["["[\n1cos(4πx-πtDπt

2020-05-24View Original

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选择题：一横波沿绳子传播，波动方程为y=0.05cos（4πx-10πt）（）（A）波长0.05m （B）波长0.5m （C）波速25 m/s （D）波速5 m/s 答案：B 提示：能阐述出解题过程的给予额外奖励，答案回复可见！

Reply #22020-05-25

本帖最后由 h20090630 于 2020-5-25 12:14 编辑 B 根据波动方程为y=0.05cos（4πx-10πt） cosφ=cos（-φ） y=0.05cos(4sx-10πt)=0.05cos(10πt-4πx)=0.05cos，故w=10π=2πν，波速u=2.5m/s，波长λ=0.5m

Reply #32020-05-25

B将波动方程化为标准形式，再比较计算。并注意到cosφ=cos(-φ)，y=0.05cos(4nx-10πt)=0.05cos(10πt-4πx)=0.05cos，故w=10π=2πν，波速u=2.5m/s，波长0.5

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