题目内容
如图所示为t=0时刻的波形图,波的传播方向平行于x轴。质点A位于xA=2m处,质点B位于xB=3m处。t=2s时,质点B第一次出现在波峰位置;t=3s时,质点A第一次出现在波峰位置。则
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250001179043730.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250001179043730.png)
A.波速为1m/s |
B.波的周期为4s |
C.波沿x轴正方向传播 |
D.t=1s时,质点A的振动速度小于质点B的振动速度 |
ABD
试题分析:由波形图知,波长λ=4m。若波沿X轴正方向传播,则A出现波峰的时刻为t=(n+
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000117920303.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000117935342.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000117951385.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000117935342.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000117998427.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000117920303.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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