Fisica 4 Eqmaxwell 1

1) Mostrar que y(x,t)=ysin(kx-wt) e solucao da eq d’Alembert, onde y, k e w sao constantes. [ \nabla^2 - \frac{1}{v^2} \frac{\partial^2}{\partial t^2} ] \Psi(x,t) = 0 \rightarrow \left[ \frac{\partial^2}{\partial x^2} - \frac{1}{v^2} \frac{\partial^2}{\partial t^2} \right] \Psi(x,t) = 0 \frac{dy}{dx} = k \gamma \cos(kx-wt) \rightarrow \frac{d^2 y}{dx^2} = -k^2 y sin(kx-wt) \frac{d}{dt} = -W y \gamma \cos(kx-wt) \rightarrow \frac{d^2 y}{dt^2} = - w^2 y \gamma sin(kx-wt) v^2 = \frac{w}{k} \rightarrow \frac{w^2}{k^2} \rightarrow \frac{1}{v^2} = \frac{k^2}{w^2} \rightarrow -k^2 y \gamma sin(kx-wt) + \frac{k^2}{w^2}( - w^2 \gamma sin(kx-wt) ) = 0 \ v. Inducao. 2) Verificar que y(x,t) satisfaz a eq differential onde \left[ \frac{\partial^2}{\partial x^2} - \frac{1}{v^2} \frac{\partial^2}{\partial t^2} \right] \Psi(x,t) = 0, y(x,t) = ysin(kx-wt) e v = \frac{w}{k} \frac{\partial^2 \Psi(x,t)}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2 \Psi}{\partial t^2} - k^2 y \gamma sin(kx-wt) = \frac{1}{v^2} ( - w^2 y \gamma sin(kx-wt) ) v^2 = \frac{w^2}{k^2} \rightarrow v = \frac{w}{k}