Forced Mechanical Vibrations
Today we use methods for solving nonhomogeneous second order linear differential
equations to study the behavior of mechanical systems.
1. Forcing: Transient and Steady State Solutions
2. Forcing: Resonance
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Transient and Steady State Solutions
Let us now consider
my ′′ + by ′ + ky = g(t)
where g(t), a forcing function, is non-zero. The forcing function is just an outside
force applied to the system. (It might represent the platform the spring is on being
shaken, for example.) We are usually interested in periodic forcing functions.
Example:
Let us return to the spring-mass system. We had a 32 pound mass, an equilibrium
stretch of L = 2 feet, and a resistance which was four times velocity. For the homogeneous problem
y ′′ + 4y ′ + 16y = 0,
we found the general solution to be
√
√
y(t) = c1 e−2t cos(2 3t) + c2 e−2t sin(2 3t).
Let us now introduce an external force g(t) = 10 cos(3t) which is applied to the
system.
If the spring is stretched two feet down and then released, solve for the position of
the spring and describe the motion.
Our initial value problem is now
y ′′ + 4y ′ + 16y = 10 cos(3t),
y(0) = 2,
y ′ (0) = 0.
This is non-homogeneous, so we will need the complementary solution to the homogeneous equation which we found above.
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Next, we need a particular solution to the non-homogeneous equation. Since g(t) =
10 cos(3t), we can use the method of undetermined coefficients. (Note that g(t) does
not duplicate any part of the complementary solution.) We get:
yp (t) = A cos(3t) + B sin(3t)
yp′ (t) = −3A sin(3t) + 3B cos(3t)
yp′′(t) = −9A cos(3t) − 9B sin(3t)
We plug this into our non-homogeneous equation and get
yp′′ + 4yp′ + 16yp
= [−9A cos(3t) − 9B sin(3t)] + 4 [−3A sin(3t) + 3B cos(3t)] + 16 [A cos(3t) + B sin(3t)]
= [−9A + 12B + 16A] cos(3t) + [−9B − 12A + 16B] sin(3t)
= [7A + 12B] cos(3t) + [7B − 12A] sin(3t)
Since we want yp′′ + 4yp′ + 16yp = 10 cos(3t), we have the equations
7A + 12B = 10
−12A + 7B = 0
Solving this system yields A = 70/193, B = 120/193, so our particular solution is
yp (t) =
120
70
cos(3t) +
sin(3t)
193
193
and the general solution is
√
√
120
70
cos(3t) +
sin(3t)
y(t) = c1 e−2t cos(2 3t) + c2 e−2t sin(2 3t) +
193
193
From our initial conditions, we know y(0) = 2, so since to see that
y(0) = c1 +
70
,
193
136
√ .
then c1 = 316/193. By plugging in y ′ (0) = 0, we determine c2 = 193
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Thus the solution is
√
√
70
136 −2t
316 −2t
120
√ e sin(2 3t) +
e cos(2 3t) +
cos(3t) +
sin(3t)
193
193
193
193 3
Notice that the first two terms (which originate in the solution to the homogeneous
problem) have a negative exponential and therefore decay to zero. Such a part of a
solution is said to be transient.
The last two terms (which originate from the forcing function) represent simple harmonic motion, and do not decay with time. The portion of the solution which remains
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as t → ∞ is referred to as the steady state solution. We can see in this case that the
steady state solution has period 2π/3 and amplitude
R=
s
70
193
2
120
+
193
2
10
=√
≈ 0.72
193
The graphs of the transient and steady state solutions are shown below from t = 0
to t = 10:
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0.4
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0.2
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1
2
3
4
2
3
4
5
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5
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Transient
Solution
Steady State Solution
√
√
136
70
−2t
√
cos(2 3t) + 193 3 e sin(2 3t)]
cos(3t) + 120
sin(3t)
193
193
You can see the effects of the transient solution vanishing as t increases in the plot of
the solution:
316 −2t
e
193
2
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2
3
4
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Solution y(t) = yc (t) + yp (t)
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Forcing Functions: Resonance
Again, we will consider the differential equation
my ′′ + by ′ + ky = g(t)
for a spring with mass m, damping coefficient b, and forcing function g(t). Let us
return to the case in which there is no damping, and consider the corresponding
homogeneous equation:
my ′′ + ky = 0
q
We then have characteristic equation mr 2 + k = 0 which has roots r = ±i k/m.
Thus, our solution has angular frequency
q
k/m.
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If we introduce a forcing function, we expect our solution to change. An interesting
case
q occurs when the forcing function has the same frequency as the natural frequency
k/m of the system.
Example:
Suppose we have a mass of 12 pounds on a spring which stretches by 1/2 foot
from the weight. If we have a forcing function g(t) = cos(8t), no damping, and
y(0) = y ′(0) = 0, describe the motion of the system.
Our mass is 12/32 = 3/8, and k = mg/L = 12/(1/2) = 24, so our differential equation
is
3 ′′
8
y + 24y = cos(8t) or y ′′ + 64y = cos(8t)
8
3
2
The homogeneous equation has characteristic equation r + 64 = 0, so r = ±8i. Thus
the complementary solution is
yc (t) = c1 cos(8t) + c2 sin(8t)
Note that the angular frequency here is 8, which is the same as the angular frequency
of the forcing function.
Now we solve the non-homogeneous part by undetermined coefficients. For the forcing
function 38 cos(8t), we would normally try A cos(8t) + B sin(8t), but since cos(8t) and
sin(8t) duplicate the homogeneous solution, we must multiply by t:
yp (t) = At cos(8t) + Bt sin(8t)
yp′ (t) = A cos(8t) − 8At sin(8t) + B sin(8t) + 8Bt cos(8t)
yp′′ (t) = −8A sin(8t) − 8A sin(8t) − 64At cos(8t) + 8B cos(8t) + 8B cos(8t) − 64Bt sin(8t)
= −16A sin(8t) − 64At cos(8t) + 16B cos(8t) − 64Bt sin(8t)
Plugging into y ′′ + 64y = 38 cos(8t) yields
−16A sin(8t) − 64At cos(8t) + 16B cos(8t)
8
cos(8t)
3
8
−16A sin(8t) + 16B cos(8t) =
cos(8t)
3
−64Bt sin(8t) + 64At cos(8t) + 64Bt sin(8t) =
So finally we have A = 0 and B = 1/6. Thus our particular solution is
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yp (t) = t sin(8t)
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and the general solution is
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y(t) = yc (t) + yp (t) = c1 cos(8t) + c2 sin(8t) + t sin(8t)
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Solving so that y(0) = y ′ (0) = 0 gives c1 = c2 = 0, so we have
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y(t) = t sin(8t)
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Notice the behavior: as t → ∞, y(t) oscillates and grows:
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10
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The above is an example of resonance. When the forcing function has a frequency
which matches the natural frequency of the system, we will duplicate the complementary solutions and will require a particular solution of the form
At cos(ω0 t) + Bt sin(ω0 t)
This solution will grow and oscillate as t → ∞, just as in the above example.
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