Ordinary Differential Equations. Section 3.8
Dr. Marco A Roque Sol
Section 3.8: Forced Vibrations.
The equation of motion of a mass-spring system subject to an external force F(t) i
my00+ γy0+ ky = F (t),
where m is the mass, γ is the damping coefficient, and k is the stiffness of the spring.
This is the full blown case where we consider every last possible force that can act upon the system. The differential equation for this case is,
mu00+ γu0+ ku = F (t) The displacement function this time will be,
where the complementary solution will be the solution to the ( free, damped) homogeneous case and the particular solution will be found using undetermined coefficients or variation of parameters. There are a couple of things to note here about this case. First, from our work back in the free, damped case we know that the complementary solution will approach zero as t → ∞. Because of this, the complementary solution is often called the transient solution in this case.
Also, because of this behavior the displacement will start to look more and more like the particular solution as t increases and so the particular solution is often called the steady state solution or forced response.
Suppose that the external force is periodic, F (t) = F0cos(ωt),
where F0 > 0 is the amplitude and ω > 0 is the frequency of the
force. Then the equation of motion is
Undamped Forced Vibrations.
If there is no damping ( γ = 0), then the equation of motion is my00+ ky = F0cos(ωt)
The form of the particular solution depends on whether the forcing frequency ω is the same as the natural frequency ω0 =pk/m of
the corresponding unforced system or not . Case 1: ω 6= ω0
In this case our initial guess is okay since it wont be the complementary solution. Upon differentiating the guess and plugging it into the differential equation and simplifying we get,
m (Acos(ωt) + Bsin(ωt))00+ k (Acos(ωt) + Bsin(ωt)) = F0cos(ωt)
m −ω2Acos(ωt) − ω2Bsin(ωt) + k (Acos(ωt) + Bsin(ωt)) = F0cos(ωt)
−mω2A + kA cos(ωt) + −mω2B + kB sin(ωt) = F
0cos(ωt)
Setting coefficients equal gives us,
cos(ωt) −mω2+ k A = F0 =⇒ A =
F0
k − mω2
The particular solution is then F0 k − mω2cos(ωt) = F0 m (k/m − ω2)cos(ωt) = F0 m ω02− ω2 cos (ωt ) Note that we rearranged things a little. Depending on the form that you’d like the displacement to be in we can have either of the following.
u(t) = c1cos(ω0t) + c2sin(ω0t) +
F0
m ω02− ω2 cos (ωt ) u(t) = Rcos(ω0t − δ) +
F0
m ω02− ω2 cos (ωt ) If we used the sine form of the forcing function we could get a similar formula.
Case 2: ω = ω0
In this case we will need to add in a t to the guess for the particular solution.
UP(t) = Atcos(ω0t) + Btsin(ω0t)
Differentiating our guess, plugging it into the differential equation and simplifying gives us the following.
−mω20+ k Atcos(ω0t) + −mω02+ k Btsin(ω0t) + ...
... + 2mω0Bcos(ω0t) − 2mω0Asin(ω0t) = F0cos(ωt)
but
−mω02+ k = m −ω2
So, the first two terms actually drop out and this gives us cos(ω0t) 2mω0B = F0 =⇒ B =
F0
2mω0
sin(ω0t) 2mω0A = 0 =⇒ A = 0
In this case the particular will be, F0
2mω0
tsin(ω0t)
the displacement for this case is then u(t) = c1cos(ω0t) + c2sin(ω0t) +
F0
2mω0
tsin(ω0t)
depending on the form that you prefer for the displacement. u(t) = Rcos(ω0t − δ) +
F0
2mω0
So, what was the point of the two cases here? Well in the first case, our displacement function consists of two cosines and is nice and well behaved for all time.
In contrast, the second case, will have some serious issues at t increases. The addition of the t in the particular solution will mean that we are going to see an oscillation that grows in amplitude as t increases. This case is called resonance and we would generally like to avoid this at all costs.
In this case resonance arose by assuming that the forcing function was,
We would also have the possibility of resonance if we assumed a forcing function of the form.
F (t) = F0sin(ω0t)
We should also take care to not assume that a forcing function will be in one of these two forms. Forcing functions can come in a wide variety of forms. If we do run into a forcing function different from the one that used here you will have to go through undetermined coefficients or variation of parameters to determine the particular solution.
Example: A 3 kg object is attached to spring and will stretch the spring 392mm by itself. There is no damping in the system and a forcing function of the form
F (t) = 10 cos(ωt)
is attached to the object and the system will experience resonance. If the object is initially displaced 20cm downward from its
equilibrium position and given a velocity of 10cm/sec upward, find the displacement at any time t.
Since we are in the metric system we wont need to find mass as its been given to us. Also, for all calculations we will be converting all lengths over to meters. The first thing we need to do is find k.
k = mg
L =
(3)(9.8) 0.392 = 75
Now, we are told that the system experiences resonance so let’s go ahead and get the natural frequency.
ω0 = r k m = r 75 3 = 5 The IVP for this is then
3u00+ 75u = 10cos(5t); u(0) = 0.2, u0(0) = −0.1 The complementary solution is the free undamped solution which is easy to get and for the particular solution we can just use the formula that we derived above. The general solution is then,
u(t) = c1cos(5t) + c2sin(5t) +
1
3tsin(5t) Applying the initial conditions gives
u(t) = 1 5cos(5t) − 1 50sin(5t) + 1 3tsin(5t)
The last thing that we’ll do is combine the first two terms into a single cosine. R = s 1 5 2 + −1 50 2 = 0.201 δ1 = tan−1 −1/50 1/5 = −0.099 δ2= δ1+ π = 3.042
In this case c1 > 0 is positive and c2 < 0 . This means that the
phase shift needs to be in Quadrant IV and so the first one is the correct phase shift this time.
The displacement then becomes, u(t) = 1 5 r 101 100cos 5t + tan−1 −1 10 +1 3tsin(5t)
Example: Solve the initial value problem and plot the solution. u00+ u = 0.5cos(0.8t), u(0) = 0, u0(0) = 0
and discuss the result.
The general solution, of the associate homogeneous equation, is u = c1cos(ω0t) + c2sin(ω0t) +
F0
m(ω02− ω2)cos(ωt)
Applying initial conditions, we obtain c1 = −
F0
m(ω20− ω2); c2 = 0
and the particular solution of the IVP is UP =
F0
m(ω2 0− ω2)
This is the sum of two periodic functions of different periods but the same amplitude. Making use of the trigonometric identities for cos(A ± B) with A = (ω0+ ω)t/2 and B = (ω0− ω)t/2, we can
write the above equation in the form u = F0 m(ω02− ω2)sin (ω0− ω)t 2 sin (ω0+ ω)t 2
If |ω0− ω| is small, then ω0+ ω is much greater than it.
Consequently, sin(ω0+ ω)t/2 is a rapidly oscillating function
compared to sin(ω0− ω)t/2. Thus the motion is a rapid oscillation
with frequency (ω0+ ω)/2 but with a slowly varying sinusoidal
amplitude. F0 m|ω2 0− ω2| sin (ω0− ωt) 2
This type of motion, possessing a periodic variation of amplitude, exhibits what is called a beat. In this case ω0= 1, = 0.8, and
F0= 0.5, so the solution of the given problem is
u(t) = [2.77sin(0.1t)] sin(0.9t) Here is a sketch of the displacement for this example.
