1.第一颗松子落下的波形:y1=-2sintπ/6
第二颗松子落下的波形:y2=-2sin(t-2)π/6,
所以两列波叠加得:y=-2sintπ/6-2sin(t-2)π/6
=-2*2sin[(tπ/6)+(t-2)π/6]cos[tπ/6)-(t-2)π/6]
=-4sin[(t-1)π/3]cos(π/3)
=-2sin[(t-1)π/3]
2.即求t=9(s)时,y=-2sin[(9-1)π/3]=-2sin(2π/3)= 负的根号3
1.第一颗松子落下的波形:y1=-2sintπ/6
第二颗松子落下的波形:y2=-2sin(t-2)π/6,
所以两列波叠加得:y=-2sintπ/6-2sin(t-2)π/6
=-2*2sin[(tπ/6)+(t-2)π/6]cos[tπ/6)-(t-2)π/6]
=-4sin[(t-1)π/3]cos(π/3)
=-2sin[(t-1)π/3]
2.即求t=9(s)时,y=-2sin[(9-1)π/3]=-2sin(2π/3)= 负的根号3