题目内容
如图所示为一列简谐横波某时刻的波形图,波传播的速度为20m/s,若以此时刻之后0.1s开始计时,则a点的振动图象应是图中的( )![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195727134877877/SYS?20131029195727134877002_ST/images0.png)
A.
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195727134877877/SYS?20131029195727134877002_ST/images1.png)
B.
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195727134877877/SYS?20131029195727134877002_ST/images2.png)
C.
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195727134877877/SYS?20131029195727134877002_ST/images3.png)
D.
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195727134877877/SYS?20131029195727134877002_ST/images4.png)
【答案】分析:由波动图象读出波长,求出周期,根据0.1s时间与周期的关系分析此时刻之后0.1s时质点a的状态,选择振动图象.
解答:解:由波动图象读出波长λ=8m,周期T=
=
,时间t=0.1s=
,则此时刻之后0.1s时,质点到达平衡位置且向上运动,速度最大,位移为零,所以以此时刻之后0.1s开始计时,a点的振动图象应是图中的A.
故选A
点评:简谐运动的位移图象是正弦或余弦函数,区别在于t=0时刻质点的状态.
解答:解:由波动图象读出波长λ=8m,周期T=
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195727134877877/SYS?20131029195727134877002_DA/0.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195727134877877/SYS?20131029195727134877002_DA/1.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195727134877877/SYS?20131029195727134877002_DA/2.png)
故选A
点评:简谐运动的位移图象是正弦或余弦函数,区别在于t=0时刻质点的状态.
![](http://thumb2018.1010pic.com/images/loading.gif)
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