![振动力学(第二版)](https://wfqqreader-1252317822.image.myqcloud.com/cover/176/31729176/b_31729176.jpg)
2.1 运动方程的建立
【例2.1】 已知弹簧—质量系统如图2.4所示,质量为m,弹簧刚度为k,弹簧原长为l。试确定系统的振动方程。
【解】 图2.4是最简单的单自由度系统。考察弹簧—质量系统沿铅垂方向的自由振动。设x1向下为正,由牛顿第二定律知系统的运动方程为
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00014001.jpg?sign=1739355079-IapzOXn9mew7JBBsWGcywNsCxpx3uv9I-0-167db30f299564826a03d5e8b23f4191)
若设偏离静平衡位置的位移为x,则因x1=x+l+mg/k,故上式变为
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00014002.jpg?sign=1739355079-RSea7dGCKFJKcbef9Q30eKeEYrEEA6gn-0-239539d6d65ba0d33821938ca9a8cd4f)
因此,当像重力一类的不变力作用时,可只考虑偏离系统静平衡位置的位移,那么运动方程中不会再出现重力这类常力,使方程形式变得简洁。现约定,以后若无特别指明,一律以系统稳定的静平衡位置作为运动(或广义)坐标的原点。
【例2.2】 如图2.5所示扭摆,已知扭轴的切变模量为G,极惯性矩为Ip,转动惯量为J,轴长为l。试求扭摆的振动方程。
【解】 如图2.5所示,相对于固定轴x发生扭动,以θ为广义坐标建立系统的转动运动方程。经分析知有两力矩作用在圆盘上,即惯性力矩和恢复力矩
。由动静法原理得
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00014005.jpg?sign=1739355079-jGEK3ijgDikDj1qJHUq6GcI22Q1jhphq-0-4c1530709f21296fabc6b2a96e09e3d1)
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00014006.jpg?sign=1739355079-soEaV9OrfqPxWkEJTTWN2b8YNmhtnWit-0-12fa07c1c214baaa9cfb14ab4f5951ba)
图 2.4
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00014007.jpg?sign=1739355079-8usaJLEuRyGmRBQ5VQRIe6zIqRfM1FQi-0-62118478155984018c96a12d8557c755)
图 2.5
其中 为轴的扭转刚度,设为kt,故
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00014009.jpg?sign=1739355079-jQKfqU3bkQhc4aqSI4mjMMoz8rKPpX3Q-0-142926b16ec1bf0e3bfbfaeb7fdbebc9)
【例2.3】 一质量为m的重物附加在简支梁上,系统参数及截面尺寸如图2.6(a)、(b)所示。试将系统简化为单自由度系统,并求其振动方程。
【解】 梁的质量与m相比可略去。弹簧常数k取决于质量m在梁上的位置。对图2.6(a)所示的简支梁,由材料力学得
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00014010.jpg?sign=1739355079-hGfDRVnGpyBlEzxO24l69bEQbhTXhj7T-0-b4ea72a6c5508df95299e3a535418291)
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00014011.jpg?sign=1739355079-O3aV0lWm5ml0htulnlbG6c3niFm8XGW9-0-848f8d073c2a9a19a35f85c60e723d7a)
因矩形横截面惯性矩 ,所以
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00014013.jpg?sign=1739355079-qDh6zY1yDVh2aRz3g6485ofMfFbeJl36-0-cbb5e3b37d821ed1fba3e303f7483aad)
将带重物的简支梁简化为图2.6(c)所示的相当系统,惯性力与弹性恢复力相平衡,则有
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00015001.jpg?sign=1739355079-gy5HWXfCfRppJ6XM7Bc8W7WJmKZIZ9vz-0-c87f4a9086d78ba8b7580c4a18e2ca2b)
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00015002.jpg?sign=1739355079-uAe1JdBo4owyjM7loGTSLxVl5833AZYB-0-63eae5ac91596750f95549b438425c60)
如果梁的两端不是简支,则Δ应有不同数值。
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00015003.jpg?sign=1739355079-n8SlXgivvVENmJ2MMSPEvjjIOzsUu7hu-0-fb5518ec143228b466464a9dd8869143)
图 2.6
【例2.4】 如图2.7所示系统,相关参数已在图上标出。试求系统的振动方程。
【解】 求解时可以选择任意坐标x1、x2、θ作为变量,但它们相互关联,只有一个是独立的。现取绕固定轴O的转角θ为独立坐标,则等效转动惯量
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00015004.jpg?sign=1739355079-bdcitkYe1xy5i85zx2gFx3hFrmVrtUpB-0-2db06170cb03adddbcd9f3254b50839b)
图 2.7
Jc=m1a2+m2b2+m3r2
其中,r是m3的惯性半径。系统的等效角刚度
ke=k1a2+k2b2+k3c2
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00015005.jpg?sign=1739355079-alxrSEKX1G3yhHTiISigNCpz6bqFUCXL-0-2a53df5e345e4c91dc32a787997b1c0f)
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00015006.jpg?sign=1739355079-Pnb1Z9cC9j9TP3T3OTJimLedZRCaog2m-0-a262b62c3a665f26e555a5fb6ebc8537)
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00015007.jpg?sign=1739355079-k0vhqs1aIZIYIVexmPJ4lbcbEk1HRMHk-0-974b9aca854359609542426672f84799)
可以取x1为独立坐标,于是
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00015008.jpg?sign=1739355079-uYBXMoZGjuK0hNw3YdMgKU0rABxOLYj0-0-93932c7d3e84fc4847c67a736939a615)
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00015009.jpg?sign=1739355079-siyITHUvEh3YRVI7K3UfdByIt0SO9Q8Z-0-8cb8e07f169a370ac97e7bfcff7f5b75)
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00015010.jpg?sign=1739355079-qHQkEQesFYx4XSnAzA2o1vC37rcEbEAe-0-d3d457e6b4478b83e3b5f9b96d8a0a03)
经推导可得系统运动方程
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00015011.jpg?sign=1739355079-mkQJwYGRElBUvHAXCBOq7NLx56JDWngw-0-5805d0b45a768f0cfa34089a96031694)
同理以x2为独立坐标,可得
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00015012.jpg?sign=1739355079-WkesOzltUGD3YahIrsQGlkXkBSQH4GlL-0-947c50f852d4b1dd8cb184d147d25285)
![](https://epubservercos.yuewen.com/B43255/17180246604477406/epubprivate/OEBPS/Images/img00015013.jpg?sign=1739355079-pgEl4lq2EHpwzRd8TcaCg9JjNrDf2fuc-0-49da125f236d6fb5b743aaec36314271)
不难验证A=B=C。
可见,对结构较复杂的单自由度系统(其中有些元件作平移,另一些作转动),不管选择哪一个坐标变量作为独立坐标,其振动方程形式不变。这说明系统固有振动规律与坐标选择无关。