题目

一列横波沿直线传播，计时开始时的波形如图所示，质点A此时位于平衡位置沿y轴正方向运动,质点A的平衡位置与原点O相距0.5m。已知经过0.02s,A质点第一次达到最大位移处，由此可知（ ）A. 这列波的波长为1mB. 这列波的频率为50HzC. 这列波向右传播D. 这列波的波速为12.5m/s 答案：【答案】ACD【解析】由题，波长λ=2×0.5m=1m。故A正确。由题，T=4×0.02s=0.08s，f=1/T=12.5Hz．故B错误；根据，故D正确；此时质点A沿正方向运动，波向右传播，故C正确。听下面一段材料，回答第1-3题。 1. What are they going to do tomorrow morning according to the talk? A. To have a meeting. B. To have a performance. C. To visit the sick speaker. 2. Why does the chairman have to change today's plan? A. Because the first speaker is sick. B. Because Diana wants to speak first C. Because they want to have coffee earlier. 3. When are they going to have coffee break? A. At 9:00. B. At 9:30. C. At 10:00.