1 prova 1 parte 4 - Fernando Brito

3º A(w) = \frac{F_0}{m[(w_0^2-w^2)^2 + (\beta w)^2]^{\frac{1}{2}}} F_0 = 5 N w_0 = 10 rad/s m = 0,1 kg \beta = 2 rad/s w = 6 rad/s A(w) = \frac{5}{\sqrt{0,1(100-36) + 4.36}} = \frac{5}{6,51} = 0,77 m A\sqrt{2} = A\sqrt{}\frac{0,875}{n} \psi(w) = -\text{arc tg}\left[\frac{6 . 2}{100 - 36}\right] = -\text{arc tg}[0,196] = -\text{arc tg}[0,875] x(t) = 0,52 \cos(6t - 0,785) . e^{-\beta t} \, e^{\psi = - 0,385 \, rad} D \text{ A max } -> w = w_0 A(w_0) = \frac{F_0}{m \beta w_0} = \frac{5}{0,1. 2. 10} = 2,5 m -> \text{ Amp. máxima} Amp. de baixas freq.: w -> 0 \text{ A(0) = } \frac{F_0}{m w_0} = \frac{5}{0,1.100} = 0,5 m -> \text{ Amp. de baixas freq.} Meia largura = \gamma = 2 rad/s Fator de Qualidade: Q = \frac{w_0}{\beta} = \frac{10}{2} = 5 A(w) \frac{A_{\text{max}}}{\sqrt{2}} \text{Amp.} 0,7 \frac{1}{2} \text{freq. de ressonância} 9 -> 10 -> 11 0,5 de baixas freq.