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1、TheforcedmotionequationofTwo-Degree-Of-FreedomSystemsisIfω1=ω2=ω,thentheequationturnstobeOtherwiseChapter3VibrationofTwo-Degree-OF-FreedomSystems§3.2ForcedVibrationofTwo-Degree-OF-FreedomSystemsAccordingtosuperpositionprinciple,wemaysolvetwoequationsasbelowandaddthesolutionstogethertogettheto
2、talresponseofthesystem.Chapter3VibrationofTwo-Degree-OF-FreedomSystems§3.2ForcedVibrationofTwo-Degree-OF-FreedomSystemsNowassumingthatω1=ω2=ω,thenThesolutioncanbewrittenasInsertingthisintotheaboveequation,weobtainChapter3VibrationofTwo-Degree-OF-FreedomSystems§3.2ForcedVibrationofTwo-Degree-O
3、F-FreedomSystemsThesolutionisWhere[Zij]isimpedancematrixChapter3VibrationofTwo-Degree-OF-FreedomSystems§3.2ForcedVibrationofTwo-Degree-OF-FreedomSystemsTheamplitudeoftheforcedvibrationareForundampedTwo-Degree-of-FreedomSystemsandF2=0Chapter3VibrationofTwo-Degree-OF-FreedomSystems§3.2ForcedVib
4、rationofTwo-Degree-OF-FreedomSystemsm1=m2=m=1kg;k1=0,k2=2k,k3=3k,;k=1000N/m.Derivethesteady-stateresponseofthesystem,anddrawtheamplitude–frequencycurve.Example3-5Solution:TheequationofthesystemisChapter3VibrationofTwo-Degree-OF-FreedomSystems§3.2ForcedVibrationofTwo-Degree-OF-FreedomSystemsBe
5、causek11=2k;k12=-2k;k21=-2k;k22=5k,sotheequationchangedtoChapter3VibrationofTwo-Degree-OF-FreedomSystems§3.2ForcedVibrationofTwo-Degree-OF-FreedomSystemsTheamplitudeofthemassesareSubstitutem1=m2=m=1kg,k1=0,k2=2k,k3=3k,k=1000N/mintotheequations,wehaveChapter3VibrationofTwo-Degree-OF-FreedomSys
6、tems§3.2ForcedVibrationofTwo-Degree-OF-FreedomSystemsHerearethetwonaturalfrequenciesofsystem.Aswesolvedinexample3.1,theyareequaltoChapter3VibrationofTwo-Degree-OF-FreedomSystems§3.2ForcedVibrationofTwo-Degree-OF-FreedomSystemsExample3-6m1=m=1kg,m2=2m,k1=k;k2=k;k3=2k;k=100N/m,F=F1eiωt.Deriveth
7、esteady-stateresponseofthesystem,anddrawtheamplitude–frequencycurve.Chapter3VibrationofTwo-Degree-OF-FreedomSystems§3.2ForcedVibrationofTwo-Degree-OF-FreedomSystemsSolution:TheequationofthesystemisChapter3VibrationofTwo-Degree-OF-FreedomSyste
