Dynamics cheat sheet

8.3 Harmonic forced input F sin (ϖt)
8.3.1 Undamped forced vibration
u
K
M
ReF e it
mu ′′ + ku = F sin ϖt
Since there is no damping in the system, then there is no steady state solution. In other words, the particular
solution is not the same as the steady state solution in this case. We need to find the particular solution using
method on undetermined coefficients.
Let u = u h + u p . By guessing that u p = c 1 sin ϖt then we find the solution to be
u = A cos ω n t + B sin ω n t + F 1
sin ϖt
k 1 − r2 Applying initial conditions is always done on the full solution. Applying initial conditions gives
u (0) = A
u ′ (t) = −Aω sin ω n t + Bω cos ω n t + ϖ F 1
cos ϖt
k 1 − r2 u ′ (0) = Bω n + ϖ F 1
k 1 − r 2
B = u′ (0)
− F r
ω n k 1 − r 2
Where r = ϖ ω n
The complete solution is
( u ′ (0)
u (t) = u (0) cos ω n t + − F k
ω n
r
1 − r 2 )
sin ω n t + F k
1
sin ϖt (1)
1 − r2 Example: Given force f (t) = 3 sin (5t) then ϖ = 5 rad/sec, and ˆF = 3. Let m = 1, k = 1, then ω n = 1
rad/sec. Hence r = 5, Let initial conditions be zero, then
( )
5
1
u = −3
1 − 5 2 sin t + 3 sin 5t
1 − 52 = 0.625 sin t − 0.125 sin 5.0t
Resonance forced vibration When ϖ ≈ ω we obtain resonance since r → 1 in the solution given in Eq (1)
above and as written the solution can not be used for analysis. To obtain a solution for resonance some calculus
is needed. Eq (1) is written as
( u ′ (0)
u (t) = u (0) cos ωt +
ω
− F k
ωϖ
ω 2 − ϖ 2 )
sin ωt + F k
ω 2
ω 2 sin ϖt
− ϖ2 (1A)
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