MATH 312 Section 5.1: Linear Models: IVPsmath.wallawalla.edu/~duncjo/courses/math312/spring08/notes/5-1... · Recalling the Problem ... Solve d2x dt2 +ω 2x = 0. Solution Using an

Free Undamped Motion Free Damped Motion Driven Motion Conclusion

MATH 312Section 5.1: Linear Models: IVPs

Prof. Jonathan Duncan

Walla Walla University

Spring Quarter, 2008


Outline

1 Free Undamped Motion

2 Free Damped Motion

3 Driven Motion

4 Conclusion


Recalling the Problem

In section 1.3 we first examined the model for the motion of aspring/mass system.

Example

In our spring/mass system, x(t) isthe displacement of the mass fromits equilibrium position at time tand s is the displacement of thespring and mass at equilibrium fromthe unstretched spring’s position.


Constructing the Model

We now construct a differential equation for this system.

Known Relations in the Spring/Mass System

F = k(s + x) (Hook’s Law)

F = ma (Newton’s 2nd Law)

a =d2x

dt2(Acceleration)

mg = ks (Equilibrium)

Example

Putting this all together, we get:

md2x

dt2= −kx or

d2x

dt2+ ω2x = 0 where ω2 =

k

m


Solving the Equation

The resulting differential equation is a second order linear equationwhich we can solve using the methods learned in chapter 4.

Example

Solve d2xdt2 + ω2x = 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 simpleharmonic motion, also called free undamped motion.


Understanding Initial Conditions

In order to apply this model solution to a real-world situation, wewill need initial conditions.

Initial Conditions

Two obvious initial conditions are:

Initial Position: x(0) = x0

Initial Velocity: x ′(0) = v0

Example

Here are two examples of possible initial conditions.

Starting below the equilib-rium point, with an upwardvelocity:

x0 > 0

v0 < 0

Starting above the equilib-rium point, with a down-ward velocity:

x0 < 0

v0 > 0


Characteristics

There are several characteristics of free undamped motion whichcan be found from the solution.

Period

The period is the time it takes the system to make one completecycle. 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 isrelated to period by frequency = 1

period = 12π/ω = ω

2π .

Amplitude

The amplitude of the system is how far above and below theequilibrium point the system travels. We explore this further onthe next slide.


Finding the Amplitude

Because our solution is the sum of two trigonometric functions, wemust make use of some identities to find the amplitude.

x(t) = A

(C1

Acos ωt +

C2

Asin ωt

)Next we find a phase angle ϕ for which

sin ϕ =C1

Aand cos ϕ =

C2

A

x(t) = A (sinϕ cos ωt + cos ϕ sin ωt)

x(t) = A sin(ωt + ϕ)


Finding A

We know that such an A will be the amplitude since it is thecoefficient of a lone trigonometric function.

Finding A

sin ϕ =C1

A=

opposite

hypotenuse

cos ϕ =C2

A=

adjacent

hypotenuse

⇒ A =√

C 21 + C 2

2


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 isreleased from rest 6 inches below the equilibrium position.

1 Find the position of the weight at times t = π12 and t = π

4 .

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

d2x

dt2+ 64x = 0


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) = k1e−αt or k(t) = k1t

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 dxdt yielding:

md2x

dt2= −kx − β

dx

dt


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 givenby:

md2x

dt2= −kx − β

dx

dt⇒ d2x

dt2+ 2λ

dx

dt+ ω2x = 0

Where ω2 = km as before, and λ = β

2m

Solving:

Solving this last equation yields the auxiliary equation:

m2 + 2λm + ω2 = 0

So that:m = −λ±

√λ2 − ω2


Case I: λ2 − ω2 > 0

The first of the three solution cases we study is when we have twodistinct real roots.

Over-damped Systems

If λ2 − ω2 > 0 then the auxiliary equation has two distinctsolutions yielding the following solution to the differential equation.

x(t) = e−λt(C1e

√λ2−ω2t + C2e

−√

λ2−ω2t)

The system is calledover-damped because β ismuch bigger than k. Noticethat the mass can pass throughthe equilibrium point at mostonce.


Case II: λ2 − ω2 = 0

Next we look at the case where the discriminant is zero.

Critically Damped Systems

If λ2 − ω2 = 0 then the auxiliary equation has a single real solutionwith multiplicity two, yielding the solution:

x(t) = C1e−λt + C2te

−λt

The system is called criticallydamped since the mass β andk are in balance. Note that themass can still pass through theequilibrium point at most once.


Case III: λ2 − ω2 < 0

The final case is that of two complex solutions.

Under-Damped Systems

If λ2 − ω2 < 0 then the auxiliary equation has a pair of complexconjugate solutions yielding the solution:

x(t) = C1e−λt cos

√ω2 − λ2t + C2e

−λt sin√

ω2 − λ2t

This system is calledunder-damped as the mass canpass through the equilibriumpoint multiple times. However,as t increases, the amplitudewill go to zero.


An Example

Consider the following example.

Example

A 4 foot spring measures 8 feet in length after an 8 pound weight isattached to it. The medium through which the weight moves offers aresistance equal to

√2 times its velocity. Find and describe the solution

to this problem if the weight is released from equilibrium with adownward velocity of 5 ft/sec.

x(t) =640

7e−

√2

16t sin

√7

128t

This is underdamped motion resulting inthe following graph. The mass passesthrough equilibrium when:

t =nπ√

128√7


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 thespring mass system, the equation becomes:

md2x

dt2= −kx − β

dx

dt+ F (t)

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 dampingmedium is:

d2x

dt2+ 2λ

dx

dt+ ω2x = F (t)


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. Themass is released 1 foot below equilibrium, with downward velocity of 5ft/sec. A viscus medium dampens with a force numerically equivalent to2 times the instantaneous velocity. Find the equation of motion if thesystem is driven by an extremal force equal to:

f (t) = 12 cos 2t + 3 sin 2t

x(t) = e−t cos 2t + 6et sin 2t + 3 sin 2t

The steady-state is shown in green, thetransient terms in yellow, and the sum isred.


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 − ω2 > 0Case II: λ2 − ω2 = 0Case III: λ2 − ω2 < 0

3 Model and solution process for driven motion.

Documents

MATH 312 Section 5.1: Linear Models: IVPsmath.wallawalla.edu/~duncjo/courses/math312/spring08/notes/5-1... · Recalling the Problem ... Solve d2x dt2 +ω 2x = 0. Solution Using an