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)−18sin(3t) N, determine the steady-state response in the form Rcos(ωt−δ).

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

when heated, the valance electrons increased their speed and transferred the heat from electron to electron.   i just took the test.