physics 430: lecture 12 2-d oscillators and damped oscillations dale e. gary njit physics department
DESCRIPTION
October 12, 2010 Isotropic Oscillator If we redefine the origin of time to coincide with the time that, say, the x position is at its maximum, this becomes where is the relative phase = ( y x ). Consider a ball bearing in a bowl. It may oscillate in only one direction, i.e. in the x direction or the y direction. This motion would correspond to the above equations when the constant A y = 0 or A x = 0, respectively. The ball could go in a straight line at an angle to the x axis, i.e. in both x and y. That would correspond to A x = A y, and = 0. The ball could go in a circle about the bottom of the bowl, which would correspond to A x = A y, and = /2, in one direction, or = /2 in the other direction. Some other possibilities: x y x y x y TRANSCRIPT
Physics 430: Lecture 12 2-D Oscillators and Damped
OscillationsDale E. Gary
NJIT Physics Department
October 12, 2010
5.3 Two-Dimensional Oscillators
It is trivial to extend our idea of oscillators to other dimensions. For example, the spring arrangement in the figure at right oscillates in two dimensions.
In general, the springs in the x and y directions could have a different spring constants
Note that these springs may represent binding forces of an atom in a molecule or crystal.
If the spring constants are the same, the oscillator is called isotropic, and there is a single frequency of oscillation
There are two equations of motion, one for each dimension, given by
Although the solutions are the same for x and y, the constants of integration, which depend on the initial conditions, and not the same (in general):
.yx kk
.2
2
yyxx
./ mk
.)cos()()cos()(
yy
xx
tAtytAtx
October 12, 2010
Isotropic Oscillator If we redefine the origin of time to coincide with the time that, say,
the x position is at its maximum, this becomes
where is the relative phase = ( y x). Consider a ball bearing in a bowl. It may oscillate in only one
direction, i.e. in the x direction or the y direction. This motion would correspond to the above equations when the constant Ay = 0 or Ax = 0, respectively.
The ball could go in a straight line at an angle to the x axis, i.e. in both x and y. That would correspond to Ax = Ay, and = 0.
The ball could go in a circle about the bottom of the bowl, which would correspond to Ax = Ay, and = /2, in one direction, or = /2 in the other direction.
Some other possibilities:
,)cos()(
)cos()(
tAty
tAtx
y
x
x
y
x
y
x
y
October 12, 2010
Anisotropic Oscillator As noted before, in general, the springs in the x and y directions
could have a different spring constants (How could we do this in the bowl and marble case?)
In that case, the oscillation frequencies would be different in the two directions
and the oscillator is called anisotropic (differs depending on direction). We can easily write down the solution as:
You can play with a java applet to see the “orbits” for this case..
)cos()()cos()(
tAty
tAtx
yy
xx
.yx kk
,/
,/
mk
mk
yy
xx
Click here
October 12, 2010
5.4 Damped Oscillations Recall when we were discussing the drag force, that we
characterized it as either being proportional to v, or to v2. A drag force, or other resistive force in an oscillator leads to the oscillations dying out after awhile, a phenomenon we call damped oscillations.
Let’s investigate a damped oscillator whose damping is proportional to v, or
For a damped spring, for example, our equation of motion becomes
Writing it to emphasize that it is homogeneous: or… For later convenience, we will substitute
where is called the damping constant. Large => large damping. As usual, we will also write
.x
.kxxbxm
springforce
resistiveforce
.0 kxxbxm
,2mb
./o mk
0.b kx x xm m
October 12, 2010
Damped Oscillator Equation With these substitutions, our damped oscillator equation of motion
becomes
This is the starting point for our complete discussion, which will be based on the solutions to this equation in various limits. You may already know how to solve such an equation in the general case.
The solution to such a linear equation is to assume a solution of the form
which, when substituted into the equation, givesand after cancelling the common term, we have what is sometimes called the auxiliary equation:
This reduces the solution to that of solving a quadratic in r, which calls for use of the quadratic equation. The two solutions are:
The general solution is found by a linear combination of and , i.e.
,)( rtetx
.02 2o xxx
.02 2o
2 rtrtrt ereer
.02 2o
2 rr
2o
21 r 2
o2
2 rtre 1 tre 2
ttttrtr oo eCeCeeCeCtx
222221
2121)(
October 12, 2010
Undamped and Weakly Damped To understand the physics captured in the general solution
let’s look at some limits. For no damping at all ( = 0), we recover the usual solution for
simple harmonic motion:
Now consider the case of weak damping ( < o). This case is easiest to visualize if we write where
When the damping is small, we can think of 1 as a small correction to the undamped oscillation frequency o. The complete solution is
Graphically, this looks like the plot at right. The oscillation damps with an envelope
given by the leading term et. Thus, here acts as a decay parameter.
,12o
2 i
,)(2222
21
ttt oo eCeCetx
.)( oo22
2121tititt eCeCeCeCtx oo
.22o1
)( .)( o2111 titit eCeCetx
et
note, oscillation frequency is slightly lower
October 12, 2010
Strong Damping The general solution
has a qualitatively different behavior in the limit of strong damping ( > o), sometimes called overdamping. In this case, the radicalis purely real, so we may as well leave the solution in its original form
The lack of a complex exponential is a clue that there is no real oscillation involved. In fact, both terms decrease exponentially and the motion looks like:
Decay parameter (slowest decay term) is
,2o
2
,)(2222
21
ttt oo eCeCetx
.)(2222
212121
tttrtr oo eCeCeCeCtx
t
x(t)
initial conditionsxo = 0, vo 0
t
x(t)
initial conditions
xo 0, vo = 0
long-term behaviordecays as
toe
22
.22o
October 12, 2010
Critical Damping The last limit we want to discuss is critical damping, when = o. In this
case, there is a mathematical issue that arises. Now our two solutions
become one solution, r1 = r2. Mathematically, we have a problem, since with only one solution, we have
only one arbitrary constant, which is not sufficient—it does not give a complete solution.
Fortunately (and in general), when the auxiliary equation
gives a repeated root, we can find another solution (as you can easily check)
The general solution is then a linear combination of our two solutions:
The graph of the solution qualitatively looks like the overdamped case, but the decay parameter ( = o) is larger (i.e. the decay is faster). In fact, in the critical damping case the decay is faster than in any other case. Obviously, if you want to keep something stable against oscillations you want to arrange for it to be critically damped.
2o
21 r 2
o2
2 r
0)(22 2222o
2 rrrrr
.)( rttetx
)( .)( o2121 ttrtrt teCeCteCeCtx
October 12, 2010
A Closer Look at Decay Parameters The decay parameters that govern the drop in amplitude at long
times are:
This dependence can be graphed as below:
damping decay parameter
none = 0 0
under < o
critical = o
over > o .22o
o
decayparameter