbrace{\alpha + \dfrac{\varphi_1+\varphi_2}{2}}_{\color{blue}{=\;\alpha '}} + \dfrac{\varphi_1-\varphi_2}{2}\Big) \\
&\quad\quad\quad\quad + \,cos\Big(\underbrace{\alpha + \dfrac{\varphi_1+\varphi_2}{2}}_{\color{blue}{\text{nous avons posé }\\ \alpha + (\varphi_1+\varphi_2)/2\; = \;\alpha '}} - \dfrac{\varphi_1-\varphi_2}{2}\Big)\,\Big]\\
&\\
&=A\;\big[\,cos\Big(\alpha ' + \dfrac{\varphi_1-\varphi_2}{2}\Big) \\
&\quad\quad\quad\quad + \,cos\Big(\alpha ' - \dfrac{\varphi_1-\varphi_2}{2}\Big)\,\Big]\\
&\\
&=A\;\big[\,\underbrace{cos(\alpha ')\,cos\Big(\dfrac{\varphi_1-\varphi_2}{2}\Big)\,-\,sin(\alpha ')\,sin\Big(\dfrac{\varphi_1-\varphi_2}{2}}_{\color{blue}{\text{car }cos(a+b)\;=\;cos\,a\,cos\,b\;-\;sin\,a\,sin\,b}}\big)\\
&\quad + \underbrace{cos(\alpha ')\,cos\Big(\dfrac{\varphi_1-\varphi_2}{2}\Big)\,+\,sin(\alpha ')\,sin\Big(\dfrac{\varphi_1-\varphi_2}{2}}_{\color{blue}{\text{car }cos(a-b)\;=\;cos\,a\,cos\,b\;+\;sin\,a\,sin\,b}}\big)\,\Big]\\
&\\
&=2\,A\cdot cos(\alpha ')\,cos\Big(\dfrac{\varphi_1-\varphi_2}{2}\Big)
\end{align}`
$
<br>
$
`\quad\boldsymbol{\mathbf{=\color{blue}{\underbrace{\color{brown}{2\,A\cdot cos\Big(\dfrac{\varphi_1-\varphi_2}{2}\Big)}}_{\text{amplitude de l'onde résultante}}} \color{brown}{\cdot cos\Big(}\color{blue}{\underbrace{\color{brown}{kx - \omega t + \dfrac{\varphi_1+\varphi_2}{2}}}_{\text{pulsation }\omega\text{ inchangée}}}\color{brown}{\Big)}}}`
$
RÉSUMÉ
RÉSUMÉ
: ---
: ---
...
@@ -525,14 +495,18 @@ Le modèle mathémait
...
@@ -525,14 +495,18 @@ Le modèle mathémait


*
Soient deux ondes harmoniques synchrones, d'amplitudes égales, se propageant vers les $
`x`
$ positifs :
*
Les deux ondes harmoniques sont :
*
synchrones
*
d'amplitudes égales
*
et se propagent vers les $
`x`
$ croissants.
<br>
$
`U_1(x,t) = A\cdot cos(kx - \omega t + \varphi_1)`
$.
$
`U_1(x,t) = A\cdot cos(kx - \omega t + \varphi_1)`
$.
$
`U_2(x,t) = A\cdot cos(kx - \omega t + \varphi_2)`
$.
$
`U_2(x,t) = A\cdot cos(kx - \omega t + \varphi_2)`
$.
*
Calcul de l'onde résultante :
*
Calcul de l'onde résultante :
<br>
<br>
$
`\color{brown}{\mathbf{U(x,t)}}\; = U_1(x,t) + U_2(x,t)`
$
$
`\color{brown}{\mathbf{U(x,t)}}\; = U_1(x,t) + U_2(x,t)`
$
<br>
$
`\begin{align} \quad &=A\;\big[\,cos(\underbrace{kx - \omega t}_{\color{blue}{\text{ posons }\\ kx - \omega t \,=\, \alpha}} + \varphi_1) + cos(\underbrace{kx - \omega t}_{\color{blue}{=\; \alpha}} + \varphi_1)\,\big]
$
`\begin{align} \quad &=A\;\big[\,cos(\underbrace{kx - \omega t}_{\color{blue}{\text{ posons }\\ kx - \omega t \,=\, \alpha}} + \varphi_1) + cos(\underbrace{kx - \omega t}_{\color{blue}{=\; \alpha}} + \varphi_1)\,\big]
&\\
&\\
&=A\;\big[\,cos\Big(\alpha + \dfrac{\varphi_1+\varphi_1}{2} + \dfrac{\varphi_2-\varphi_2}{2}\Big) \\
&=A\;\big[\,cos\Big(\alpha + \dfrac{\varphi_1+\varphi_1}{2} + \dfrac{\varphi_2-\varphi_2}{2}\Big) \\
...
@@ -552,10 +526,13 @@ $`\begin{align} \quad &=A\;\big[\,cos(\underbrace{kx - \omega t}_{\color{blue}{\
...
@@ -552,10 +526,13 @@ $`\begin{align} \quad &=A\;\big[\,cos(\underbrace{kx - \omega t}_{\color{blue}{\
<br>
<br>
$
`\quad\boldsymbol{\mathbf{=\color{blue}{\underbrace{\color{brown}{2\,A\cdot cos\Big(\dfrac{\varphi_1-\varphi_2}{2}\Big)}}_{\text{amplitude de l'onde résultante}}} \color{brown}{\cdot cos\Big(}\color{blue}{\underbrace{\color{brown}{kx - \omega t + \dfrac{\varphi_1+\varphi_2}{2}}}_{\text{pulsation }\omega\text{ inchangée}}}\color{brown}{\Big)}}}`
$
$
`\quad\boldsymbol{\mathbf{=\color{blue}{\underbrace{\color{brown}{2\,A\cdot cos\Big(\dfrac{\varphi_1-\varphi_2}{2}\Big)}}_{\text{amplitude de l'onde résultante}}} \color{brown}{\cdot cos\Big(}\color{blue}{\underbrace{\color{brown}{kx - \omega t + \dfrac{\varphi_1+\varphi_2}{2}}}_{\text{pulsation }\omega\text{ inchangée}}}\color{brown}{\Big)}}}`
$
*
Je remarque que :
*
Je remarque que l'
*onde résultante*
*
L'onde résultante est harmonique.
*
est
**harmonique**
.
*
Elle a la même fréquence que les deux ondes initiales
*
a la
**même fréquence**
$
`\nu\,=\,\dfrac{\omega}{2\pi}que les deux ondes initiales
*
Amplitude :
* Son amplitude est :
$`
A_{onde//résult.} =
\l
eft| 2
\,
A
\c
dot cos
\B
ig(
\d
frac{
\v
arphi_1-
\v
arphi_2}{2}
\B
ig)
\r
ight|
`$
...
...
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