Vibra: probs c5

Main cap.5

5.1

5.2A system with $m=0.010kg$ , $s=36Nm^{-1}$ and $b=0.50kgs^{-1}$ is driven by a harmonically varying force of amplitude 3.6N. Find the amplitude A and the phase constant $\varphi$ of the steady-state motion when the angular frecuency is $a)8s^{-1}b)80s^{-1}andc)800s^{-1}$

solución:

a)

la relacion correspondiente entre la amplitud $F_{o}$ de la fuerza y la amplitud del desplazammiento es

$F_{o}=SA_{o}\left[1-\left(w^{2}/w_{o}^{2}\right)\right]$

o sea que

$A_{o}={\frac {F_{o}/S}{1-\left({\frac {W}{W_{o}}}\right)^{2}}}$

sutiyuyendo valores para encontrar la amplitud:

$w_{o}={\sqrt {S/m}}={\sqrt {36/0.010}}=60s^{-1}$

$A_{o}={\frac {3.6/36}{1-\left({\frac {8}{60}}\right)^{2}}}=0.1m=100mm$

la constate de fase esta dada por:

$\tan \varphi ={\frac {-\gamma w}{\left(w_{o}^{2}-w^{2}\right)}}={\frac {bw/m}{w_{o}^{2}-w^{2}}}=-{\frac {bw}{m\left(w_{o}^{2}-w^{2}\right)}}$

sutituyendo valores:

$\tan \varphi ={\frac {-\gamma w}{\left(w_{o}^{2}-w^{2}\right)}}={\frac {bw/m}{w_{o}^{2}-w^{2}}}=-{\frac {bw}{m\left(w_{o}^{2}-w^{2}\right)}}={\frac {\left(0.50\right)\left(8\right)}{\left(0.010\right)\left(60^{2}-8^{2}\right)}}={\frac {4}{35.36}}=-0.113$

$\arctan \left(-0.113\right)=\varphi =-0.112\ldots \left(rad\right)$

pasando el resultado a grados

$\varphi =-0.112=-6.447^{o}\approx -6.5^{o}$

el mismo procedimiento se utliza para calcular el resto de los incisos.

--MISS (discusión) 01:12 21 may 2013 (CDT)

5.3 De la Ecuacion (5.6)

$A(w)={\frac {F_{0}}{m}}\left[(w_{0}^{2}-w^{2})^{2}+\gamma ^{2}w^{2}\right]^{-{\frac {1}{2}}}$

Sacamos la derivada e igualamos a 0 para obtener el valor maximo

$0={\frac {dA}{dW}}=A(w)=-{\frac {1}{2}}{\frac {F_{0}}{m}}\left[(w_{0}^{2}-w^{2})^{2}+\gamma ^{2}w^{2}\right]^{-{\frac {3}{2}}}\left[2(w_{0}^{2}-w^{2})(-2w)+2w\gamma ^{2}\right]$

Por lo tanto

$2(w_{0}^{2}-w^{2})(-2w)+2w\gamma ^{2}=0$

$w\left[-2(w_{0}^{2}-w^{2})+\gamma ^{2}\right]=0$

Entonces $w=0$ ó $-2(w_{0}^{2}-w^{2})+\gamma ^{2}=0$

$-2(w_{0}^{2}-w^{2})+\gamma ^{2}=0$

$w_{0}^{2}-w^{2}={\frac {1}{2}}\gamma ^{2}$

$w^{2}=w_{0}^{2}-{\frac {1}{2}}\gamma ^{2}$

--Mario Moranchel (discusión) 12:14 19 jun 2013 (CDT)

5.16 For the motion

$\psi =\left(7.5mm\right)\cos \left[\left(6.28s^{-1}\right)t+27{\text{º}}\right]-\left(23mm\right)\sin \left[\left(6.20s^{-1}\right)t+121{\text{º}}\right]$