dfrac (1)(6)m({v)_(0)}^2-|||-5.一质点作一维谐振动,方程为 dfrac ({d)^2x}(d{x)^2}+4x=0 则它的振动周期为 ()-|||-A.dfrac (pi )(2) B.π C.2π D.dfrac (3)(2)pi

(sin x)=dfrac (1)({cos )^2x} in (0,dfrac (pi )(2)),则(sin x)=dfrac (1)({cos )^2x}

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

(int )_(0)^dfrac (pi {4)}dfrac (x)(1+cos 2x)dx=( ) .(int )_(0)^dfrac (pi {4)}dfr

A、 dfrac (pi )(6) B、 dfrac (2pi )(3) C、 dfrac (4pi )(3) D、 dfrac (pi )(3)-|||-2、

6.5178:一质点沿x轴作简谐振动,振动方程为 =4times (10)^-2cos (2pi t+dfrac (1)(3)pi )(S1) t=-|||-0

设=dfrac (2x)({x)^2-(y)^2} ，则 =dfrac (2x)({x)^2-(y)^2}=dfrac (2x)({x)^2-(y)^2}___

(3) (int )_(0)^dfrac (pi {4)}dfrac (x)(1+cos 2x)dx= () .-|||-(A) dfrac (pi )(8)+

一质点沿 x 轴 作简谐振动振动范围的中心 点 x 轴的原点已知 周期 T 振幅为 A 若 t = 0 时刻质点位于正最大位移处则振动方程为=Acos [ df

5.设函数 (x)=cos x, 则 (int )_(0)^dfrac (pi {2)}xf(x)dx= ()-|||-【】-|||-A.0 B 1 C -1

[题目]一物体作简谐振动,振动方程为-|||-=Acos (omega t+dfrac (1)(2)pi ) 则该物体在 t=0 时刻的动能-|||-与 =df