题目内容
将一木板从一沙摆(可视为简谐运动的单摆)下面以a=0.2 m/s2的加速度匀加速水平抽出,板上留下的沙迹如图9-4-11所示,量得
="4" cm,
="9" cm,
="14" cm,y="2" cm,试求:该沙摆的振幅、周期和摆长.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241205575641657.jpg)
图9-4-11
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120557486282.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120557533285.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120557549320.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241205575641657.jpg)
图9-4-11
0.02 m 1 s 0.25 m
用匀加速直线运动的加速度公式a=
求出Δt,由T=2Δt得到单摆周期,再由周期公式求摆长.
从图中可知沙摆的振幅为A=y=0.02 m.
由a=
=
得振动周期为
T=2
=2×
s="1" s
再由T=2π
g得沙摆摆长为
l=
×9.8 m≈0.25 m.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120557596404.gif)
从图中可知沙摆的振幅为A=y=0.02 m.
由a=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120557596404.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120557627514.gif)
T=2
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120557642428.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120557658402.gif)
再由T=2π
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120557674304.gif)
l=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120557689565.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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