Download MATH 312 Section 5.1: Linear Models: IVPs

Free Undamped Motion
Free Damped Motion
Driven Motion
MATH 312
Section 5.1: Linear Models: IVPs
Prof. Jonathan Duncan
Walla Walla University
Spring Quarter, 2008
Conclusion
Free Undamped Motion
Free Damped Motion
Outline
1
Free Undamped Motion
2
Free Damped Motion
3
Driven Motion
4
Conclusion
Driven Motion
Conclusion
Free Undamped Motion
Free Damped Motion
Driven Motion
Recalling the Problem
In section 1.3 we first examined the model for the motion of a
spring/mass system.
Example
In our spring/mass system, x(t) is
the displacement of the mass from
its equilibrium position at time t
and s is the displacement of the
spring and mass at equilibrium from
the unstretched spring’s position.
Conclusion
Free Undamped Motion
Free Damped Motion
Driven Motion
Conclusion
Constructing the Model
We now construct a differential equation for this system.
Known Relations in the Spring/Mass System
F = k(s + x)
F = ma
(Hook’s Law)
(Newton’s 2nd Law)
2
d x
dt 2
mg = ks
a=
(Acceleration)
(Equilibrium)
Example
Putting this all together, we get:
m
d 2x
= −kx
dt 2
or
d 2x
+ ω2 x = 0
dt 2
where ω 2 =
k
m
Free Undamped Motion
Free Damped Motion
Driven Motion
Conclusion
Solving the Equation
The resulting differential equation is a second order linear equation
which we can solve using the methods learned in chapter 4.
Example
Solve
d 2x
dt 2
+ ω 2 x = 0.
Solution
Using an auxiliary equation yields the solution
x = C1 cos ωt + C2 sin ωt
Note:
Functions of this form represent the position of a system in simple
harmonic motion, also called free undamped motion.
Free Undamped Motion
Free Damped Motion
Driven Motion
Conclusion
Understanding Initial Conditions
In order to apply this model solution to a real-world situation, we
will need initial conditions.
Initial Conditions
Two obvious initial conditions are:
Initial Position: x(0) = x0
Initial Velocity: x 0 (0) = v0
Example
Here are two examples of possible initial conditions.
Starting below the equilibrium point, with an upward
velocity:
Starting above the equilibrium point, with a downward velocity:
x0 > 0
x0 < 0
v0 < 0
v0 > 0
Free Undamped Motion
Free Damped Motion
Driven Motion
Conclusion
Characteristics
There are several characteristics of free undamped motion which
can be found from the solution.
Period
The period is the time it takes the system to make one complete
cycle. From our solution function, this can be shown to be T = 2π
ω .
Frequency
The frequency is the number of oscillations per unit time. It is
1
1
ω
related to period by frequency = period
= 2π/ω
= 2π
.
Amplitude
The amplitude of the system is how far above and below the
equilibrium point the system travels. We explore this further on
the next slide.
Free Undamped Motion
Free Damped Motion
Driven Motion
Conclusion
Finding the Amplitude
Because our solution is the sum of two trigonometric functions, we
must make use of some identities to find the amplitude.
x(t) = A
C1
C2
cos ωt +
sin ωt
A
A
Next we find a phase angle ϕ for which
sin ϕ =
C1
A
and
cos ϕ =
C2
A
x(t) = A (sin ϕ cos ωt + cos ϕ sin ωt)
x(t) = A sin(ωt + ϕ)
Free Undamped Motion
Free Damped Motion
Driven Motion
Finding A
We know that such an A will be the amplitude since it is the
coefficient of a lone trigonometric function.
Finding A
sin ϕ =
C1
opposite
=
A
hypotenuse
C2
adjacent
=
A
hypotenuse
q
⇒ A = C12 + C22
cos ϕ =
Conclusion
Free Undamped Motion
Free Damped Motion
Driven Motion
An Example
We now apply this knowledge to an example from your text.
Example
A 20 point weight stretches a spring 6 inches. The weight is
released from rest 6 inches below the equilibrium position.
π
12
and t = π4 .
1
Find the position of the weight at times t =
2
What is the velocity at t = π4 ?
3
When does the weight pass through the equilibrium point?
4
Find the period, frequency, and amplitude of the motion.
All of the above can be found from the differential equation
d 2x
+ 64x = 0
dt 2
Conclusion
Free Undamped Motion
Free Damped Motion
Driven Motion
Assumptions of Free Undamped Motion
The free undamped model makes several naive assumptions.
Assumption #1
The spring constant remains constant. It may instead vary with age,
temperature, or other factors.
We May Wish to Use:
k(t) = k1 e −αt
or
k(t) = k1 t
Assumption #2
There is no resisting or damping force acting on the spring.
We May Wish to Use:
If β is the constant of resistance, force is reduced by β dx
dt yielding:
m
d 2x
dx
= −kx − β
dt 2
dt
Conclusion
Free Undamped Motion
Free Damped Motion
Driven Motion
Modeling Free Damped Motion
The equation for free damped motion is as follows.
Free Damped Motion
The position of the mass in a free damped spring/mass system is given
by:
d 2x
dx
d 2x
dx
m 2 = −kx − β
⇒ 2 + 2λ
+ ω2 x = 0
dt
dt
dt
dt
Where ω 2 =
k
m
as before, and λ =
β
2m
Solving:
Solving this last equation yields the auxiliary equation:
m2 + 2λm + ω 2 = 0
So that:
m = −λ ±
p
λ2 − ω 2
Conclusion
Free Undamped Motion
Free Damped Motion
Driven Motion
Conclusion
Case I: λ2 − ω 2 > 0
The first of the three solution cases we study is when we have two
distinct real roots.
Over-damped Systems
If λ2 − ω 2 > 0 then the auxiliary equation has two distinct
solutions yielding the following solution to the differential equation.
√
√
2
2
2
2
x(t) = e −λt C1 e λ −ω t + C2 e − λ −ω t
The system is called
over-damped because β is
much bigger than k. Notice
that the mass can pass through
the equilibrium point at most
once.
Free Undamped Motion
Free Damped Motion
Driven Motion
Conclusion
Case II: λ2 − ω 2 = 0
Next we look at the case where the discriminant is zero.
Critically Damped Systems
If λ2 − ω 2 = 0 then the auxiliary equation has a single real solution
with multiplicity two, yielding the solution:
x(t) = C1 e −λt + C2 te −λt
The system is called critically
damped since the mass β and
k are in balance. Note that the
mass can still pass through the
equilibrium point at most once.
Free Undamped Motion
Free Damped Motion
Driven Motion
Case III: λ2 − ω 2 < 0
The final case is that of two complex solutions.
Under-Damped Systems
If λ2 − ω 2 < 0 then the auxiliary equation has a pair of complex
conjugate solutions yielding the solution:
p
p
x(t) = C1 e −λt cos ω 2 − λ2 t + C2 e −λt sin ω 2 − λ2 t
This system is called
under-damped as the mass can
pass through the equilibrium
point multiple times. However,
as t increases, the amplitude
will go to zero.
Conclusion
Free Undamped Motion
Free Damped Motion
Driven Motion
Conclusion
An Example
Consider the following example.
Example
A 4 foot spring measures 8 feet in length after an 8 pound weight is
attached to it. The√medium through which the weight moves offers a
resistance equal to 2 times its velocity. Find and describe the solution
to this problem if the weight is released from equilibrium with a
downward velocity of 5 ft/sec.
640 − √2 t
x(t) =
e 16 sin
7
r
7
t
128
This is underdamped motion resulting in
the following graph. The mass passes
through equilibrium when:
√
nπ 128
t= √
7
Free Undamped Motion
Free Damped Motion
Driven Motion
Conclusion
Modeling Driven Motion
Our last modification to the spring/mass system is to add a driving force.
Driven Motion
If an external force is given by the function f (t) and is applied to the
spring mass system, the equation becomes:
m
dx
d 2x
= −kx − β
+ F (t)
dt 2
dt
We rewrite this equation to put it into our more standard form.
Standard Form
The standard equation for a driven spring/mass system in a damping
medium is:
d 2x
dx
+ 2λ
+ ω 2 x = F (t)
2
dt
dt
Free Undamped Motion
Free Damped Motion
Driven Motion
Conclusion
An Example
We end with another example involving driven motion.
Example
A mass of 1 slug is attached to a spring with constant 5 = lb/ft. The
mass is released 1 foot below equilibrium, with downward velocity of 5
ft/sec. A viscus medium dampens with a force numerically equivalent to
2 times the instantaneous velocity. Find the equation of motion if the
system is driven by an extremal force equal to:
f (t) = 12 cos 2t + 3 sin 2t
x(t) = e −t cos 2t + 6e t sin 2t + 3 sin 2t
The steady-state is shown in green, the
transient terms in yellow, and the sum is
red.
Free Undamped Motion
Free Damped Motion
Driven Motion
Conclusion
Important Concepts
Things to Remember from Section 5.1
1
Model and solution process for free undamped motion.
2
Model and solution process for free damped motion with cases
Case I: λ2 − ω 2 > 0
Case II: λ2 − ω 2 = 0
Case III: λ2 − ω 2 < 0
3
Model and solution process for driven motion.