此时,式中x与L自身为负值。 1) y=Acos T(4t+2x) Acos(4 =Acos4丌(t+ =-2m/ 2 r=2HZ (2)y=AcoS 4T(t+ 波 4(t+-)=2knk=0,±1,+2, t=4.2s代入(42+=-) x=k-8.4m x=-8.4m,-74m2…-0.4m,0.6m 0.6 0.3 6-7y=3cos(4m-丌) (1)y=3cos4x(t+-)-丌 丌(t+ 丌 yB=3cos 4T( 3cos 4rt-=-=3cos 4m-1g 9 丌=3cos4m--丌55 cos ( ) u L u x y = A  t + + 此时，式中 x 与 L 自身为负值。 6-6 (1) y = Acos (4t + 2x) = Acos(4t + 2x) ) 2 cos 4 ( x = A  t + 2Hz 0.5 2 4 2 / = = = = = −      T s u m s  =1m （2） ) 2 cos 4 ( x y = A  t + 波峰： ) 1 2 cos 4 ( + = x  t ) 2 0, 1, 2, 2 4 ( + = k k =   x  t  t=4.2s 代入( 2 2 4.2 x k + = ) 8.4 , 7.4 ,, 0.4 ,0.6 , 8.4 x m m m m x k m = − − − = − 0.3 2 0.6 = = = u x t 6-7 y = 3cos(4t − ) (1)       = 3cos 4 ( + ) − u x y t       =  + ) − 20 3cos 4 ( x t       =  − ) − 20 9 y 3cos 4 (t B       = −       = −       =  −  −     5 4 3cos 4 5 14 3cos 4 5 9 3cos 4 t t t