题目内容
如图所示为水平放置的两个弹簧振子A和B的振动图像,已知两个振子质量之比为mA :mB=2:3,弹簧的劲度系数之比为kA:kB=3:2,则它们的周期之比TA:TB= ;它们的最大加速度之比为aA:aB= 。![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412425582610225.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412425582610225.png)
2:3;9:2
分析:根据弹簧振子的周期公式T="2π"
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124255841544.png)
解:已知两个振子质量之比为mA:mB=2:3,弹簧的劲度系数之比为kA:kB=3:2,根据弹簧振子的周期公式T=2π
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124255841544.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124255919628.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124255966622.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124255997647.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124256075648.png)
故答案为:2:3,9:2.
![](http://thumb2018.1010pic.com/images/loading.gif)
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