[题目]一平面简谐波沿ox正方向传播,波动表达-|||-式为 =0.10cos [ 2pi (dfrac (t)(2)-dfrac (x)(4))+dfrac (pi )(2)] (st), 该波在 t=0.5s 时刻的-|||-波形图是 ()-|||-0.1 0.1-|||-2-|||-C 寓-|||-A B-|||-y y-|||-2-|||-C x x-|||-0.1 0.11-|||-C D

一列平面简谐波,其波函数为=-0.5cos (dfrac (pi )(2)t-pi x+dfrac (pi )(2)),则该平面简谐波的传播方向_,坐标原点处质

3.一平面简谐波沿Ox轴正方向传播,波的表达式为 =Acos 2pi (vt-x/lambda ), 而-|||-另一平面简谐波沿Ox轴负方向传播,波的表达式为

πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T

两列平面简谐波在一很长的弦上传播,设其方程为_(1)=0.03cos (pi t-dfrac (pi x)(2))和_(1)=0.03cos (pi t-dfr

把复数=((cos dfrac {2pi )(9)+isin dfrac (2pi )(9))}^3转化为三角形式A =((cos dfrac {2pi )(9

9.沿着相反方向传播的两列相干波,其波动方程为 _(1)=Acos 2pi (vt-dfrac (x)(lambda )) _(2)=-|||-cos 2pi

设=(int )_(-dfrac {pi )(2)}^dfrac (pi {2)}dfrac (sin x)(1+{x)^2}(cos )^4xdx, =(in

(5200)已知波长为的平面简谐波沿x轴负方向传播.x =  /4处质点的振动方程为=Acos dfrac (2pi )(lambda )cdot ut

lim _(xarrow dfrac {pi )(2)}dfrac (cos x)(pi -2x)..

(int )_(-dfrac {pi )(2)}^dfrac (pi {2)}((cos )^2x+dfrac (xcos x)(1+{cos )^2x})dx