饮冰水大使
幼苗
共回答了20个问题采纳率:70% 举报
(1)10cm;0.24cm (2)0.9m/s;0.27s (3)0.3m/s;0.8s
解析:(1)振幅A=10cm;波长λ=0.24cm
(2)波向右传播:波速v
1 =
=
m/s=0.9m/s
T
1 =t,周期T
1 =
t=
×0.2s≈0.27s
(3)波向左传播:速度v
2 =
=
m/s=0.3m/s
T
2 =t,周期T
2 =4t=4×0.2s=0.8s
本题考查机械波的多解问题,由图像可以直接读出振幅和波长,由图像可以看出波向右传播时,由实线到虚线相差四分之三个波长,时间差四分之三个周期
1年前
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