题目内容
图甲为一列横波在t=1.0时的波动图象,图乙为该波中x=2m处质点P的振动图象,下列说法正确的是
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241341163783075.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241341163783075.png)
A.波速为4m/s; | B.波沿x轴正方向传播; |
C.再过1.0s,P点的动能最大; | D.再过1.25s,P点振动路程为0.5cm |
C
由甲图得
=4m,由图乙得T=2s,
,A错;t=1.0s时,由图乙可看出p点在平衡位置向下振动,结合图甲可判断出波沿x轴负方向传播,B错;再过1.0s,即t=2.0s时,p点在平衡位置向上振动,速度最大,动能增大,C对;再过1.0s,P点振动路程为x1=2A=2
0.2=0.4cm
再过0.25s,
,P点振动路程为x2=0.2sin(
)=0.2![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134116503214.png)
所以P点振动总路程为x=x1+x2=0.4+0.14=0.54cm,D错。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134116425323.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134116472917.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134116503214.png)
再过0.25s,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241341165661054.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134116612536.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134116503214.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134116722792.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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