【单选题】两个同方向同频率的简谐振动,其振动表达式分别为 x1=0.06cos(5t+π/2) x2=0.02cos(π/2-5t) 则它们合振动的振幅和初相分别为A. 0.08 π/2B. 0.04 π/2C. 0.04 -π/2D. 0.08 π

步骤 1：将两个简谐振动表达式转换为相同形式
x1=0.06cos(5t+π/2)
x2=0.02cos(π/2-5t) = 0.02cos(5t-π/2)
步骤 2：计算合振动的振幅
A = √(A1^2 + A2^2 + 2A1A2cos(φ1-φ2))
其中 A1 = 0.06, A2 = 0.02, φ1 = π/2, φ2 = -π/2
A = √(0.06^2 + 0.02^2 + 2*0.06*0.02*cos(π/2-(-π/2)))
A = √(0.0036 + 0.0004 + 2*0.06*0.02*cos(π))
A = √(0.004 + 0.0024*(-1))
A = √(0.004 - 0.0024)
A = √(0.0016)
A = 0.04
步骤 3：计算合振动的初相
φ = arctan((A1sinφ1 + A2sinφ2) / (A1cosφ1 + A2cosφ2))
φ = arctan((0.06sin(π/2) + 0.02sin(-π/2)) / (0.06cos(π/2) + 0.02cos(-π/2)))
φ = arctan((0.06*1 + 0.02*(-1)) / (0.06*0 + 0.02*0))
φ = arctan((0.06 - 0.02) / 0)
φ = arctan(0.04 / 0)
φ = arctan(∞)
φ = π/2