题目内容
已知一列横波沿x轴的方向传播,周期T=0.8s,图中实线、虚线分别为时刻t1和t2的波形图线,已知t2-t1=4.6s,则此波在这段时间传播的方向和距离为( )![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195405551035035/SYS201310291954055510350007_ST/images0.png)
A.x轴的正方向,46m
B.x轴的负方向,46m
C.x轴的正方向,2m
D.x轴的负方向,6m
【答案】分析:由图读出波长λ,由波速公式v=
求出波速v,即可由x=vt求出4.6s内波传播的距离x,分析波传播的距离与波长的关系,结合波形平移法,可判断出波的传播方向.
解答:解:由图读出波长λ=8m,则波速为v=
=
m/s=10m/s
4.6s内波传播的距离x=vt=10×4.6m=46m
n=
=
=5
,即x=5
,利用波形的平移法,可知波的传播方向沿x轴负方向.
故选B
点评:本题知道两个时刻的波形,求出波传播的距离,判断波的传播方向是关键.
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195405551035035/SYS201310291954055510350007_DA/0.png)
解答:解:由图读出波长λ=8m,则波速为v=
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195405551035035/SYS201310291954055510350007_DA/1.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195405551035035/SYS201310291954055510350007_DA/2.png)
4.6s内波传播的距离x=vt=10×4.6m=46m
n=
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195405551035035/SYS201310291954055510350007_DA/3.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195405551035035/SYS201310291954055510350007_DA/4.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195405551035035/SYS201310291954055510350007_DA/5.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195405551035035/SYS201310291954055510350007_DA/6.png)
故选B
点评:本题知道两个时刻的波形,求出波传播的距离,判断波的传播方向是关键.
![](http://thumb2018.1010pic.com/images/loading.gif)
练习册系列答案
相关题目