MAT 239 (Differential Equations, Prof. Swift)§3.7 and 3.8: Oscillators and Complex Numbers

Topics:Oscillators without external force.The R, δ description of oscillator solutions.The polar form of complex numbers.Driven oscillators.

Oscillators: For most of you, the reason you are required to take this class is so thatyou can solve the DEs describing oscillators.

Suppose that m is the mass of an oscillator, γ is the damping constant, and k isthe spring constant. If there is no external force, then the DE for the position of themass, u(t), is

mu′′ + γu′ + ku = 0, or u′′ +γ

mu′ + ω2

0u = 0,

where ω0 =√

k/m is the natural frequency of the oscillator. It turns out that ω0

is the frequency of the undamped oscillations when γ = 0, but remember that ω0 isdefined even if the oscillator is damped (i.e. γ > 0).

Recall that the weight is not the mass. The weight is W = mg. In English units,g = 32ft/sec2, so the mass m = W/g has units [lb sec2/ft]. Furthermore, γ has units[lb sec/ft], and k has units [lb/ft]. Things are more straightforward in metric units.

Assuming that m and k are positive, and γ ≥ 0, the roots of the characteristicequation, and the resulting motion of the oscillators, fall into four classes:

γ = 0, Undamped: roots ±iω0.

0 < γ < 2mω0, Underdamped: roots − γ2m

± iµ, where µ =√ω20 − ( γ

2m)2 < ω0

γ = 2mω0, Critically Damped: roots − γ2m

= −ω0, with multiplicity 2

γ > 2mω0, Overdamped: two distinct real, negative roots − γ2m

±√( γ2m

)2 − ω20.

One goal of this handout is to get a better understanding of the motion in theundamped and underdamped cases.The R, δ description of oscillator solutions. The general solution of the un-damped oscillator can be written in two equivalent ways:

u(t) = C1 cos(ω0t) + C2 sin(ω0t) = R cos(ω0t− δ), (1)

where the constants C1, C2, R and δ satisfy the equations

C1 = R cos δ, C2 = R sin δ. (2)

The constants C1 and C2 are usually better for initial value problems, but thesecond form of the solution in equation (1) is easier to understand. The amplitude ofthe oscillation is R, and δ gives a phase shift. Every solution is periodic with periodT = 2π/ω0.

In the underdamped case, the general solution is

u(t) = e−γ2m

t (C1 cosµt+ C2 sinµt) = Re−γ2m

t cos(µt− δ).

The envelope of the decaying oscillation is u(t) = ±Re−γ2m

t. (Recall that −1 ≤cos(µt − δ) ≤ 1.) We call µ the quasi frequency, and Td = 2π/µ is the quasi period,since the motion “sort of” repeats after time Td.The Polar Form of Complex Numbers: A complex number z = C1+ i C2 can bewritten in polar form z = Reiδ, where R is the radius and δ is the phase of z. SinceC1 + i C2 = Reiδ = R(cos δ + i sin δ), the relationship between the rectangular formand the polar form is given by equation (2): C1 = R cos δ and C2 = R sin δ.

I will now prove equations (1) and (2). Recall from §3.4 that when the roots of thecharacteristic equation are r = ±i ω0 the general solution to can be written as a linearcombination of the real and imaginary parts of the complex solution u(t) = eiω0t:

u(t) = C1Re(eiω0t

)+ C2Im

(eiω0t

)= C1 cosω0t+ C2 sinω0t

= Re [(C1 − i C2)(cosω0t+ i sinω0t)] Note the minus sign!

= Re((C1 − i C2)e

iω0t)

Now, if we choose R and δ so that

C1 + i C2 = Reiδ (3)

then C1 − i C2 = Re−iδ, and continuing from the last equation we get

u(t) = Re(Re−iδeiω0t

)= RRe

(ei(ω0t−δ)

)= R cos(ω0t− δ)

This completes the proof of equations (1) and (2).Equations (2) and (3) say the same thing. They relate rectangular coordinates

(C1, C2) to polar coordinates (R, δ). Given R and δ, it is easy to find C1 and C2. Thereverse procedure is not so easy. If we know C1 and C2, and want R and δ, we havethe formulas

R =√C2

1 + C22 , tan(δ) =

C2

C1

.

The formula for R is wonderful. But, the formula for tan(δ) does not determine δ.It is not always true that δ = arctan(C2/C1). If δ is in the second or third quadrant,then δ = arctan(C2/C1) + π. If C1 = 0, then tan(δ) is undefined. Draw a picture ofthe point (C1, C2) in the plane to help you determine δ.

Example: Write u = − cos 3t+ sin 3t in the form u = R cos(ω0t− δ).First of all, we must have ω0 = 3. Then observe that C1 = −1 and C2 = 1. Now

draw a picture of the point (−1, 1) in the (C1, C2) plane. The polar coordinates ofthis point are easily obtained from geometry: R =

√2 and δ = 3π/4. (We have also

shown that −1 + i =√2ei3π/4.) Therefore,

u(t) = − cos(3t) + sin(3t) =√2 cos(3t− 3π/4).

The amplitude of the oscillation is R =√3. This identity can, and should, be checked

with a calculator. Plot the two expressions as y1 and y2, with either y1 dotted or y2thick. You should see the graph of y2 drawn on top of the graph of y1.

Driven Oscillators: The most important problems involving oscillators have a forc-ing (or driving) term on the right hand side: We will focus on the problem

mu′′ + γu′ + ku = F0 cos(ωt), (4)

where m > 0, γ ≥ 0, and k > 0.This problem can be solved by the method of undetermined coefficients. This is

quite easy when γ = 0, but it’s a mess when γ > 0. We separate the problem intothree cases:Case I: γ = 0, ω = ω0 =

√k/m.

Case II: γ = 0, ω = ω0.Case III: γ > 0Examples of Case I and II: Solve the IVPs u′′+ku = cos(3t), u(0) = 0, u′(0) = 0,with k = 10 (Case I) and k = 9 (Case II).

The solutions are u(t) = − cos(√10t) + cos(3t), and u(t) = 1/6 t sin(3t), respec-

tively. Here are some graphs of solutions.

1 2 3 4t

-0.6

-0.4

-0.2

0.2

0.4

0.6

u

k = 10

1 2 3 4t

-0.6

-0.4

-0.2

0.2

0.4

0.6

u

k = 9

Figure 1: The solutions to the IVPs for a short time interval. The solutions lookalmost identical.

10 20 30 40t

-2

-1

1

2u

k = 10

10 20 30 40t

-6-4-2

246

u

k = 9

Figure 2: The solutions to the IVPs for a longer time interval.

The envelope of the solution with k = 10 can be obtained from the trig identities

cos(A±B) = cos(A) cos(B)∓ sin(A) sin(B).

We can combine these together, anticipating that we want an identity for u(t) =− cos(

√10t) + cos(3t), where

√10 > 3:

− cos(A+B) + cos(A−B) = 2 sin(A) sin(B)

Applying this to our problem with A + B =√10 and A − B = 3, so that A =

(√10 + 3)/2 ≈ 3.08 and B = (

√10− 3)/2 ≈ 0.0811:

u(t) = − cos(√10t) + cos(3t) = 2 sin(A t) sin(B t) = [2 sin(B t)] sin(A t)

The “carrier” wave is sin(A t) which has period 2π/A ≈ 2.04 and the “envelope” isu(t) = ±2 sin(B t). Now, 2 sin(B t) has period 2π/B ≈ 77.4. The envelope is plottedas a dotted line in Figures 1 and 2. Figure 2 (k = 10) shows the phenomenon of“beats,” wherein the amplitude of the oscillation varies with a period of half of 77.4,which is 38.7.

Case III: This is the most important case. Any oscillators you will run into will havesome damping. The general solution to equation (4) is

u(t) = uh(t) + Up(t)

where uh(t) → 0 as t → ∞, since the oscillator is damped (γ > 0). The steady statesolution, also called the forced response, has the same frequency as the driving term:

Up(t) = A cos(ωt) +B sin(ωt) = R cos(ωt− δ)

If the A, B form of the particular solution is substituted into equation (4), we get asystem of two linear equations for A and B which have the solution

A =F0m(ω2

0 − ω2)

m2(ω20 − ω2)2 + γ2ω2

, B =F0γω

m2(ω20 − ω2)2 + γ2ω2

.

Note that these expressions use ω0 =√

k/m, even though the oscillator is damped.The “rectangular coordinates” A and B can be converted to “polar coordinates” Rand δ, as described in in §3.7. Note that B > 0, so δ is in quadrant I or II. Theinverse cosine function gives an angle in quadrant I or II, so the best expressions are

R =F0√

m2(ω20 − ω2)2 + γ2ω2

, δ = arccos

(m(ω2

0 − ω2)√m2(ω2

0 − ω2)2 + γ2ω2

)

These expressions for the forced response are quite complicated. Nonetheless, we canget a feeling for them by considering limits and drawing figures.

Note that as ω → 0, the response has A → F0/k and B → 0. This makes sense,because the right-hand side of (4) is F0 when ω = 0, and a particular solution in thiscase is Up(t) = F0/k. Furthermore, R → 0 as ω → ∞.

All of the constantsm, γ, etc. make the expressions for A and B look complicated.However, things are much simpler if we define two dimensionless quantities:

x =ω

ω0

, and Q =mω0

γ

Here, x is the ratio of the driving frequency (ω) to the natural frequency (ω0). Theamount of damping is measured by the so-called oscillator Q, or quality factor. Ahigh Q oscillator has very little damping. With these two variables x and Q, theexpression for R is more understandable:

R =F0

k

1√(1− x2)2 + (x/Q)2

If we treat Q as a constant, then R as a function of x is a maximum at xm, with amaximum value of Rm, given by

x2m = 1− 1

2Q2, Rm =

F0

k

Q√1− 1/(2Q)2

≈ F0

k

(Q+

1

8Q

)provided thatQ2 ≥ 1/2. The approximate expression for Rm is good forQ large. Notethat x2

m is half way between 1 (when ω = ω0) and the square of the quasifrequencyµ, since it can be shown that (µ/ω0)

2 = 1− 1/Q2.

0.25 0.5 0.75 1 1.25 1.5 1.75 2x

1

2

5

10Amp

Figure 3: The scaled amplitude (Amp = Rk/F0) of the steady state response as afunction of x = ω/ω0 for Q = 10, 5, 2, 1√

2, and 1

3. The dotted line goes through the

maxima of the curves, (xm, Rm).

0.25 0.5 0.75 1 1.25 1.5 1.75 2x

����4

����2

3 ��������4

Π

∆

Figure 4: The phase δ of the steady state response as a function of x = ω/ω0 forQ = 10, 5, 2, 1/

√2, and 1/3. As Q → ∞, the phase approaches the step function

δ = 0 if x < 1 and δ = π if x > 1.

Finding the Steady State Oscillator Response Using Complex Numbers:The standard method of calculating of A and B in the particular solution is trulygruesome. However, it is quite easy using (what else?) complex numbers. The ODE(4) has the form

L[u(t)] = mu′′ + γu′ + ku = Re(F0 e

iωt)

where L is a linear operator. We look for a solution of the form

Up(t) = A cos(ωt) +B sin(ωt) = Re((A− i B)eiωt

)When we plug this into the ODE to find Up(t), we just have to solve

L[(A− i B)eiωt] = F0 eiωt.

Example: Find a particular solution to

L[u] = u′′ + u′ + 4u = cos(ωt)

To find the particular solution, we need to solve L[(A− i B)eiωt] = eiωt for A and B.The operator L[(A− i B)eiωt] is easily computed, since A and B are constants. Theresult is

(A− i B)(−ω2 + i ω + 4)eiωt = eiωt.

We can divide both sides by eiωt to get

(A− i B)(−ω2 + i ω + 4) = 1,

or

(A− i B) =1

−ω2 + i ω + 4=

1

4− ω2 + i ω=

4− ω2 − i ω

(4− ω2)2 + ω2.

Therefore A =4− ω2

(4− ω2)2 + ω2and B =

ω

(4− ω2)2 + ω2. The steady state solution is

Up(t) =4− ω2

(4− ω2)2 + ω2cos(ωt) +

ω

(4− ω2)2 + ω2sin(ωt)

Alternative Method: The way to really do these problems is to let A be a complexamplitude and write

Up(t) = Re(Aeiωt) = Re(A) cos(ωt)− Im(A) sin(ωt)

Then the previous problem becomes:

A(−ω2 + i ω + 4)eiωt = eiωt.

or

A =1

−ω2 + i ω + 4=

1

4− ω2 + i ω=

4− ω2 − i ω

(4− ω2)2 + ω2.

This gives the same solution as before. (These two methods are almost the same.)Extra Credit: Use either of these complex number methods to justify the general

formulas (10), (11), and (12) in §3.8 of the book. This is worth 5 class points, andit’s not really very hard. Due in class on the review day for Exam 2.