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设平面横波I沿BP方向传播,它在B点的振动方程为y1=0.2x10-2cos2πt(m),平面横波2沿CP
设平面横波I沿BP方向传播,它在B点的振动方程为y1=0.2x10-2cos2πt(m),平面横波2沿CP
方向传播,它在C点的振动方程为y2=0.2x10-2cos(2πt+π) (m),如图所示.P处与B相距0.4m,与C相距0.5m,波速为0.2m·s-1.求:(1)两波传到P处的相位差;(2)在P处合振动的振幅.
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