土動(dòng)力學(xué)基礎(chǔ)習(xí)題解答第2章

第二章振動(dòng)理論第一部分：課堂思考和證明一、課堂思考110個(gè)周期測(cè)得的幅值比為2D解：由對(duì)數(shù)遞減率公式將??=

δ=ln??=2=0.2代入上式，解得10

??√1???2??=√

(ln0.2)2(ln0.2)2+4??2

=0.0112、如阻尼比??=0.05和0.2，分別評(píng)價(jià)阻尼比對(duì)自振頻率的影響。解：弱阻尼振動(dòng)的振動(dòng)方程可表示為??=?????????????sin(????√1???2??)其自振圓頻率為

????

=

√1???2當(dāng)??=0.05時(shí)，????=0.9987????當(dāng)??=0.2時(shí)，????=0.9798????對(duì)于弱阻尼振動(dòng)系統(tǒng)，若阻尼比變?yōu)樵瓉淼?倍，自振頻率僅減小為原來的0.98倍?？梢?，阻尼比對(duì)自振圓頻率的影響不大。3、如阻尼比??=0.05和0.2，分別評(píng)價(jià)阻尼比對(duì)振幅的影響。解：對(duì)數(shù)遞減率公式δ=ln??=2?? ??√1???2??=0.05??=r=exp(??)=1.36

??√1???2

=0.314 當(dāng)??=時(shí)???????? 其對(duì)數(shù)遞減率為??=3振幅比r=exp(??)=3.60可見，阻尼比對(duì)振幅的衰減影響很大。二、證明題：1、證明在以下初始條件下的單質(zhì)點(diǎn)有阻尼振動(dòng)體系的自由振動(dòng)解。??=0初始條件{??=??=??(??)=??????????sin??????+cos??????)=????????????sin(??????+??0)?? =

+,

=????=√??02+

11????2

+

2 0]2,tan??0=

??????0??0+????????0證：將??=0代入單質(zhì)點(diǎn)有阻尼振動(dòng)體系的自由振動(dòng)通解??(0)==對(duì)??(??)求導(dǎo)得振動(dòng)體系速度函數(shù)??(??)=???? ????????????sin??????+cos??????)+??????????cos?????? sin??????)將??=0代入上式得

??(0)=???? +=由三角函數(shù)關(guān)系可得：

∴

=????????0+??0????tan??

=??2=

??????00 +∵????????????sin(??????+??0)=????????????(sin??????cos??0+cos??????sin??0)=??????????sin??????+cos??????)∴??cos??0=??1,??sin??0=??21??又∵??12+??22=??2=??02+????

2[??0+????????0]21??∴??=√??02+????

2[??0+????????0]2得證。2、已知單質(zhì)點(diǎn)有阻尼振動(dòng)體系在簡(jiǎn)諧荷載下??=??0sin(????)的強(qiáng)迫振動(dòng)解為：??=??????

????

sin

??+

cos

??)+??sin(???? ??)??=1√[1 (??/??????=1√[1 (??/????)2]2+??/????tan??0=2??1 如果初始條件為??(??=0)=??0;??(??=0)=??0證明系數(shù)??1和??2的值為??1

=1

+

??0????cos(????

)],

=??0

+??0??sin(??)?? 0證：將??=0代入上述通解，得??(??=0)=??2

??0??sin(????

)=??0∴??2

=??0

+??0??sin(??)?? 0求導(dǎo)得強(qiáng)迫振動(dòng)速度方程??=???? ????????????sin??????+cos??????)+??????????cos?????? sin??????)+????cos(???? ??)?? 0將??=0代入上式，得整理得

??(??==

+

+??0????cos(????

)=??0得證。

??1

=1

+

??0????cos(??)]?? 0第二部分課后習(xí)題1、下面給出了兩個(gè)簡(jiǎn)諧振動(dòng)得位移方程tszm：簡(jiǎn)諧振A，??=0.01sin(????)B，??=0.01sin(2????)。二者具有相同的振幅加速度幅值。解：兩個(gè)振動(dòng)的速度和加速度方程分別為A:??=0.01??cos(????) ??=?0.01??2sin(????)B:??=0.02??cos(2????) ??=?0.04??2sin(2????)B振動(dòng)的速度為A振動(dòng)的2倍，加速度為4倍。2300kN2-7所示的質(zhì)量-彈簧-??=??=1000????·??/??D、對(duì)數(shù)遞減率δ及有阻尼的自振頻率。解：????=2√????=2√20000×(300/9.8)=1564.92????·??/????=??=

=0.639??=2?? ??

1564.92=2??×

=5.220√1???2 √1?0.6392?? =

????

=????√1???2

=√20000/(300/9.8)√1?0.6392

=3.129H???? 2?? 2?? 2??32中的質(zhì)量-彈簧-??=解：該系統(tǒng)振動(dòng)方程為??(??)=

???????sin

2???2??=??0???????sin????????√ 2?????

?? ???? ??其中：????=25.56??????/??,??=16.33 ????=19.66rad/s位移方程為速度方程為

??(??)=0.1017???16.33??sin19.66????(??)=1.661???16.33??sin19.66??+1.999???16.33??cos19.66??初始時(shí)刻，??=0??,??(0)=0??,??(0)=2??/??，則自由振動(dòng)的總能量為1 1 300E=2????(0)2當(dāng)t=1min=60s時(shí)：

=2×

×4=61.22??????(60)≈0??,??(60)≈0??/??此時(shí)，自由振動(dòng)停止，系統(tǒng)動(dòng)能、彈簧能及總能量均為0。420cm，求其產(chǎn)生的動(dòng)荷載。解：總質(zhì)量動(dòng)荷載

????=2×5=10??????(??)=????????2sin(????)=10×0.2×31.42sin(31.4??)=1971.92sin(31.4??)單位為??542礎(chǔ)傳給地基的最大動(dòng)作用力。解：基礎(chǔ)豎向振幅為??=???????? √[1?

(???????)2??)2]2+4??2(???????)2= 10×0.2

(31.4?25.56)2300×1000?9.8+10√[1?(31.4?25.56)2]2+4×0.6392(31.4?25.56)2=0.067×10?3×0.91=0.061×10?3??=0.061????支座的最大動(dòng)作用力為=??√??2+(????)2=0.061×103 ×+(1000×31.4)2=2.26????6、題4的動(dòng)荷載作用在題2的振動(dòng)系統(tǒng)上產(chǎn)生強(qiáng)迫振動(dòng)，求：產(chǎn)生共振時(shí)的轉(zhuǎn)速。共振時(shí)的動(dòng)力放大系數(shù)和位移振幅。D解：1）共振轉(zhuǎn)速????

=

?√1 2??2=2×=60??????/??2) 共振放大系數(shù)和振幅????

= 12??√1 ??2

= 12×0.639×√1

=1.017?? =??

=1.017×

10×0.2

=1.017×0.067×103?? ??3)

?? +10=0.068×103 ??=0.068??????=1