Oscillations
•
SHM review
•
– Analogy with simple pendulum
•
– Break down of auxiliary equation
method and how to fix it
– General solution
– Solution in terms of initial
conditions
– Over damping as ideal damping
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
Over damping
– General solution
– Solution in terms of initial
conditions
Phase diagrams
– Un-damped phase diagram
– Obtaining phase equations directly
– The under-damped logarithmic
spiral
– Critical damping example
– General solution
•
Critical Damping
•
Harmonic oscillations in two
dimensions
– Lissajous figures
1
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.
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.
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.
2
Simple harmonic motion
Oscillations about equilibrium
x  xeq  A sin  t   
F  k ( x  xe )  k x '
A is the amplitude
k
m
U ( x)  x ' 2 ; T  x ' 2
2
2
 is the angular frequency
  k/m
 is the phase
U  mgy   mgl cos 
y
x
l cos 
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.

l
U  mgl 1   2 / 2 
mg
m
2
l


mgl
T

 
2l
2
Mass-spring analogies
U
m  m x'  l
mg
k
We also show that the
2 simple pendulum put in
l
small oscillation has an
analogous potential and
kinetic energy expression
and hence will have the
same formal solution as the
g spring-mass system.
3
l
 
mg
l
k
mg


  pend 
m
ml
SHM solution by DE methods
F  k x  m x  x   2 x  0
This is a homogenous, linear DE
We solve it in general by inserting
a solution of the form: x  Bexp( rt )
x   2 x  r 2 B exp(rt )   2 B exp( rt )  0
This leads to an auxillary equation of
One way around this problem is to insist
that x is real by demanding x* = x
x  B1 exp(i t )  B2 exp(-i t )
x*  B1 *exp(i t )  B2 *exp(i t )
x* = x if and only if B2  B1 *
This really means that rather than
form r    0 which has two roots: having arbitrary B and B we really
1
2
r1,2   - 2  i . We construct the only have one arbitrary number say B1.
Fortunately B1 is complex and thus has
general solution as the sum of terms
an arbitrary amplitude and phase. We
need two arbitrary real quantities in
x(t )  B1 exp(i t )  B2 exp(-i t )
order to match the initial position and
phase.
Complex numbers snuck in our solution!
Indeed B1,2 should be taken as complex
numbers as well. But x(t) must be real?
4
2
2
DE methods (continued)
x(t )  B1 exp(i t )  B1 *exp(-i t )
Choose B1   exp(i ) where  and
 are real numbers corresponding to the
phase and modulus of B1.
x(t )   e  i exp(i t )   ei exp(-i t )
  exp i  t       exp  i  t    
 2
exp i  t      exp  i  t    
2
x  2 cos  t     x  A cos  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.
m x   k x  bx or writing
 = k/m and  =b/2m we have
x  2 x   2 x  0
Auxillary eq by substituting
x  B exp(rt )
B exp(rt )  r 2  2  r   2   0
r 2  2 r   2  0
r1,2      2   2
5
Damping cases
Solutions will depend quite a bit on
relationship between  and .
r1,2      2   2
x  B1 exp(r1t )  B2 exp(r2t )
There are three cases:
Under damping
Define 1   2   2
r1,2     i1
x  e   t  B1 exp(i1t )  B2 exp( i1t ) 
We force a real x via x*=x which
x*  e   t  B1 *exp(i1t )  B2 *exp(i1t ) 
    0 under damped
 x  e   t  B1 exp(i1t )  B1 *exp( i1t ) 
 2   2  0 critically damped
Choose B1 
2
2
 2   2  0 over damped
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.
xe
 t

A
exp( i )
2
real amp & phase
A i1t   i1t  
e
e
2

x  Ae   t cos 1t    or e   t sin 1t   
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. 6
An under-damped case
1.5
x  Ae  t cos 1t   
x
1
 0
  /4
  /2
0.5
t
0
0
-0.5
0.5
1
1.5
2
2.5
3
1  4 
3.5
-1
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 frequency 1 so you can see a few wiggles before the
oscillation fades out.
7
Over damped case
r1,2      2   2
x  B1 exp(r1t )  B2 exp(r2t )
2   2   2
r1,2      2
x  B1 exp       2  t   B2 exp       2  t 
Defining 1 =    2 and 2 =    2
x  B1 exp(1t )  B2 exp(2t )
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 2   both 1=   2 and 2 =   2 are
both positive and thus both terms correspond to exponential decay.
We can re-arrange the blue form to match initial conditions.
  exp(1t )  1 exp(2t ) 
 exp(1t )  exp(2t ) 
x(t )  x0  2

v
 0








2
1


2
1

The initial condition form is easy to confirm by expanding to 1st order in t.
8
Critical damping
r1,2      2   2
but  2   2
It is interesting to note that the critically
damped case actually dies away faster than
Hence r1,2    and our usual expression the over damped case. Consider the case
where v0=0.
x  B1 exp(rt
1 )  B2 exp( r2t ) becomes
x   B1  B2  exp(  t )  B3 exp(  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!
The critically damped case will fall off
according to exp(- t)
The over damped case will have a
exp[-(2 t] piece which dies off faster
than the critically damped case. But it will
also contain a exp[-(2 t] piece which
Appendix C of M&T says in the case of
degenerate roots, the full solution is of the dies off slower than the critically damped
following form which can be confirmed by case.
direct substitution.
For many applications: vibration
abatement, shocks, screen door dampers,
x  A1  A2 t exp(  t )
one strives for critical damping.
In terms of initial conditions:


x   x0 1   t   v0t  exp( t )
9
Example of Critical Damping
1.2
1
0.8
0.6
  1.5 0
0.4
  0
0.2
  3 0
0
0
1
2
3
4
5
6
7
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 exp(1t )  1 exp(2t ) 
x(t )  x0 
 over damp
2  1


x   x0 1   t   v0t  exp(  t )
critical damp
10
Phase Diagram
It 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.
This curve is an
p2 k 2
 x  E ellipse with an area
2m 2
monotonic in energy.
Without damping the particle will execute
endless elliptical orbits of fixed energy in
this phase space p
x
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   kx
Essentially 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.
11
Phase Diagrams (continued)
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:
dx
dx
m  kx 
  2 x
dt
dt
dx
dx / dt  2 x
dx  2 x
 x dividing :




dt
dx / dt
x
dx
x
2
2
2
2
x

v
x

x
0
0
x dx   2 x dx  x ' dx '   2  x ' dx ' 
 2
v0
x0
2
2
x
x
 x 2   2 x 2  x 2 0   2v 2 0 This gives the phase diagram ellipse
We 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. x  B1 exp(1t )  B2 exp(2t )
The text shows a way of getting the spiral
by a variable transformation.
2
p
1.5
1
0.5
x
0
-1
-0.5
0
0.5
1
1.5
-0.5
12
-1
Critically damped phase diagram
It is very straight-forward to draw the
x   x0 1   t   v0t  exp(  t )
phase diagram once one has x(t) and its
x     x0 1   t   v0t  exp(  t )  derivative. We plot plot x and v for 50
times for 3 sets of initial conditions.
You can easily visualize the motion
  x0  v0  exp(  t ) 
x   v0   v0t  x0  t  exp(  t )
including the maximum displacement
and velocity and retrograde motion
2
1.5
1.5
Whoops I messed it up. How did I know?
What’s wrong with this picture?
x
1
x
x (0)  1 x (0)  0
1
x (0)  1 x(0)  1
x (0)  1 x(0)  0
x (0)  1 x(0)  2
0.5
x
0.5
0
-0.5
0
0
0.5
1
x
1.5
0
0.5
1
1.5
-0.5
-0.5
-1
-1
13
-1.5
Harmonic motion in two directions
F  k x x x  k y y y  m x x  k y y y
The equations of motion just decouple
x   x2 x 2  0 and y   x2 y 2  0 as do the
solutions and initial conditions
x  Ax cos  x t   x 
y  Ay cos  y t   y 
If  y   / 2; x=A cos  t and y=A sin t
and x 2  y 2  A2  cos 2 t+sin 2 t   A2
which is a circle
Lissajous figure’s are often a great way to
measure a phase difference on an
oscilloscope.
 y  1700
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
 y  450
conventions, the figure maximally opens
up into a circle when y =0.
-1.5
-1
Many cases are easy to see.
If  y  0 then x=y straight line.
If  y   then x=-y straight line.
1.5
 y  100
1
0.5
0
-0.5
0
-0.5
-1
0.5
1
1.5
 y  900
14
Lissajou figure with unequal frequency
Case where Ax  Ay  1
 y  100
1.5
 x   y  0 and  y  2 x  2
1
0.5
x  cos  t ; y  cos 2 t
0
-1.5
-1
-0.5
0
0.5
1
1.5
-0.5
y  cos 2  t  sin 2  t 
-1
cos 2  t  (1  cos 2  t )
-1.5
y  2 cos 2  t  1  2 x 2  1
 y  170
0
1.5
1.5
1
1
0.5
0.5
0
0
-1.5
-1
-0.5
0
-0.5
-1
-1.5
 y  900
0.5
1
1.5
-1.5
-1
-0.5
0
0.5
1
-0.5
-1
-1.5
15
1.5
A cool trig identity

1
n
n  n  3
n  n  4
n
n2
n4
n 6
cos nt   2cos t    2cos t   
2cos

t

2cos

t

...






2
1
2 1 
3 2 

Hence if Ax  Ay  1  x   y  0 and  y  n x  n
x  cos  t ; y  cos n t
y will be a polynomial in x of the form
y  a1 x n  a2 x n  2  a3 x n  4  ...
for  x  0  y    y  - cos n t
and the curve will flip over.
16
Lissajous figures with different frequencies
To the right is are
Lissajous figures
for the case Ax= Ay
and y = 3x and
x=0 for 3 different
y phases. The
figure maximally
opens up when y
=90.
 y  45
1.5
 y  100
0
1
1
0.5
0.5
0
0
-1.5
-1
-0.5
 y  90
0
0.5
1
1.5 -1.5
-1
-0.5
-0.5
-1
-1
-1.5
-1.5
 y  170
1.5
0
1
1
1.5
1.5
1
0
0
-0.5
0.5
0.5
0.5
-1
0
-0.5
0
-1.5
1.5
0
0.5
1
1.5
-1.5
-1
-0.5
0
-0.5
-0.5
-1
-1
-1.5
-1.5
0.5
1
17
1.5
Lissajous with irrational frequency ratios
 y  2.515  x
1.5
1
To the right are a Lissajous
figure for the case Ax= Ay , and
x =y =0 with the indicated
frequency ratios
0.5
0
-1.5
-1
-0.5
0
0.5
1
1.5
-0.5
-1
When the frequency ratio is not
a simple fraction (or irrational)
the path does not close and
finally fills up the whole screen.
-1.5
1.5
 y  2.506  x
1
0.5
How would this figure look if
0
 y  2.5 x ?
-1.5
-1
-0.5
0
0.5
1
-0.5
-1
-1.5
18
1.5