(1)由伯努力方程p+ρgh+(1/2)*ρv^2=恒量p1=200000pa,ρ=1000kg/m^3,v2=4v1=8m/s,h1=h2+5得出:p2=220000pa(2)T=2π/5π=0.4S,角频率为5π,频率为2.5Hz,振幅为0.5m,初相位为π/3(3)利用萨法尔定律得答案为:4倍根号2*μ0/πL(4)振幅为A,角频率w为b,k=c,利用公式波速v=w/k,波速为b/c,周期T=2π/w频率f=1/T为:b/2π波长为2π/k=2π/c