Physique, lisible pour tous
Equa Diff

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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

Le 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

Revised on February 20, 2022 17:18:10 by Anonymous Coward