oscillations (1)

1

Oscillations• SHM review

– Analogy with simple pendulum

• SHM using differential equations– Auxiliary Equation

• Complex solutions

• Forcing a real solution

• The damped harmonic oscillator– Equation of motion

– Auxiliary equation

– Three damping cases

• Under damping– General solution

• Over damping– General solution

– Solution in terms of initial conditions

• Critical Damping– Break down of auxiliary equation

method and how to fix it

– General solution

– Solution in terms of initial conditions

– Over damping as ideal damping

• Phase diagrams– Un-damped phase diagram

– Obtaining phase equations directly

– The under-damped logarithmic spiral

– Critical damping example

• Harmonic oscillations in two dimensions

– Lissajous figures

2

What will we do in this chapter?

This is the first of several lectures on the the harmonic oscillator. We begin by reviewing our previous solution for SHM and use similar techniques to solve for a simple pendulum. We next solve the SHM using the auxiliary equation technique from linear differential equation theory. This allows us to extend our treatment to the case of a damped harmonic oscillator with a damping force proportional to drag.

Three damping cases are considered: under damped , over damped, and critically damped. The critically damped case -- besides being very practical -- brings a new wrinkle to the auxiliary equation technique.

We discuss the phase diagram which is a plot of trajectories in a phase space consisting of p and x. Several methods for computing these trajectories are discussed and the under damped, un- damped, and critically damped examples are drawn.

We conclude with a brief discussion of harmonic motion in two dimensions and Lissajou figures.

3

Simple harmonic motion

2 2( ) ' ;

( '

2

)

'2

e

k mU x x T x

F k x x k x

We already are familiar with this problem applied to the mass spring problem. We write the force law and potential and the solution which we obtained using energy conservation.

is the ampli

sin

A

tude

is the angu

Oscillations about eq

lar frequency

is the pha

ui

s

librium

e

/

eqx x A t

k m

cosl

l

mg

xy

2

22

cos

1 / 2

2 2 Mass-spring analogies

m m x'

pend

U mgy mgl

U mgl

mg mU l mgl T l

l

mgl k

l

k mg g

m ml l

We also show that the simple pendulum put in small oscillation has an analogous potential and kinetic energy expression and hence will have the same formal solution as the spring-mass system.

4

SHM solution by DE methods2

2

2 2 2

0

This is a homogenous, linear DE

We solve it in general by inserting

a solution of t

auxillary equatio

he form: Bexp( )

exp( ) exp( ) 0

This leads to an of

form

n

F k x m x x x

x rt

x x r B rt B

r

rt

21,

1 2

2

2 which has two roots:

- . We construct the

general solution as the sum of terms

( ) exp( ) ex

(-

0

p )x t B

r i

i t B i t

Complex numbers snuck in our solution! Indeed B1,2 should be taken as complex numbers as well. But x(t) must be real?

One way around this problem is to insist that x is real by demanding x* = x

2 1

1 2

1 2

exp( ) exp(- )

* *exp( ) *exp( )

x* = x if and only if *

x B i t B i t

x B i t B i

B

t

B

This really means that rather than having arbitrary B1 and B2 we really only have one arbitrary number say B1. Fortunately B1 is complex and thus has an arbitrary amplitude and phase. We need two arbitrary real quantities in order to match the initial position and phase.

5

DE methods (continued)

1 1

1

1

( ) exp( ) *exp(- )

Choose exp( ) where and

are real numbers corresponding to the

phase and modulus of B .

( ) exp( ) exp(- )

exp exp

exp2

i i

x t B i t B i t

B i

x t e i t e i t

i t i t

i t

exp

22 c cos osx A

i t

x t t

If we had chosen B1=-i exp(-i) we would have obtained an equally valid solution x = 2 sin(t - ).

As another example we could include a linear drag force acting on the mass along with the spring.

2

2

2 21,2

2

2 2

or writing

we have

2 0

Auxillary eq by substituting

e

= k/m

xp(

and =b/2

)

exp( ) 2 0

2 0

m

m x k x bx

x x x

x B rt

B rt r r

r

r r

6

Damping casesSolutions will depend quite a bit on relationship between and .

2

2 2

21,2

1 1 2 2

2 2

2 2

exp( ) exp( )

There are three

0 under damped

0 critically damped

0 over d

cas

amped

es:

r

x B rt B r t

The under damped case will have complex auxiliary roots and will have oscillatory behavior. The over damped case will have real roots and thus have a pure exponential time evolution. The critically damped case, with a single root, has some non-intuitive aspects to its solution.

Under damping

2 21 1,2 1

1 1 2 1

1 1 2 1

1 1 1 1

1

Define

exp( ) exp( )

We force a real x via x*=x which

* *exp( ) *exp( )

exp( ) *exp( )

Choose B exp( ) real amp & phase2

2

t

t

t

t

r i

x e B i t B i t

x e B i t B i t

x e B i t B i t

Ai

Ax e e

1 1

11cos or sin

i t i t

t tx t

e

Ae e t

The under damped oscillator has two constants (phase and amplitude) to match initial position and velocity. The solution dies away while oscillating but with a frequency other than = (k/m)1/2.

7

An under-damped case

-1

-0.5

0

0.5

1

1.5

0 0.5 1 1.5 2 2.5 3 3.5

x

t

/ 4

0

/ 2

1costx Ae t

1 4

Here is a plot of x(t) for the under-damped case for three different choices of phase. We have chosen the damping coefficient to be 1/4 of the effective frequencyso you can see a few wiggles before the oscillation fades out.

8

Over damped case

2 21,2

1 1

1 2

2 2

2 22 1,2

2 2

1 1 2

2

2

1 2 2 2

Defining = and =

e

exp( ) exp( )

exp

xp( ) exp( )

exp

r

x B rt B r t

r

x B t

x

B t

B t B t

In this case both the B1 and B2 coefficients must be real so that x is real. The two terms are both there to allow us to match the initial position and velocity boundary conditions. Since both 1= and 2 = are both positive and thus both terms correspond to exponential decay.

We can re-arrange the blue form to match initial conditions.

02 1 1 2 1 2

2 1 2 10

exp( ) exp( ) exp( ) exp( )( )x x

tv

t t tt

The initial condition form is easy to confirm by expanding to 1st order in t.

9

Critical damping

2 2 2 21,2

1,2

1 1 2 2

1 2 3

but

Hence and our usual expression

exp( ) exp( ) becomes

exp( ) exp( )

r

r

x B rt B r t

x B B t B t

To get a real x, B3 must be real. But this can’t possibly be right! With only 1 real coefficient we can’t match the initial position and velocity!

Appendix C of M&T says in the case of degenerate roots, the full solution is of the following form which can be confirmed by direct substitution.

1 2 exp( )x A A t t In terms of initial conditions:

0 01 exp( )x x t v t t

It is interesting to note that the critically damped case actually dies away faster than the over damped case. Consider the case where v0=0.

The critically damped case will fall off according to exp(- t)

The over damped case will have a exp[-( t] piece which dies off faster than the critically damped case. But it will also contain a exp[-( t] piece which dies off slower than the critically damped case.

For many applications: vibration abatement, shocks, screen door dampers, one strives for critical damping.

10

0

0.2

0.4

0.6

0.8

1

1.2

0 1 2 3 4 5 6 7

03

0 01.5

Example of Critical Damping

We use solutions with x0=1 and v0=0 and consider the case of critical damping and two cases of over damping. The critical damped case dies out much faster and the over damped case dies out more slowly as the over damping ratio increases.

2 1 1 20

2 1

0 0

exp( ) exp( )( ) over damp

1 exp( ) critical damp

t tx t x

x x t v t t

11

Phase DiagramIt is fashionable to view the motion of mechanical systems as trajectories in a phase space which is a plot of p (or v) versus x. Oscillations are a perfect example of such plots. A first “phase diagram” is for an un-damped oscillator. Since we have a purely position dependent one dimensional force we know that energy is conserved.

22

2 2

p kx E

m

This curve is an ellipse with an area monotonic in energy.

Without damping the particle will execute endless elliptical orbits of fixed energy in this phase space

x

p

In Newtonian mechanics, we specify the initial conditions as x(0) and p(0) (or v(0)). This corresponds to one point in the phase diagram which specifies the initial energy or the phase space ellipse. We note that with this choice of coordinates, we have a clockwise phase space orbit. This is because of the negative sign in the equation of motion: p kxEssentially this means that p must decrease when x is positive and must increase when x is negative.

The particular phase space orbit will depend on the nature of the forces but in general we know that no two phase space orbits can intersect. Otherwise several motions would be possible for the same set of initial conditions.

12

Phase Diagrams (continued)

-1

-0.5

0

0.5

1

1.5

2

-1 -0.5 0 0.5 1 1.5

x

pWe show the phase diagram for an under damped oscillation. Again we have a clockwise motion but this time the energy continuously decreases with time and eventually disappears. This sort of curve is called a logarithmic spiral. We could in principle obtain this directly from out solution. 1 1 2 2exp( ) exp( )x B t B t

The text shows a way of getting the spiral by a variable transformation.

One can construct the phase diagram directly from x(t) and its derivative or (in the absence of dissipation) from the conservation of energy. Often the phase diagram can be obtained through a clever separation of variables:

0 0

2 22

2 2 2 20 02 2 2

2 2 2 2 2 20 0

/ dividing :

/

' ' ' '2 2

This gives the phase diagram ellipse

x x

v x

dx dx dx dx dt x dx xm kx x xdt dt dt dx dt x dx x

x v x xx dx x dx x dx x dx

x x x v

13

Critically damped phase diagram

0 0

0 0

0 0

20 0 0

1 exp( )

1 exp( )

exp( )

exp( )

x x t v t t

x x t v t t

x v t

x v v t x t t

-1

-0.5

0

0.5

1

1.5

0 0.5 1 1.5x

x What’s wrong with this picture?

It is very straight-forward to draw the phase diagram once one has x(t) and its derivative. We plot plot x and v for 50 times for 3 sets of initial conditions. You can easily visualize the motion including the maximum displacement and velocity and retrograde motion

Whoops I messed it up. How did I know?

x

x (0)

(0)

(0) 1

1 (0

(0) 0

(0) 1 (0) 1

0

0

)

1 ( ) 2

x

x

x

x

x

x

x

x

14

Harmonic motion in two directions

2 2 2 2

The equations of motion just decouple

0 and y 0 as do the

solutions and initial condition

co

s

s

s

co

x x x

x y y

x x

y y y

F k x x k y y m x x k y y

x

x

y A t

x y

A t

Lissajous figure is a plot of y versus x as t is varied. To the right is a Lissajous figure for the case Ax= Ay and x= y and x=0 for 4 different y phases. With these conventions, the figure maximally opens up into a circle when y =0.

010y 0170y

090y

045y

Many cases are easy to see.

If 0 then x=y straight line.

If then x=-y straight line.

y

y

y

2 2 2 2 2 2

If / 2; x=A cos t and y=A sin t

and cos t+sin t

which is a circle

x y A A

Lissajous figure’s are often a great way to measure a phase difference on an oscilloscope.

15

Lissajou figure with unequal frequency

-1.5

-1

-0.5

0

0.5

1

1.5

-1.5 -1 -0.5 0 0.5 1 1.5

-1.5

-1

-0.5

0

0.5

1

1.5

-1.5 -1 -0.5 0 0.5 1 1.5

-1.5

-1

-0.5

0

0.5

1

1.5

-1.5 -1 -0.5 0 0.5 1 1.5

010y

090y 0170y

2 2

2 2

2 2

Case where 1

0 and 2 2

cos ; cos 2

cos sin

cos (1 cos )

2cos 1 2 1

x y

x y y x

A A

x t y t

y t t

t t

y t x

16

A cool trig identity

2 4 63 41cos 2cos 2cos 2cos 2cos ...

1 22 1 2 3n n n nn nn n n

n t t t t t

2 41 2 3

Hence if 1 0 and

cos ; cos will be a polynomial in x of the form

...

for 0 -cos

and the curve will flip over.

x y x y y x

n n n

x y

A A n n

x t y n t y

y a x a x a x

y n t

17

Lissajous figures with different frequencies

-1.5

-1

-0.5

0

0.5

1

1.5

-1.5 -1 -0.5 0 0.5 1 1.5

010y

090y 0170y

045y To the right is are Lissajous figures for the case Ax= Ay and y = 3x and x=0 for 3 different y phases. The figure maximally opens up when y

=90.

18

Lissajous with irrational frequency ratios

-1.5

-1

-0.5

0

0.5

1

1.5

-1.5 -1 -0.5 0 0.5 1 1.5

To the right are a Lissajous figure for the case Ax= Ay , and x =y =0 with the indicated frequency ratios

When the frequency ratio is not a simple fraction (or irrational) the path does not close and finally fills up the whole screen.

How would this figure look if

-1.5

-1

-0.5

0

0.5

1

1.5

-1.5 -1 -0.5 0 0.5 1 1.5

2.506y x

2.515y x

2.5y x ?