(3)相位的意义: x(t)=Acos(@t+p) =-Asin(ot a=-@-Acos(ot+p) ·相位已知则振动状态已知,相位没改变2π振动重复一次, ·相位2π范围内变化,状态不重复, A 4=2π 0 ·相位差 x=Ac0s(0t+9)14p=(0t+2)-(@t+0) x2=Acos(0,t+2)J(当02-o,时)40=-0(3)相位的意义: x(t) = Acos(ωt +) cos( ) 2 a = − A t + v = −Asin(t +) 相位已知则振动状态已知,相位没改变2 振动重复一次. 相位 2 范围内变化,状态不重复. t x O A -A = 2 相位差 cos( ) 1 = 1 1 +1 x A t cos( ) 2 = 2 2 +2 x A t ( ) ( ) = 2 +2 − 1 +1 t t (当2 −1时) =2 −1