
Respuesta :
Answer:
The answer to the question
The steady state response is uā(t) = -[tex]\frac{3\sqrt{2} }{2}[/tex]cos(3t + Ļ/4)
of the form RĀ·cos(ĻtāĪ“) with R = [tex]-\frac{3\sqrt{2} }{2}[/tex], Ļ = 3 and Ī“ = -Ļ/4
Explanation:
To solve the question we note that the equation of motion is given by
m·u'' + γ·u' + k·u = F(t) where
m = mass = 2.00 kg
γ = Damping coefficient = 1
k = Spring constant = 3 NĀ·m
F(t) = externally applied force = 27Ā·cos(3Ā·t)ā18Ā·sin(3Ā·t)
Therefore we have 2Ā·u'' + u' + 3Ā·u = 27Ā·cos(3Ā·t)ā18Ā·sin(3Ā·t)
The homogeneous equation 2Ā·u'' + u' + 3Ā·u is first solved as follows
2Ā·u'' + u' + 3Ā·u = 0 where putting the characteristic equation as
2·X² + X + 3 = 0 we have the solution given by [tex]\frac{-1+/-\sqrt{23} }{4} \sqrt{-1}[/tex] =[tex]\frac{-1+/-\sqrt{23} }{4} i[/tex]
This gives the general solution of the homogeneous equation as
uā(t) = [tex]e^{(-1/4t)} (C_1cos(\frac{\sqrt{23} }{4}t) + C_2sin(\frac{\sqrt{23} }{4}t)[/tex]
For a particular equation of the form 2Ā·u''+u'+3Ā·u = 27Ā·cos(3Ā·t)ā18Ā·sin(3Ā·t) which is in the form uā(t) = AĀ·cos(3Ā·t) + BĀ·sin(3Ā·t)
Then uā'(t) = -3Ā·AĀ·sin(3Ā·t) + 3Ā·BĀ·cos(3Ā·t) also uā''(t) = -9Ā·AĀ·cos(3Ā·t) - 9Ā·BĀ·sin(3Ā·t) from which Ā 2Ā·uā''(t)+uā'(t)+3Ā·uā(t) = (3Ā·B-15Ā·A)Ā·cos(3Ā·t) + (-3Ā·A-15Ā·B)Ā·sin(3Ā·t). Comparing with the equation 27Ā·cos(3Ā·t)ā18Ā·sin(3Ā·t) Ā we have
3Ā·B-15Ā·A = 27
3Ā·A +15Ā·B = 18
Solving the above linear system of equations we have
A = -1.5, B = 1.5 and Ā uā(t) = AĀ·cos(3Ā·t) + BĀ·sin(3Ā·t) becomes 1.5Ā·sin(3Ā·t) - 1.5Ā·cos(3Ā·t)
uā(t) = 1.5Ā·(sin(3Ā·t) - cos(3Ā·t) = [tex]-\frac{3\sqrt{2} }{2}[/tex]Ā·cos(3Ā·t + Ļ/4)
The general solution is then Ā u(t) = uā(t) + uā(t)
however since uā(t) = [tex]e^{(-1/4t)} (C_1cos(\frac{\sqrt{23} }{4}t) + C_2sin(\frac{\sqrt{23} }{4}t)[/tex] ā 0 as t ā ā the steady state response = uā(t) = [tex]-\frac{3\sqrt{2} }{2}[/tex]Ā·cos(3Ā·t + Ļ/4) which is of the form RĀ·cos(ĻtāĪ“) where
R = [tex]-\frac{3\sqrt{2} }{2}[/tex]
Ļ = 3 and
Ī“ = -Ļ/4