物理：波的描述~
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对了给我加分,错了当我没说,嘿嘿
由题意可得：T=4/10=0.4
由于四秒时间内A质点完成8次全振动 B完成10次
所以有AB之间相距2个波长
波长=1.2/2=0.6m
V=波长/T=0.6/0.4=1.5m/s
波长0.6米
波速1.1m/s
波长为X米
速度为Y米
A、B两质点之间的距离是nλ吗？
质点A完成8.5个λ算不算完成了8个全振动？8.6个呢？
题出的不严谨，不能做！呵呵！！
如果题目是A质点刚好完成8次，B质点刚好完成10次，就可以算了
波由B向A传播,A、B相隔2个λ，10-8=2个波长,所以波长=1.2/2=0.6米,波速=10*0.6/4=1.5米/秒
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