（2013?房山区二模）图甲为一列简谐横波在t=0.10s时刻的波形图，P是平衡位置为x=1m处的质点，Q是平衡位置 （2014?武清区三模）图甲为一列简谐横波在t=0.10s时...
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A、由质点Q的振动图象，读出t=0.15s时，质点Q的位移为负向最大值，则加速度达到正向最大值．故A正确．B、由振动图象读出周期T=0.20s．由振动读出t=0.15s时，Q点处于波谷，在波动图象看出x=2m处质点此时通过平衡位置向下，则质点P正沿y轴负方向运动．故B错误．
C、由振动图象可知，t=0.10s时刻，Q点经过平衡位置向下运动，可判断出来波沿x轴负方向传播．故C错误．
D、从t=0.10s到t=0.25s，经过时间为△t=0.15s=
| 3 |
| 4 |
故选A
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