Question

已知一平面简谐横波的波动表达式为y=0.002cos(400πt-20πx)(SI),k=0,±1,±2...,则t=1

Answer 1

已知波动方程y=A*cos(w*(t-x/v)),将cos(w*(t-x/v))=-1代入,求出所对应的时刻t=1s的位置坐标x,x为波谷对应的位置坐标.

将波动方程写成y=A*cos[400Pi*(t-x/20)]可知,波速v=20m/s,圆频率w=400Pi/s,周期T=2Pi/w=0.005s.

数学知识,当400Pi*(t-x/20)=(2k+1)*Pi (k=0,1,2,...)时,y=-A为波谷（A=0.002m),将t=1s代入解得：x=1-(2k+1)/400,将k=0,1,2,...依次代入解出x即为所求.

求波峰对应的位置,类似的,只需令cos[400Pi*(t-x/20)]=+1