如图所示,波源S从平衡位置y=0处开始振动,振动方向为竖直向上(y轴的

如图所示,波源S从平衡位置y=0处开始振动,振动方向为竖直向上(y轴的正方向),振动周期T=0.01 s,产生的简谐波向左、右两个方向传播,波速均为v=80 m/s。经过一段时间后,P、Q两点开始振动。已知距离=1.2 m,=2.6 m。若以Q点开始振动时刻作为计时的零点,则在下图的振动图象中,能正确描述P、Q两点振动情况的是

A.甲为Q点的振动图象                     B.乙为Q点的振动图象

C.丙为P点的振动图象                      D.丁为P点的振动图象

答案

AD     

解析:由波源的起振方向知,P、Q两点起振方向为竖直向上(y轴正方向)。

因为λ=Tv=0.8 m

所以SQ=2.6 m=

SP=1.2 m=

波传到P、Q所用时间tP=

tQ=

Q点开始振动时P点已振动。时间为

Δt=tQ-tP=

故Q起振时,P在负的最大位移处。

故甲为Q点的振动图象,丁为P点的振动图象。

故A、D两项正确。


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