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démonstrations formelles, exam I

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Énergie totale d’un MHS

E tot=K(t)+U(t) E_{tot} = K(t) + U(t)
E tot=12mv 2+12kx 2 E_{tot} = \frac{1}{2}m v^{2} + \frac{1}{2}k x^{2}
E tot=12m[Aωcos(ωt+ϕ)] 2+12k[Asin(ωt+ϕ)] 2 E_{tot} = \frac{1}{2}m [A\omega \cos(\omega t + \phi)]^{2} + \frac{1}{2}k [A \sin(\omega t + \phi)]^{2}
E tot=12mA 2ω 2cos 2+12kA 2sin 2 E_{tot} = \frac{1}{2}m A^{2}\omega^{2} \cos^{2} + \frac{1}{2}k A^{2} \sin^{2}
E tot=12mA 2kmcos 2+12kA 2sin 2 E_{tot} = \frac{1}{2}m A^{2}\frac{k}{m} \cos^{2} + \frac{1}{2}k A^{2} \sin^{2}
E tot=12A 2kcos 2+12kA 2sin 2 E_{tot} = \frac{1}{2}A^{2} k \cos^{2} + \frac{1}{2}k A^{2} \sin^{2}
E tot=12kA 2(cos 2+sin 2) E_{tot} = \frac{1}{2}k A^{2}( \cos^{2} + \sin^{2} )
E tot=12kA 2(1) E_{tot} = \frac{1}{2}k A^{2}(1)
E tot=12kA 2 E_{tot} = \frac{1}{2}k A^{2}

Équation différentielle du système masse-ressort

∑F x=ma x\sum F_{x} = m a_{x}
−kx=ma x -kx = m a_{x}
−kx=md 2xdt 2 -kx = m \frac{d^2 x}{dt^2}
d 2xdt 2=−kmx\frac{d^2 x}{dt^2} = -\frac{k}{m}x

On vérifie qu’une solution de cette équation est x(t)=Asin(ωt+ϕ)x(t)=A\sin(\omega t + \phi) avec ω 2=k/m\omega^{2} = k/m

Équation différentielle du pendule simple

Layer 1 θ \theta
∑F y=ma y\sum F_{y} = m a_{y}
T y−mg=ma y=0 T_{y}- mg = m a_{y} = 0
Tcos(θ)=mg T\cos(\theta) = mg
T=mgcos(θ) T = \frac{mg}{\cos(\theta)}
∑F x=ma x\sum F_{x} = m a_{x}
T x=−Tsin(θ)=ma xT_{x} = -T\sin(\theta) = m a_{x}
−mgcos(θ)sin(θ)=−mgtan(θ)=ma x-\frac{mg}{\cos(\theta)}\sin(\theta) = -mg \, \tan(\theta) = m a_{x}
−gtan(θ)=a x-g \tan(\theta) = a_{x}
−gθ=a x-g\theta = a_{x}
−gθ(t)=d 2x(t)dt 2-g\theta(t) = \frac{d^2 x(t)}{dt^2}
−gθ(t)=d 2[θL]dt 2-g\theta(t) = \frac{d^2 [\theta L]}{dt^2}
−gLθ(t)=d 2θ(t)dt 2-\frac{g}{L} \theta(t) = \frac{d^2 \theta(t)}{dt^2}

On vérifie qu’une solution de cette équation est θ(t)=θ maxsin(ωt+ϕ)\theta(t)=\theta_{max}\sin(\omega t + \phi) avec ω 2=g/L\omega^{2} = g/L

Vitesse d’une onde progressive sur une corde tendue

∑F y=ma y\sum F_{y} = m a_{y}
2T y=ma c2T_{y} = m a_{c}
2Tsinθ=mv 2R2T \sin \theta = m \frac{v^{2}}{R}
2Tsinθ=μ2Lv 2R2T \sin \theta = \mu 2 L \frac{v^{2}}{R}
Tθ=μLRv 2T \theta = \mu \frac{L}{R} v^{2}
Tθ=μθv 2T \theta = \mu \theta v^{2}
v 2=Tμv^{2} = \frac{T}{\mu}
Created on February 20, 2022 18:09:21 by Anonymous Coward