图中实线和虚线分别是x轴上传播的一列简谐横波在t=0和t=0.03 s 时刻

图中实线和虚线分别是x轴上传播的一列简谐横波在t=0和t=0.03 s 时刻的波形图,x=1.2 m处的质点在t=0.03 s时刻向y轴正方向运动,则(    )

A.该波的频率可能是125 Hz

B.该波的波速可能是10 m/s

C.t=0时x=1.4 m处质点的加速度方向沿y轴正方向

D.各质点在0.03 s内随波迁移0.9 m

答案

解析:由t=0.03 s时x=1.2 m处质点向y轴正向运动知波沿x轴正向传播,由此知C对.由波形图变化的周期性知:虚线波形应为实线波沿x轴正向移动λ而成,所以有T=0.03 s,T=s(n=0,1,2,3……),f=Hz,n=3时,f=125 Hz,可见A对.又从波形图知波长λ=1.2 m,所以波速v=fλ=10(4n+3) m/s,当n=0时vmin=30 m/s,可知选项B错.又在波动中,各质点只是在各自的平衡位置附近振动,并不随波迁移,故知选项D错.

答案:AC

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