=
12
m(
km
)A2 sin2(t + )
=
12
kA2 sin2(t + ) (854)
The total energy is thus:
E=
12
kA2 sin2(t + ) + 1 2
kA2 cos2(t + )
=
12
kA2 (855)
and is constant in time! Kinda spooky how that works out.
Note that the energy oscillates between being all potential at
also have: d2xI dt2 + 20 dxI dt + 2 0xI = F0 m cos(t) (900) (the
inhomogeneous ODE for a damped harmonically driven SHO) with
solution xI(t) yet to be determined. Suppose also that we have found at
least one solution to the inhomogenous ODE xp(t),
189Wiki
(or general complex) position!. So we take the real part of either of
these solutions: x+(t) = |A+|e+i(t+) = |A+|(cos(t + +) +
isin(t + +) = |A+|cos(t + +) (814) and x(t) = |A|
ei(t+) = |A|(cos(t + )isin(t + ) = |A|cos(t + )
(815)
These two solutions are
x+(t) = A+e+it (805) x(t) = Aeit (806) that follow, one for each
possible value of . Note that we will always get n independent
solutions for an nth order linear ODE, because we will always have
solve for the roots of an nth order characteristic equation,
of sharp bumps could leave your shocks still compressed and unable to
absorb the impact of the last one and keep your tires still on the
ground.
Damped oscillation is ubiquitous. Pendulums, once started, oscillate
for only a while before coming to rest. G
experimental results in an understandable if approximate way, but
hardly exact.
The exact (relevant) laws of nature are those of electromagnetism and
quantum mechanics, where the many electron problem must be solved,
which is very dicult. The problem rapi
and the moment of inertia (found using the parallel axis theorem) to
obtain the frequency of the oscillator183. 183I know, I know, you had
hoped that you could nally forget all of that stressful stu we
learned back in the weeks we covered torque. Sorry. N
In this course we will study simple harmonic oscillators, both with and
without damping (and harmonic driving) forces. The principle
examples we will study will be masses on springs and various penduli.
Week 9: Oscillations 397
m
k
x
equilibrium Fx = kx
F
This equation is maximum when = 0, at resonance. At that point
the peak average power can be written: Pavg(0) = Pmax = F2 0
4m0 = F2 0 2b (912)
d) This shape of Pavg() is very important to semi-quantitatively
understand. It is (for weak damping) a sharply
cosine or sine solution instead of an exponential solution at the
beginning of the previous section.
Note Well! An unfortunately commonly made mistake for SHO
problems is for students to take Fx = ma = kx, write it as: a = k m x
(820) and then try to subs
= x(0) = 0.1 m and v0 = 1 m/sec. Then A = rx2 0 + v2 0 2 =
p(0.01 + 0.01) = 0.12 m. Also = tan11 1= tan1(1) = 4
(calculator result) or 5 4 or 3 4 . In this case either: x(t) =
+0.12cos(10t + 4 ) (830) or x(t) = 0.1 2cos(10t + 5 4 ) (831) or
even x(t) = 0.
xed standard pace, would consistently cover twenty miles in ve
summer hours, or by increasing the length of their pace slightly,
twenty four miles in the same number of hours. Note well that the mile
was originally a decimal quantity a multiple of ten uni
where:
A=
F0/m p(2 2 0)2 + (20)2
(904)
and the phase angle is:
= tan1 20 2 2 0 (905) b) The average power delivered to
the mass by the driving force is a quantity of great interest. This is the
rate that work is being done on the mass by the driving forc
410 Week 9: Oscillations
m
l
Figure 123: A simple pendulum is just a point like mass suspended on
a long string and displaced sideways by a small angle. We will assume
no damping forces and that there is no initial velocity into or out of
the page, so tha
point is that one can always make cos() and sin() change sign by
adding or subtracting , so that = + (for example), and as long
as you change the sign of A as well, you will get exactly the same
physical solution!
That is:
x(t) = Acos(t + ) = Acos(t + ()
Since we are going to quite a bit with harmonic oscillators from now
on, we should take a few moments to plot x(t), v(t), and a(t). We
remarked above that omega had to have units of t1. The following
are some True Facts involving that You Should Know:
=
2
-0.5
0
0.5
1
1.5
t
x
Figure 126: Two identical oscillators, one undamped (with 0 = 2, or
if you prefer with an undamped period T0 = 1) and one weakly
damped (b/m = 0.3).
Several of these properties are illustrated in gure 126. In this gure
the exponential
time almost precisely matches the human heartbeat. Unfortunately,
the normal human heartbeat varies by a factor of around three as
one moves from resting to heavy exercise, a range that is further
increased by the abnormal heartbeat of people with cardiac
You should now be able to see that in fact, the total energy of the
oscillator is constant in the small angle approximation.
Of course, it is actually constant even for large oscillations, but
proving this requires solving the exact ODE with the sin() in
build up much amplitude. If you apply the torques with the wrong
phase you will not manage to get the same amplitude that youd get
pumping in phase with the motion.
Thats really it, qualitatively. Resonance consists of driving an
oscillator at its natural
constants of integration, just as one can for ordinary 1D kinematic
solutions).
Lets see how to accomplish this algebraically. Suppose we are given
x(0) = x0 and v(0) = v0 the initial position and velocity of the mass
on a spring (or other oscillator), an
In principle, if you have satised the prerequisites for this course you
have almost certainly studied imaginary numbers and complex
numbers176 in a high school algebra class and perhaps again in
college level calculus. Unfortunately, because high school m
times that you can immediately recognize them when they are given in
a problem and just write down the answer. They are the four
possibilities:
Week 9: Oscillations 407
Example 9.1.5: Given: x0 > 0, v0 = 0
The best way to write this solution is (in my opi
9.5.1: Simple Models for Molecular Bonds
Consider, then a very crude microscopic model for a solid. We know
that this solid is made up of many elementary particles, and that those
elementary particles interact to form nucleons, which bind together to
form
U(r)
Figure 130: Two generic classical potential energy functions
associated with atomic bonds on a common scale Umin = 1, rb = 1.0,
and a = 6.2, the latter a value that makes the two potential have
roughly the same force constant for small displacements.
As before, we exclude the trivial solution x(t) = 0 as being too boring
to solve for (requiring that A 6= 0, that is) and are left with the
characteristic equation for : 2 + b m + k m = 0 (885) This
quadratic equation must be satised in order for our gues
solution, not necessarily initial conditions. For example, you might be
given x(t1) and x(t2) for two specied times t1 and t2 and be required
to nd the particular solution that satised these as a constraint.
However, this problem is much more dicult and c
with
2 =mg 2 + Mg I
(869)
I left the result in terms of I because it is simpler that way, but of
course we have to evaluate I in order to evaluate 2. Using the
parallel axis theorem (and/or the moment of inertia of a rod about a
pivot through one end) we
v(t) =
2sin(10t 4
) (837)
or
x(t) = 0.1 2cos(10t + 3 4
) (838)
paired with
v(t) = 2sin(10t + 3 4
) (839)
or there is a solution that uses = 5/4 as well.
Example 9.1.3: = 10 rad/sec, x0 = 0.1 m, v0 = 1 m/sec All weve
done is ip the signs so that both x0 a
Without recapitulating the entire argument, it should be fairly obvious
that can take the real part of their sum, get formally identical terms,
and combine them to get the general real solution:
x(t) = Ae
bt 2m cos(t + ) (890)
where A is the real initial