Unforced Mechanical Vibrations
Today we begin to consider applications of second order ordinary differential equations.
1. Spring-Mass Systems
2. Unforced Systems: Damped Motion
1
Spring-Mass Systems
We consider a mass m which hangs from a spring whose rest length is l. When the
mass is attached, the spring is stretched an additional length L, so the total length
is l + L.
l
L
m
m
Above, we see the spring in rest state (far left), where it has
length l; at rest with the mass m attached (middle), where it
stretches an additional length L; and at right, the spring in
motion, where it may be above or below the rest length L.
We will set up our problem as follows:
• Let y(t) be the position of the mass at time t. Let y(0) be the rest position of
the spring-mass system. In other words, y(0) will be the location of the dotted
base line in our picture above, when the spring is stretched by length L.
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• Let positive y(t) correspond to a stretched spring, and negative y(t) correspond
to a compressed spring. In other words, the positive direction is down.
• Note that at equilibrium, there is a net force of zero on the system. (Here, the
spring is stretched by a length L.)
Now let us describe the forces acting on the system:
1. Gravity: F1 = mg (Remember, the positive direction is downward.)
2. Restoring force of the spring: Hooke’s law says that the force exerted by a spring
is proportional to the length it is stretched beyond its rest length. In this case,
at time t the spring is stretched
y(t) + L
beyond its rest length, so we have force
F2 (t) = −k[y(t) + L] = −ky(t) − kL
We have a negative sign since when the spring is stretched (when y(t) + L > 0),
the restoring force is trying to compress the spring, i.e., move it upward. When
the spring is compressed (when y(t) + L < 0), the restoring force is pushing it
downward. Either way, we get −k(y(t) + L), where k is a positive constant.
Note that when y(t) = 0, the spring is in equilibrium, so F1 + F2 = 0:
mg − kL − k · 0 = mg − kL = 0
In other words, mg = kL, and we can rewrite F2 as follows:
F2 (t) = −ky(t) − kL = −ky(t) − mg
3. Resistance: We may have a resistive force (such as friction) which acts in the
direction opposite to the current motion. We will assume a resistive force which
is proportional to the velocity, so we will get
F3 (t) = −by ′
where b is a positive constant. The book describes the resistive force in terms
of a dashpot, which is usually described as a piston moving in a cylinder filled
with oil (or something similar).
4. Forcing: We may also have an external force F4 (t) = g(t) which pushes or pulls
on the system. We will discuss forcing functions later.
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Now according to Newton’s second law, we know that mass times acceleration is equal
to the sum of all forces acting on the object. In other words,
my ′′ (t) = F1 (t) + F2 (t) + F3 (t) + F4 (t)
or
my ′′ = mg + (−ky − mg) − by ′ + g(t)
which simplifies to
my ′′ + by ′ + ky = g(t)
where
m
b
k
Note: We can calculate
= the mass of the object
= damping coefficient (to be given)
= the spring constant
k from the equilibrium information, since
mg − kL = 0
Example:
Suppose we have an 8 pound object on a spring, and at equilibrium, we have a stretch
of 6”, or 1/2 foot. Suppose at time t = 0, we start with the spring pulled down by
3” (or 1/4 foot), and an initial (downward) velocity of 1 foot/sec.
Find the amplitude, period, and angular frequency, assuming there is no resistance
or forcing.
First, let us find the mass m. We are given a weight of 8 pounds, so the mass is
m=
8 lb
1
2 = slugs
4
32 ft/sec
Next, we find k: Given that mg = kL, we have k = 8/(1/2) = 16. So our differential
equation is
1 ′′
y + 16y = 0 or y ′′ + 64y = 0
4
(Recall we assumed b = 0.) Our initial conditions are
y(0) =
1
4
and y ′ (0) = 1
Now we solve. The characteristic equation is r 2 + 64 = 0, so r = ±8i. The general
solution is therefore
y(t) = c1 cos(8t) + c2 sin(8t)
From our first initial condition, y(0) = 1/4, we get c1 = 1/4. Then since
y ′(t) = −2 sin(8t) + 8c2 cos(8t)
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we get from y ′(0) = 1 that c2 = 1/8.
Thus, our solution is
1
1
y(t) = cos(8t) + sin(8t).
4
8
The period is 2π/8 = π/4, and the angular frequency is 8. (The frequency is 8/(2π) =
4/π.)
Then to find the amplitude, we need the constants c1 and c2 :
R=
q
c21 + c22 =
s
2
1
4
2
1
+
8
=
s
1
1
+
≈ .28 feet
16 64
The above example is what we refer to as free undamped motion, or simple harmonic
motion.
2
Unforced Systems: Damped Motion
We now turn to the case where we have a damping force. (In other words, we will have
b 6= 0.) The behavior of the spring-mass system in this case is strongly dependent on
the types of roots the characteristic equation has.
Example:
Suppose we have a 32 pound mass, an equilibrium stretch of L = 2 feet, and a resistance which is four times velocity. If at time t = 0 the mass is pulled down by 6” and
then released (i.e., initial velocity is zero), determine the motion of the spring.
Here, m = 32
= 1 slug, b = 4, and k = mg
= 32
= 16. Also, mg = kL means 32 = k 2,
32
L
2
or k = 16. So our initial value problem is
y ′′ + 4y ′ + 16y = 0 y(0) = 1/2 y ′(0) = 0
The characteristic equation is
r 2 + 4r + 16 = 0
which has roots
r=
−4 ±
q
16 − 4(1)(16)
2
=
−4 ±
√
2
−48
√
= −2 ± 2 3i.
Thus our general solution is
√
√
y(t) = c1 e−2t cos(2 3t) + c2 e−2t sin(2 3t)
Now y(0) = 1/2 gives us c1 = 1/2. We then find that
√
√
√
√
√
√
y ′(t) = −e−2t cos(2 3t) − 3e−2t sin(2 3t) − 2c2 e−2t sin(2 3t) + 2 3c2 e−2t cos(2 3t)
√
√
√
√
= (2 3 c2 − 1)e−2t cos(2 3 t) − ( 3 + 2c2 )e−2t sin(2 3 t)
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so y ′(0) = 0 means c2 =
1
√
.
2 3
So our solution is
√
√
1
1
y(t) = e−2t cos(2 3t) + √ e−2t sin(2 3t)
2
2 3
What is the behavior of this system? The system oscillates, passing above and below
y(0) as t increases, but the negative exponentials assure that the oscillations decay
to zero, and the system will come to rest as t → ∞:
0.4
0.2
0.25 0.5 0.75
1
1.25 1.5 1.75
-0.2
-0.4
We can see the solution dying out even in its first “period.”
Although the function is not periodic, we may still be interested in how often it oscillates. You may sometimes see the parameter µ described as the quasi-frequency
(or damped frequency) and 2π/µ as the quasi-period (or damped period). (Check
the particular definitions in whatever you refer to; there does not seem to be standard agreement
√on terms.) Thus in this case, we could say that the quasi-period is
√
2π/(2 3) = π/ 3 ≈ 1.8.
We will get the same general behavior as the above whenever we get complex roots
to the characteristic equation. Notice that our characteristic equation is
mr 2 + br + k = 0
where m, b, and k are all positive. Thus we have roots
√
−b ± b2 − 4mk
r=
2m
So if our roots are complex, we get real part −b/(2m), which is negative. Thus we
get a negative exponential times a sine and a cosine, which will decay but oscillate.
On the other hand, suppose we get real roots. √
We see that b2 − 4mk must be less
than b2 , since m and k are positive. Therefore, b2 − 4mk must be less than b, and
we get two negative roots. This would give us two negative exponential functions as
our fundamental solutions, which would both decay to zero but not oscillate. This
is referred to as overdamped; the damping is so strong that the system just slowly
returns to its rest position. There is no oscillation in an overdamped system.
Finally, if b2 − 4mk = 0, we have a repeated root r = −b/(2m), and get two fundamental solutions of the form e−bt/(2m) and te−bt/(2m) . This again decays to zero
without oscillating, but we say that this system is critically damped, since any change
in the damping will cause it to either be overdamped or to oscillate.
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Example:
We reconsider example above, but increase damping to 8 times the velocity. Now we
have
y ′′ + 8y ′ + 16y = 0
which has characteristic equation r 2 + 8r + 16 = 0 with repeated root r = −4. Thus,
the general solution is
y(t) = c1 e−4t + c2 te−4t
Our initial condition y(0) = 1/2 means that c1 = 1/2, and y ′ (0) = 0 gives us c2 = 2.
Thus our solution is
1
y(t) = e−4t + 2te−4t
2
The solution is shown below:
0.5
0.4
0.3
0.2
0.1
0.5
1
1.5
2
We see that y(t) → 0 as t → ∞. In this case, we are critically damped; any change in
the damping would throw our solution into either the overdamped or the oscillating
case.
Example:
Finally, we consider what happens when the damping in example above is increased
to 10 times the velocity. Here we get
y ′′ + 10y ′ + 16y = 0
which has characteristic equation r 2 + 10r + 16 = 0 with roots r = −2 and r = −8.
Then our general solution is
y(t) = c1 e−2t + c2 e−8t
Solving for c1 and c2 using the same initial conditions as before gives
2
1
y(t) = e−2t − e−8t
3
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As we see below, the solution is not much different from the critically damped case:
0.5
0.4
0.3
0.2
0.1
0.5
1
1.5
2
However, in this case small changes in the damping will not change the nature of the
solution.
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