Vibra: probs c5

Main cap.5

5.1

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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)

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5.16 For the motion

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

find (a) the frequency, and (b) the time interval separating successive beats.

Para el inciso (a) sabiendo que...

$f={\frac {1}{T}}={\frac {1}{\frac {2\pi }{\omega }}}={\frac {\omega }{2\pi }}$

con

$\omega _{1}=6.28s^{-1}$

$\omega _{2}=6.20s^{-1}$

$f_{1}={\frac {6.28}{2\pi }}=0.999Hz$

$f_{2}={\frac {6.20}{2\pi }}=0.9867Hz$

$\Rightarrow f\backsimeq 0.99Hz$

Para el inciso (b)

con

$\tau ={\frac {1}{\gamma }}$

$\gamma =2\omega$

$\gamma =2(6.28)=12.56$

$\tau ={\frac {1}{12.56}}=0.0796s$

--Leticia González Zamora (discusión) 21:11 3 jun 2013 (CDT)

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--mfg-wiki (discusión) 12:01 9 may 2013 (CDT)