How is work related to force and displacement?
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Answer:
The answer to the question
The steady state response is u₂(t) = -cos(3t + π/4)
of the form R·cos(ωt−δ) with R = , ω = 3 and δ = -π/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 N·m
F(t) = externally applied force = 27·cos(3·t)−18·sin(3·t)
Therefore we have 2·u'' + u' + 3·u = 27·cos(3·t)−18·sin(3·t)
The homogeneous equation 2·u'' + u' + 3·u is first solved as follows
2·u'' + u' + 3·u = 0 where putting the characteristic equation as
2·X² + X + 3 = 0 we have the solution given by =
This gives the general solution of the homogeneous equation as
u₁(t) =
For a particular equation of the form 2·u''+u'+3·u = 27·cos(3·t)−18·sin(3·t) which is in the form u₂(t) = A·cos(3·t) + B·sin(3·t)
Then u₂'(t) = -3·A·sin(3·t) + 3·B·cos(3·t) also u₂''(t) = -9·A·cos(3·t) - 9·B·sin(3·t) from which 2·u₂''(t)+u₂'(t)+3·u₂(t) = (3·B-15·A)·cos(3·t) + (-3·A-15·B)·sin(3·t). Comparing with the equation 27·cos(3·t)−18·sin(3·t) we have
3·B-15·A = 27
3·A +15·B = 18
Solving the above linear system of equations we have
A = -1.5, B = 1.5 and u₂(t) = A·cos(3·t) + B·sin(3·t) becomes 1.5·sin(3·t) - 1.5·cos(3·t)
u₂(t) = 1.5·(sin(3·t) - cos(3·t) = ·cos(3·t + π/4)
The general solution is then u(t) = u₁(t) + u₂(t)
however since u₁(t) = ⇒ 0 as t → ∞ the steady state response = u₂(t) =
·cos(3·t + π/4) which is of the form R·cos(ωt−δ) where
R =
ω = 3 and
δ = -π/4