Find the steadystate solution having the form x C cost fo
for the damped system x\"+ x\' + x = 2cos(3t)
Solution
Let x = Ccos(wt-)
=>x\' = -Cwsin(wt-)
=>x\'\' = -Cw2cos(wt-)
x\"+x\'+x = 2cos(3t)
=>-Cw2cos(wt-)-Cwsin(wt-)+Ccos(wt-) = 2cos(3t)
=>-Cw2cos(wt)cos()-Cw2sin(wt)sin()-Cwsin(wt)cos()+Cwcos(wt)sin()+Ccos(wt)cos()+Csin(wt)sin() = 2cos(3t)
=>((-Cw2cos())+(Cwsin())+(Ccos()))cos(wt) + ((-Cw2sin())+(-Cwcos())+(Csin())sin(wt) = 2cos(3t)
Comparing L.H.S. and R.H.S. we get,
(-Cw2cos())+(Cwsin())+(Ccos()) = 2, (-Cw2sin())+(-Cwcos())+(Csin() = 0, w = 3
Substituting w = 3 in other two equations we get,
(-9C+C)cos() + 3Csin() = 2, (-3C)cos() + (-9C+C)sin() = 0
=>-8Ccos() + 3Csin() = 2.........(1), -3Ccos()-8Csin() = 0.......(2)
Multiplying equation (1) with 8 on both sides we get, -64Ccos()+24Csin() = 16............(3)
Multiplying equation (2) with 3 on both sides we get, -9Ccos()-24Csin() = 0............(4)
Equations (3) + (4) => -73Ccos() = 16............(5)
From (4) we get, -9Ccos() = 24Csin() =>tan() = -9/24 = -3/8 => = -20.56o =>cos() = 0.936
From (5) we get, C = -0.234
Therefore, x = -0.234cos(3t+20.56o)
