A spring-mass system has a spring constant of 3 Nm. A mass of 2 kg is attached to the spring, and the motion takes place in a viscous fluid that offers a resistance numerically equal to the magnitude of the instantaneous velocity. If the system is driven by an external force of 27cos(3t)āˆ’18sin(3t) N,determine the steady-state response in the form Rcos(ωtāˆ’Ī“).

Relax

Respuesta :

Answer:

The answer to the question

The steady state response is u₂(t) = -[tex]\frac{3\sqrt{2} }{2}[/tex]cos(3t + π/4)

of the form RĀ·cos(ωtāˆ’Ī“) with R = [tex]-\frac{3\sqrt{2} }{2}[/tex], ω = 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)āˆ’18Ā·sin(3Ā·t)

Therefore we have 2Ā·u'' + u' + 3Ā·u = 27Ā·cos(3Ā·t)āˆ’18Ā·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

u₁(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)āˆ’18Ā·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)āˆ’18Ā·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)

u₂(t) = 1.5·(sin(3·t) - cos(3·t) = [tex]-\frac{3\sqrt{2} }{2}[/tex]·cos(3·t + π/4)

The general solution is then Ā u(t) = u₁(t) + uā‚‚(t)

however since u₁(t) = [tex]e^{(-1/4t)} (C_1cos(\frac{\sqrt{23} }{4}t) + C_2sin(\frac{\sqrt{23} }{4}t)[/tex] ⇒ 0 as t → āˆž 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(ωtāˆ’Ī“) where

R = [tex]-\frac{3\sqrt{2} }{2}[/tex]

ω = 3 and

Ī“ = -Ļ€/4