题目内容
一列波长大于1m的横波沿着x轴正方向传播,处在x1=1m和x2=2m的两质点A、B的振动图像如图所示。由此可知
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412393992120923.png)
A.波长为4/3m B.波速为1m/s
C.3s末A、B两质点的位移相同 D.1s末A点的振动速度大于B点的振动速度
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412393992120923.png)
A.波长为4/3m B.波速为1m/s
C.3s末A、B两质点的位移相同 D.1s末A点的振动速度大于B点的振动速度
AB
分析:波的传播方向未知,分两种情况研究:一种波的传播方向:从A传到B,另一种从B传到A.根据同一时刻两个质点的状态,结合波形,列出A、B间距离与波长的关系,求出波长.由图读出周期,求出波速.简谐波传播过程中介质中各质点在做简谐运动,加速度的大小与位移大小成正比.根据ls末两质点的位移关系,分析加速度关系.
解答:解:A、B若波从A向B传播,AB间的距离△x=(n+
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123939952385.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123939952385.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123939999373.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123940014427.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123940045390.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123940061327.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123940014427.png)
C、3s末A、B两质点的位移分别为y
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123940108262.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123940139269.png)
D、由振动图象读出,ls末A质点的位移y
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123940108262.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123940139269.png)
故选AB
点评:本题中振动图象反映质点的振动情况,根据振动图象的信息,能确定两质点在同一时刻的状态,列出距离与波长的通式是关键.
![](http://thumb2018.1010pic.com/images/loading.gif)
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