[题目]一机械波沿x轴传播,图1为 t=0 时的波动-|||-图像,图2为 x=5m 处A质点的振动图像,此时P、-|||-Q两质点的位移均为 -1m, 则 ()-|||-`d/mm /cm-|||-2 2-|||-0 1015 20 25 x/m-|||-5 0 3 6 9 2 18 t/(×0.1 s) 15-|||--1-|||--2.-|||-图1 图2-|||-A.这列波向x轴正向传播,波速为 dfrac (5)(3)m5-|||-B. t=0.2s 时,P、Q两质点加速度相同-|||-C,P质点的振动方程为 =2sin (dfrac (5pi )(3)t-dfrac (pi )(6))(cm)-|||-D.这列波的波动方程为 =2sin (dfrac (pi )(10)x+dfrac (pi )(2))(cm)