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当前位置:首页 > 机械/制造/汽车 > 机械/模具设计 > 工学]机械振动噪声学习题课
机械振动噪声学习题课212J2221)(21JrmT2)(21rkU221mrJ023km2-32-3解:平动转动T+U=常数,整理,得2221)(21JRmT2-4解:0)(2230/)(22aRkmRdtUTd可得由2221])[()2(21mRJaRkU2220111()()222TUmrJkr()0dTUdt由xr020Jmxkxr解:假定绳与轮之间无滑动结合可得2012JMr2-5求如图所示系统的振动微分方程。2221122334111()()[()]222Tmlmlmll23421[()](1cos)2Ukllmgl230()tVcldt常数VUT22222112233433422[()]()0mlmlmllclkllmgl2-6解:对之求时间的导数,得2-9解:123123123111()eekkkkkkkkkkk22221111()()()()2323nllnlTmnlmlnlll2221(133)6nnml221()2Uknl222222111(133)()0(1)()()0331111(1)0(1)33eqnnmlknlmnlknlnnmxkxmmnnnn2-12解:222211[()]0122mlmlclka222211[()]0122mlmlclka2330ckamml23nkaml2113322nkaakfmllm213323223nncccccmmckamml233cacmkl3-3解:(a)建立坐标,向下为正则有即21mm0)(21kxxmmtBtAxnnsincos21mmkn00)0(;)0(xxxx0x2m1mghVghmVm22122220.212()mvmmx0.2212122mmxghmmmmx3-5解:建立x坐标,以接触后的静平衡位置为坐标原点其解为其中解的初始条件依题意为的求解:下落至处,碰撞过程利用动量守恒0x2mkgmx202212121212()cos2sinmgmmmkkxttghtkmmmmkmm221212122cossinmgmkghkttkmmkmmmm等于导致弹簧伸长的负值。于是0202xxxkxxcxmnnsradn10101000010020xxx00()sin(10)xtxt0,01.000xmx()0.01cos10xtt3-6解:依题意,系统运动微分方程为代入已知m,c,k,得即1)初始条件可得2.022()Resin()Resin(1010.2)ntdxttt0001220020,1xxxtgxxxRndnn2)00,xx)(369.11)(0102.0112102radtgmmxR2()0.0102sin(9.7981.369)txtet1101212()()()nttxtAAteAAte001.000x,x1.001.021,AA100.010.1txtte代入后可得3)此时为临界阻尼状态由可解得2()(cos)sinmLcaakaAtamgL222()cosmLcakamgLakAtcos()t222222camLmgLkakAa22221mLmgLkacatg4-2若设则解:建立如右坐标0cosmxcxkxFtsradTTdnndd41.604.01121212222223000736.41nkmkg2.5641.090.27cos3xxxtcos()xXt222430.270.0082()41.0932.56Xm)(235.024.021221radtgtgnn8.2cos(0.235)()xtmm4-4解:运动微分方程化为设则iYeyieYy224.152Y22222121rrrYXS522202.52,0.12.80210/113nfrkmYxYXS21.021.0()ixXe2()2()()0.210.032itititxXeYee13.122)2(122231rrrtg5-7解:设基础运动则依题意,又位移传递率再设则仪器加速度为而222201212TrFFrr0/1TFF22221212rrr2211r2202rr2n8.9/102500/10800/332mkn5.2277.1nn602nmin/9.2330rn5-9①若则即对本题,又解:由P136式(5-27)0/20%TFF2.021212222rrr04.01296.01222rr02496.224rr29696.296.222r29696.296.222r57.22r546.477.157.257.2nmin/4.4330rn②若则有即即取正数0)(0)()(21210112101rkkrxkJrkxkxmrrkrxrkJrxkxm即..112..0121000()0mkkrxxJkrkkr解:建立如图坐标系统,取静平衡位置为坐标原点,有写成矩阵形式2-141233111334211000kkkkmccxxxkkkmcc2-15基础111111211221222112223222()()JkrrkrrrJkrrrkrr00220021222221krkrkrkrJJ2-17解:建立如右坐标,则矩阵形式的方程为2211122222122020JkrkrJkrkr即力法也可以刚度法来解3-82222320321410339302302410339lmxkxklmklxlklmkkxxllmklkAcos(wt-)Bcos(wt-)x设参考P66-P682222222223304123932330412393lkwmkAlBklklmwlkwmklklklmw22121112226.8250.657555.4250.0825wwBAlBAl()()42()()4422cllkxkxmxllllkxkxJ22004050416Acos(wt-)Bcos(wt-)clkkmxxJlkklx设质系动量矩定理:振动微分方程3-9则微分方程为:222222224054162405416112ccclkkmwAlkBklJwlkkmwlkklJwJml22121112221.644.11.448.4kkwwmmBAlBAl主振型:1121和11222200(1)00ttttkkJkkJtAsin2122022(2)0ttttAkJkAkkJ21220220ttttAkJkAkkJ22122222;22ttkkJJ解:系统运动方程可建立为设解代入上式,得系统特征方程为解得3-10参考P63-P6512,1211;22uu2211一阶二阶uP)3(0PuKuPuJuTT将代入(2)式得振型图为—利用坐标变换为振型矩阵代入方程(1)得展开,计算后得u002240022440042121PPkkPPJJ02244022442211kPPJkPPJ)4(0022121211PPPP000;21000;0011uPuP即即初始条件转化到P坐标下,有tPtP2211cos221cos221ttPu21cos221cos2212211于是得(4)的解为121222111uadjuu0002212210PP其中1122212020mxkxkxmxkxkx)(0022002121axxkkkkxxmm即)sin(21tBAxx)(002222bBAmkkkmkO1x1O1x2解:系统在图示坐标下的运动方程为设解代入上式,并整理得3-11mkmk3,22212221和12121;1BBAA1111u振型矩阵为解得分别将代入(b),解得BA02222mkkkmk由有非零解,知mmmmuMuT20021111001111][1122[]1122mmummxuyyux1111121111111u21212121111121xxxxxxy令正则化的振型矩阵为主坐标与原坐标的关系为:解耦11222200sin(1)0ttttkkJtkkJT11222001002ttttkkJkk
本文标题:工学]机械振动噪声学习题课
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