...传播，已知x=-5m处质点的振动方程为y=Acosπt， 波速为u=4m/s，则波动方程为：
A. y=Acos[t-(x-5)/4]
C. y=Acos[t-(x+5)/4]
B. y=Acos[t+(x+5)/4]
D. y=A...
您是不是要找： ace
A.Y2=Acos[2πt-π/2]
B.Y2=Acos[2πt-π]
C.Y2=Acos[2πt+π/2]
D.Y2=2Acos[2πt-0.1π][99%]
A.Y2=Acos[2πt-π/2]
B.Y2=Acos[2πt-π]
C.Y2=Acos[2πt+π/2]
D.Y2=2Acos[2πt-0.1π]
A.Y2=Acos[2πt-π/2]
B.Y2=Acos[2πt-π]
C.Y2=Acos[2πt+π/2]
D.Y2=2Acos[2πt-0.1π][99%]
A.Y2=Acos[2πt-π/2]
B.Y2=Acos[2πt-π]
C.Y2=Acos[2πt+π/2]
D.Y2=2Acos[2πt-0.1π]
A.Y2=Acos[2πt-π/2]
B.Y2=Acos[2πt-π]
C.Y2=Acos[2πt+π/2]
D.Y2=2Acos[2πt-0.1π][99%]
A.Y2=Acos[2πt-π/2]
B.Y2=Acos[2πt-π]
C.Y2=Acos[2πt+π/2]
D.Y2=2Acos[2πt-0.1π]
A.Y2=Acos[2πt-π/2]
B.Y2=Acos[2πt-π]
C.Y2=Acos[2πt+π/2]
D.Y2=2Acos[2πt-0.1π][99%]
A.Y2=Acos[2πt-π/2]
B.Y2=Acos[2πt-π]
C.Y2=Acos[2πt+π/2]
D.Y2=2Acos[2πt-0.1π]
A.Y2=Acos[2πt-π/2]
B.Y2=Acos[2πt-π]
C.Y2=Acos[2πt+π/2]
D.Y2=2Acos[2πt-0.1π][99%]
A.Y2=Acos[2πt-π/2]
B.Y2=Acos[2πt-π]
C.Y2=Acos[2πt+π/2]
D.Y2=2Acos[2πt-0.1π]
A.Y2=Acos[2πt-π/2]
B.Y2=Acos[2πt-π]
C.Y2=Acos[2πt+π/2]
D.Y2=2Acos[2πt-0.1π][99%]
A.Y2=Acos[2πt-π/2]
B.Y2=Acos[2πt-π]
C.Y2=Acos[2πt+π/2]
D.Y2=2Acos[2πt-0.1π]
A.Y2=Acos[2πt-π/2]
B.Y2=Acos[2πt-π]
C.Y2=Acos[2πt+π/2]
D.Y2=2Acos[2πt-0.1π][99%]
A.Y2=Acos[2πt-π/2]
B.Y2=Acos[2πt-π]
C.Y2=Acos[2πt+π/2]
D.Y2=2Acos[2πt-0.1π]
...振动方程为：A. y=Acos[w(t+l/u)+Φ0] B.y=Acos[w(t-l/u)+Φ0]C. y=Acos[wt+l/u+Φ0] D. y=Acos[wt-l...[97%]
...那么x=0处质点的振动方程为：
A. y=Acos[w(t+l/u)+Φ0] B.y=Acos[w(t-l/u)+Φ0]
C. y=Acos[wt+l/u+Φ0] D. y=Acos[wt-l/u+Φ0]
...两个波的合成： E1=acos（kx＋ωt) E2=－acos（kx－ωt)．[97%]
利用波的复数表达式求以下两个波的合成：E1=acos(kx+ωt)，E2=-acos(kx-ωt)．
