答案:(1)若波向左传播,根据波的周期性可知波传播的距离为:x=(n+14)λ=(n+14)×24cm=(24n+6)cm,(n=0,1,2,3…)因此波速的可能值为:v=xt=(24n+6)×10−20.3m/s=(0.8n+0.2)m/s,(n=0,1,2,3…)由t=(n+14)T,得:T=tn+14=1.24n+1s,(n=0,1,2,3…)(2)若波向右传播,可能传播的距离根据虚线波形判知,有关系式:x=(n+34)λ=(24n+18)cm,(n=0,1,2,3…)因此波速的可能值为:v=xt=(24n+18)×10−20.3=(0.8n+0.6)m/s,(n=0,1,2,3,…)由t=(n+34)T,得:T=4t4n+3=1.24n+3s,(n=0,1,2,3…)答:(1)如果波是向左传播的,波速是(0.8n+0.2)m/s,(n=0,1,2,3…),波的周期是1.24n+1s,(n=0,1,2,3…).(2)如果波是向右传播的,波速是(0.8n+0.6)m/s,(n=0,1,2,3,…),波的周期是1.24n+3s,(n=0,1,2,3…).