两简谐振动的振动方程分别为_(1)=3cos (2pi t-dfrac (pi )(3)), _(2)=4cos (2pi t+dfrac (pi )(6))单
质点作简谐运动,振动曲线如图所示,其振动方程的初相位φ为( )A.dfrac (pi )(2)B.dfrac (pi )(2)C.0D.πdfrac
dfrac (2pi )(sqrt {3)} D. dfrac (pi )(6)
dfrac (2pi )(sqrt {3)}-|||-D. dfrac (pi )(6)
(B) pi +dfrac (4)(3) (C) pi +2 . (D) pi +dfrac (8)(3)
dfrac (pi )(3)B . dfrac (pi )(3)C . dfrac (pi )(3)D . dfrac (pi )(3)[单选题]复数1+i对应
(B) dfrac (lambda )(2pi {varepsilon )_(0)a}. (C) dfrac (lambda )(4pi {varepsilon
3.要使函数φ(x )= ,dfrac {pi )(2)] (B)[π,2π] (C) [ 0,dfrac (pi )(2)] (D) [ dfrac
dfrac (qQ)(4pi {varepsilon )_(0)l} C. dfrac (qQ)(2pi {varepsilon )_(0)l} D. -dfr
9-6 图(a)中所画的是两个简谐振动的曲线,若这两个简谐振动可叠加,-|||-则合成的余弦振动的初相位为 ()-|||-(A) dfrac (3)(2)pi