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Click to edit Master title style,Click to edit Master text styles,Second level,Third level,Fourth level,Fifth level,同方向同频率,的简谐振动的叠加,同方向同频率的简谐振动的叠加,设两个简谐振动的频率相等为,，振动方向为,X,轴方向，以,x,1,和,x,2,分别代表两运动的质点位移：,x,1,= A,1,cos (,+ ,1,),x,2,= A,2,cos (,+ ,2,),式中,A,、,A,2,中,1,、,2,分别表示这两个简谐振动的振幅和初相位 。因此，质点的合位移为：,x = x,1,+ x,2,=A,1,cos(,+,1,) + A,2,cos(t+,2,),= Acos(,+),其中,A = A,1,2,+ A,2,2,+ 2A,1,A,2,cos(,2,- ,1,),1/2,tg =,（,A,1,sin,1,+ A,2,sin,2,）,（,A,1,cos,1,+ A,2,cos,2,）,讨论,(1),当,2,-,1,=2k,A = A,1,+ A,2,A,最大，,加强,。,(1),当,2,-,1,=(2k+1),A = |A,1,A,2,|,A,最小，,减弱,。,k,取整数,0,、,1,、,2,、,3,、,4,、,5,、等等。,1,2,A,2,A,1,A,X,x,2,x,1,x,O,例,两个同方向同频率的简谐振动，其合振动的振幅为,20 cm,，与第一个简谐振动的相位差为,-,= /6,，若第一个简谐动的振幅为,17.3 cm,，试求：,1,、第二个简谐振动的振幅,A,2,2,、第一、二两个简谐振动的相位差,1,-,2,解：已知,A = 20 cm,A,1,= 17.3 cm,A,2,=A,2,+A,1,2,-2AA,1,cos(,-,),1/2,= 10 cm,-,x,o,A,A,2,A,1,A,2,= A,1,2,+ A,2,2,+ 2A,1,A,2,cos (,1,-,2,), cos (,1,-,2,),= A,2,- A,1,2,- A,2,2, / 2A,1,A,2,= 0,1,-,2,=/2,由图分析,A,2,超前,A,1, ,2,-,1,= /2,1,-,2,=,/2,-,x,o,A,A,2,A,1,r,X,X,Y,Y,Z,Z,0,第九节 运动的相对性,K,K,r,r,X,X,Y,Y,Z,Z,P,0,第九节 运动的相对性,K,K,r,r,r,X,X,Y,Y,Z,Z,P,0,第九节 运动的相对性,K,K,r,r,r,X,X,Y,Y,Z,Z,P,0,第九节 运动的相对性,K,K,r,r,r,=,+,0,r,r,r,r,r,r,r,r,r,t,t,t,=,=,+,+,X,X,Y,Y,Z,Z,0,P,d,d,d,d,d,d,0,第九节 运动的相对性,0,K,K,r,r,r,r,r,r,r,r,r,v,v,v,t,t,t,=,=,=,+,+,+,X,X,Y,Y,Z,Z,0,0,P,d,d,d,d,d,d,0,第九节 运动的相对性,0,K,K,r,r,r,r,r,r,r,r,r,v,v,v,t,t,t,=,=,=,+,+,+,X,X,Y,Y,Z,Z,0,0,P,d,d,d,d,d,d,a,a,a,=,+,K,K,PK,PK,0,第九节 运动的相对性,0,K,K,也可写成,v,v,v,=,+,K,K,PK,PK ,v,PK,v,=,PK,v,K,K,；,v,v,v,=,+,K,K,PK,PK,r,r,r,r,r,r,r,r,r,v,v,v,t,t,t,=,=,=,+,+,+,X,X,Y,Y,Z,Z,0,0,P,d,d,d,d,d,d,v,v,车,地,雨,地,0,第九节 运动的相对性,0,K,K,r,r,r,r,r,r,r,r,r,v,v,v,v,v,v,t,t,t,=,=,=,=,+,+,+,+,车,车,车,地,地,地,X,X,Y,Y,Z,Z,0,0,雨,雨,雨,P,d,d,d,d,d,d,v,v,车,地,雨,地,0,第九节 运动的相对性,0,K,K,r,r,r,r,r,r,r,r,r,v,v,v,v,v,v,t,t,t,=,=,=,=,+,+,+,+,车,车,车,地,地,地,X,X,Y,Y,Z,Z,0,0,雨,雨,雨,雨,P,d,d,d,d,d,d,v,地,v,v,车,地,雨,地,0,第九节 运动的相对性,0,K,K,r,r,r,r,r,r,r,r,r,v,v,v,v,v,v,v,t,t,t,=,=,=,=,+,+,+,+,车,车,车,地,地,地,地,X,X,Y,Y,Z,Z,0,0,雨,雨,雨,雨,P,d,d,d,d,d,d,v,车,地,v,v,车,地,雨,地,0,第九节 运动的相对性,0,K,K,r,r,r,r,r,r,r,r,r,v,v,v,v,v,v,v,v,t,t,t,=,=,=,=,+,+,+,+,车,车,车,地,地,地,地,X,X,Y,Y,Z,Z,0,0,雨,雨,雨,雨,雨,P,d,d,d,d,d,d,v,车,地,v,v,车,地,雨,地,0,车,0,K,K,例一人骑自行车向东而行，当速度为,10 m/s,时，觉得有南风,；当速度增至,15 m/s,，觉得有东南风，求风的速度,v,风对地,。,解：当,v,人对地,1,= 10,i,时,v,风对人,1,= v,1,j,v,风对地,=,v,风对人,1,+,v,人对地,1,= v,1,j,+ 10,i,=10,i,+ v,1,j,( 1),当,v,人对地,2,=15,i,时,v,风对人,2,= -0.707v,2,i,+ 0.707v,2,j,v,风对地,=,v,风对人,2,+,v,人对地,2,= - 0.707v,2,i,+ 0.707v,2,j,+ 15,i,= ( 15 - 0.707v,2,),i,+ 0.707v,2,j,(2),45,o,i,j,v,风对人,1,v,风对人,2,o,(v,1,),(v,2,),(1),与,(2),式相等：,10,i,+ v,1,j,= ( 15 - 0.707v,2,),i,+ 0.707v,2,j,分量相等,: 10 =15 - 0.707v,2,v,2,=7.07 m/s,v,1,= 0.707v,2,v,1,= 0.707, 7.07 = 5 m/s,v,风对地,= 10,i,+ v,1,j,= 10,i,+ 5,j,m/s,v,风对地,= ( 10,2,+ 5,2,),1/2,= 11.2 m/s,tg, = 5/10 = 0.5, = 27,o,(,东偏北,),45,o,i,j,v,1,风对人,v,2,风对人,o,(v,2,),(v,1,),v,风对地,27,o,
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