MATH 2280-002
Lecture Notes: 2/01/2013-2/04/2013
Math 2280 Section 002 [SPRING 2013]
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Mass-Spring-Dashpot Systems
Let’s take a look at a useful application of linear DE’s with constant coefficients – namely, massspring-dashpot systems!
A mass m is attached to a spring. A dashpot provides resistance to the force of the spring, and x(t)
measures the distance from the mass to the equilibrium (or resting) position. According to Hooke’s
law, the force of the spring is proportional to the distance the spring is stretched/compressed:
FS = −kx.
The positive constant k is called the spring constant or Hooke’s constant.
We’ll assume the dashpot provides a force proportional to the velocity of the mass:
FR = −cv = −c
dx
.
dt
This should remind you of our discussion of air resistance earlier in the semester. The positive
constant c is called the damping constant. There may also be an external force on the object.
FE = F (t).
By Newton’s law,
Total force = FS + FR + FE = ma = mx00 .
This gives us the DE
mx00 + cx0 + kx = F (t)
We’ll consider several scenarios.
• Undamped. No dashpot (or friction), so c = 0.
• Damped. Dashpot/friction included, so c > 0.
• Free. No external force, so F (t) = 0.
• Forced. External force included, so F (t) 6= 0.
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MATH 2280-002
2
Lecture Notes: 2/01/2013-2/04/2013
Free Undamped Motion
If we have no damping or external force,
mx00 + kx = 0.
Divide by m.
k
x + x = x00 + ω02 x = 0,
m
00
r
where ω0 =
k
.
m
We know how to solve this DE using the techniques we’ve just learned!
Exercise: Verify that this DE has the solution
x(t) = A cos(ω0 t) + B sin(ω0 t)
q
k
Defining ω0 to be m
may seem arbitrary at first, until you realize that we defined ω0 so that our
answer would look nice in the end.
Define new constants C and α by
C=
p
A2 + B 2 ,
cos(α) =
A
,
C
sin(α) =
B
.
C
The picture to have in mind when defining C and α is the right triangle below.
Exercise: Verify that x(t) = C cos(ω0 t − α).
We worked this exercise out in class and I would encourage you to try it out on your own. We
A
and
needed a few ”tricks,” but here is the basic outline – factor out a C; recognize that cos(α) = C
B
sin(α) = C ; and use the adding angles formula for cosine.
We wanted our solution in form x(t) = C cos(ω0 t−α) because, in this form, it is easy to describe the
behavior of the graph. You should be able to easily sketch the graph without the aid of a computer.
At this point, I would like to introduce some useful terminology. (Many of these terms should be
familiar.)
• C is the amplitude.
• ω0 is the circular frequency.
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MATH 2280-002
Lecture Notes: 2/01/2013-2/04/2013
• α is the phase angle.
• T =
• ν=
• δ=
2π
ω0 is the period (the time to complete one oscillation).
ω0
1
T = 2π is the frequency (cycles per unit time).
α
ω0 is the time lag.
The equation
x(t) = C cos(ω0 t − α) = C cos(ω0 (t − δ))
describes simple harmonic motion. We’ve taken our standard cosine function and applied several
transformations to the graph. The δ corresponds to horizontal translation by δ units. The ω
shrinks/expands the graph in the horizontal direction, changing the period from 2π to ω2π0 . The C
shrinks/expands the graph in the vertical direction so that the maximum of the graph is C and the
minimum is −C. A sample graph is included below.
Example: A body with mass m = 0.5 kilograms is attached to the end of a spring that is stretched
2 meters by a force of 144 newtons. It is set in motion with initial position x0 = 1.5 (m) and initial
velocity v0 = −9 (m/s). (The body is displaced to the right and is moving to the left at time t.)
Find the following:
•
•
•
•
•
the position function of the body
amplitude
frequency
period
time lag
We know that it takes 144 newtons to stretch our spring 2 meters. By Hooke’s law, the force
exerted by the spring is
−144 = −k(2).
Notice that the force exerts a negative force since it opposes the force doing the stetching. The
point is k = 72. That means we want to solve the linear DE
.5x00 + 72x = 0.
In this case, the circular frequency is
q
72
.5
= 12. That means the general solution to our DE is
x(t) = A cos(12t) + B sin(12t).
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MATH 2280-002
Lecture Notes: 2/01/2013-2/04/2013
9
Use the initial conditions to get that A = 1.5 = 32 and B = − 12
= − 34 . So our particular solution
is
3
3
x(t) = cos(12t) + − sin(12t),
2
4
π
and this function has period 12
and frequency 12
π . Of course, to complete the rest of the problem,
we need to rewrite our solution in the form C cos(ω0 t − α). The amplitude is
s √
2 r
3 2
3
9
9
3 5
C=
+ −
=
+
=
.
2
4
4 16
4
A
We need to be slightly clever to get the phase angle. We know that, since cos(α)
=C
is positive and
B
A
−1
sin(α) = C is negative, α is a fourth-quadrant angle. Because cos
gives
us
a
first-quadrant
C
angle, we get
2
A
= 2π − cos−1 √
≈ 5.81954.
α = 2π − cos−1
C
5
That means the time lag is
δ=
and
α
≈ 0.485 seconds
ω0
√
3 5
cos(12t − 5.81954).
x(t) ≈
4
4