Question: The solution of the forced Mass - Spring - Damper with the following parameters m = 1 3 2 0 m = 1 3 2

The solution of the forced Mass-Spring-Damper with the following parameters
m=1320m=1320 kg, c=7920c=7920 Ns/m, k=81200k=81200 N/m, undergoing forcing function F(t)=5400cos(4t)F(t)=5400cos(4t)
takes the form
x(t)=e3t(A1cos(7.2467t)+A2sin(7.2467t))+0.0795cos(4t0.4852)x(t)=e3t(A1cos(7.2467t)+A2sin(7.2467t))+0.0795cos(4t0.4852).
Note that 0(3)0(3), d(7.2467)d(7.2467), A(0.0795)A(0.0795) and (0.4852)(0.4852) have already been determined.
Given the initial conditions:
x(0)=0.02x(0)=0.02m and x(0)=4x(0)=4m/s
find the exact solution.
Enter your answers for A1A1 and A2A2 to four decimal places in the appropriate boxes below:
A1:A1:
A2:A2:

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