Free Vibration [PDF]

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Free vibration of One Degree of Freedom Systems Free vibration of a system is vibration due to its own internal forces (free of external impressive forces). It is initiated by an initial deviation (an energy input) of the system from its static equilibrium position. Once the initial deviation (a displacement or a velocity or both) is suddenly withdrawn, the strain energy stored in the system forces the system to return to its original, static equilibrium configuration. Due to the inertia of the system, the system will not return to the equilibrium configuration in a straightforward way. Instead it will oscillate about this position  free vibration. A system experiencing free vibration oscillates at one or more of its natural frequencies, which are properties of its mass and stiffness distribution. If there is no damping (an undamped system), the system vibrates at the (undamped) frequency (frequencies) forever. Otherwise, it vibrates at the (damped) frequency (frequencies) and dies out gradually. When damping is not large, as in most cases in engineering, undamped and damped frequencies are very close. Therefore usually no distinction is made between the two types of frequencies. The number of natural frequencies of a system equals to the number of its degrees-of-freedom. Normally, the low frequencies are more important. Damping always exists in materials. This damping is called material damping, which is always positive (dissipating energy). However, air flow, friction and others may ‘present’ negative damping. 1



Undamped Free Vibration Equation of motion based on the free-body diagram k



mx   kx



mx  kx  0



m x



x   n2 x  0



n 



k m



a mass-spring system



natural frequency   2



x (t )  A sin  n t  B cos  n t



m k



period



A and B are determined by the initial



conditions. Sin or Cos ?



  ? n  ? x (0)  ? x (0)  ?



x(t ) 







x (0) sin n t  x (0) cos  n t n



2











 x (0) 



n 







 [ x (0)]2 sin(n t   )



 x(0)n    x (0) 



where



  arctan



Vibration of a pendulum How to establish the equation of motion? 2







What is its natural frequency?



l m



ml 2   mgl sin 







l  g sin   0







l  g  0



n 



g l



Systems with Rotational Degrees-of-Freedom



Equation of Motion



K



J o  K  0



Jo



natural frequency



n 



K Jo



Systems involving rotational degrees-of-freedom are always more difficult to deal with, in particular when translational degrees-offreedom are also present. Gear care is needed to identify both degrees-of-freedom and construct suitable equations of motion.



Damped Free Vibration (first hurdle in studying vibration) mx   kx  cx



k



c 3



mx  cx  kx  0



m x



1.



standard equation



x  2 n x   n2 x  0



damping factor



 



c c  2m n 2 km



oscillatory motion (under-damped   1 ) x (t )  exp(  n t )[C1 exp(  2  1  n t )  C2 exp(   2  1  n t )]



x (t )  exp(  n t )( A sin  d t  B cos  d t )  X exp(  n t ) sin( d t   )



x (t )  exp(  n t )[



x (0)   n x (0) sin  d t  x (0) cos  d t ] d



d   n 1   2



Students’ exercise



damped natural frequency



2



1 0 0



1



2



3



4



5



-1 -2



2.



nonoscillatory motion (over-damped   1 )



x (t )  exp(  n t )[ A exp(  2  1  n t )  B exp(   2  1  n t )]



4



4 3 2 1 0 0



3. critically damped motion ( 



1



0.5



1



1.5



)



2 1.5



x (t )  ( A  Bt ) exp(  n t )



1 0.5 0 -0.5



0



0.5



1



1.5



4. negative damping of   0 as a special case of   1 : x (t )  exp(  n t )[C1 exp(  2  1  n t )  C2 exp(   2  1  n t )]



positive



5



2



2



Divergent oscillatory motion (flutter) due to negative damping



Determination of Damping



x (t )  X exp(  n t ) sin( d t   )



d 2



X



Xexp(-nt)



1



x1



x2



X sin 



0 0



1



2



3



4



-1



-2



2 exp( 0.05t ) sin(0.9988t   )



two consecutive peaks: x1  X exp( n t1 ) sin(dt1   ) x2  X exp( n t 2 ) sin(dt2   )  X exp( n t2 ) sin(dt1   )



logarithm decrement



  ln



x1   n d x2



Example: 6







 



  n d



5



The 2nd and 4th peaks of a damped free vibration measured are respectively 0.021 and 0.013. What is damping factor? Solution: x (t 2 )  exp( n 2 d ) x (t 4 ) 2 n d  2 n



2π



n 1   2



 



4 π 1 2



 x(t 2 )    x(t 4 ) 



2 n d  ln  x(t 2 )    x(t 4 ) 



 ln



If a small damping is assumed,  



 x (t 2 )    x (t 4 ) 



2 n d  4 π  ln



1  x(t 2 )    0.0382  3.82% . ln 4 π  x(t 4 ) 



If such an assumption is not made, then



1 2







1  x(t 2 )   ln 4 π  x(t 4 ) 



. This leads to



1  x(t 2 )   ln 4 π  x(t 4 )   1  x(t 2 )     1  ln 4 π x ( t ) 4    







2



 1  x(t 2 )   2    ln  1   2  4π  x(t 4 )  



 



. This leads to



2



 0.0381  3.81%



. So virtually the same value.



General differential equations an



dn x d n 1 x dx  a  ......  a1 1  a0  0 n 1 n n 1 dt dt dt



first solve the characteristic equation an n  an 1n 1  ......  a1  a0  0



If all roots j are distinct, then the general solution is



7



and hence



n



x (t )   b j exp(  j t ) j 1



where bj are constants to be determined. If there are repeated roots, t (integer m  1) appears in a solution. These are not interesting cases for mechanical vibration. m



 in response to the change of a parameter reveal stability properties Im()



Re()



root locus diagram to be seen in Control



8