单选题:一平面简谐横波的波动表达式为y=0.05Cos(20πt+47πχ)(SI)。取k=0,±l,±2,…,则t=0.5s时 题目分类:公共基础 题目类型:单选题 查看权限:VIP 题目内容: 一平面简谐横波的波动表达式为y=0.05Cos(20πt+47πχ)(SI)。取k=0,±l,±2,…,则t=0.5s时各波峰所在处的位置为( )。 参考答案: 答案解析:
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