题目内容
(2009·山东高考)图12-2-10所示为一简谐波在t=0时刻的波形图,介质中的质点P做简谐运动的表达式为y=Asin5πt,求该波的波速,并画出t=0.3 s时的波形图(至少画出一个波长).![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241415383123694.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241415383123694.jpg)
10 m/s 波形图见解析
由简谐运动表达式可知ω=5π rad/s,t=0时刻质点P向上运动,故波沿x轴正方向传播.由波形图读出波长λ=4 m.
T=
①
由波速公式v=
②
联立①②式,代入数据可得v=10 m/s
t=0.3 s时的波形图如图所示.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241415388422525.jpg)
T=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141538328510.png)
由波速公式v=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141538468427.png)
联立①②式,代入数据可得v=10 m/s
t=0.3 s时的波形图如图所示.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241415388422525.jpg)
![](http://thumb2018.1010pic.com/images/loading.gif)
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