某平面简谐波在t = 1/4s时波形如图所示,则该波的波函数为:[img=231x102]1803d308279a21d.png[/img]
A: y = 0.5cos(4πt－πx/2－π/2) (cm)
B: y = 0.5cos(4πt + πx/2 + π/2) (cm)
C: y = 0.5cos(4πt + πx/2－π/2)(cm)
D: y = 0.5cos(4πt－πx/2 + π/2) (cm)
A: y = 0.5cos(4πt－πx/2－π/2) (cm)
B: y = 0.5cos(4πt + πx/2 + π/2) (cm)
C: y = 0.5cos(4πt + πx/2－π/2)(cm)
D: y = 0.5cos(4πt－πx/2 + π/2) (cm)
举一反三
- 某平面简谐波在t=0.25s时波形如图所示,则该波的波动方程为 A: y=0.5cos[4p(t－x/8)－p/2] (cm) B: y=0.5cos[4p(t+x/8)+p/2] (cm) C: y=0.5cos[4p(t+x/8)－p/2] (cm) D: y=0.5cos[4p(t－x/8)+p/2] (cm)
- 设函数$$y=y(x)$$由$$\left\{ \begin{matrix} x=a(t-\sin t), \\ y=a(1-\cos t) \\ \end{matrix} \right.$$确定，则$${y}''(x)=$$()． A: $$-\frac{1}{a(1-\cos t)}$$ B: $$-\frac{1}{a{{(1-\cos t)}^{2}}}$$ C: $$\frac{1}{a(1-\cos t)}$$ D: $$\frac{1}{a{{(1-\cos t)}^{2}}}$$
- 求微分方程[img=143x21]17da5f14490e50e.png[/img]的通解,实验命令为(). A: dsolve(D2y-2*Dy+5*y=sin(2*x),x)ans =exp(x)*sin(2*x)*C2+exp(x)*cos(2*x)*C1+1/17*sin(2*x)+4/17*cos(2*x) B: dsolve('D2y-2*Dy+5*y=sin(2*x)','x')ans =cos(2*x)*(sin(4*x)/17 - cos(4*x)/68 + 1/4) - sin(2*x)*(cos(4*x)/17 + sin(4*x)/68) + C1*cos(2*x)*exp(x) - C2*sin(2*x)*exp(x) C: dsolve(D2y-2*Dy+5*y=sin(2*x),'x','y')ans =exp(x)*sin(2*x)*C2+exp(x)*cos(2*x)*C1+1/17*sin(2*x)+4/17*cos(2*x)
- 设\(z = f(x,y)\)，\(x = \sin t\)，\(y = {t^3}\)，则全导数\( { { dz} \over {dt}} = \) A: \({f'_x} \sin t+ 3{t^2}{f'_y}\) B: \({f'_x} \cos t+ {t^2}{f'_y}\) C: \({f'_x} \cos t+ 3{t^2}{f'_y}\) D: \({f'_y} \cos t+ 3{t^2}{f'_x}\)
- 有一平面简谐波沿Ox轴的正方向传播，已知其周期为0.5 s，振幅为1 m，波长为2 m，且在t=0时坐标原点处的质点位于负的最大位移处，则该简谐波的波动方程为（ ） A: y=cos(πt-4πx+π) B: y=cos(4πt+πx+π) C: y=cos(4πt-πx-π) D: y=cos(4πt-πx)