Math 2250-4
Wed Nov 6
5.6 continued: Forced mechanical (and electrical) oscillations.
Yesterday we started to discuss the physical phenomena which arise in undamped forced oscillation
problems, and the mathematics that explains these phenomena:
m x##C k x = F0 cos w t
F0
k
x##C
x=
cos w t
m
m
F0
2
x##C w0 x =
cos w t
m
We used section 5.5 undetermined coefficients algorithms. The solutions are:
,
w s w0 undetermined coefficients implies
x t = xP C xH = A cos w t C C0 cos w0 t K a
which may or may not be periodic, depending on whether the two sinusoidal functions have a common
period.
After calculation, one verifies that for initial conditions
x 0 = x , x# 0 = v
0
the solution is
0
F
xt =
v
0
m w
2
cos w t K cos w0t
2
K w0
C x cos w0t C
0
0
w0
sin w0t .
(That's as far as we got on Tuesday's notes, which we will return to, to finish the rest of the undampedforced discussion summarized below.)
If w z w0 but w s w0 beating will occur as the difference of cosines above is sometimes in-phase and
sometimes out of phase. We can study this more carefully by using cosine angle addition formulas to
rewrite the IVP solution above as
F
xt =
,
0
2
2
m w K w0
1
2 w C w0 t sin
2 sin
v
1
0
w
K
w
t
C
x
cos
w
t
C
sin w0t
0
0
2
0
w0
w = w0 Resonance: undetermined coefficients implies there is a particular solution
x = t A cos w0 t C B sin w0 t
P
After calculation, one verifies that the IVP solution is
F
xt =
0
2 m w0
.
v
t sin w0 t C x cos w0t C
0
0
w0
sin w0t .
.
, After finishing the discussion of undamped forced oscillations, we will discuss the physics and
mathematics of damped forced oscillations
m x##C c x#C k x = F0 cos w t .
Here are some links which address how these phenomena arise, also in more complicated real-world
applications in which the dynamical systems are more complex and have more components. Our baseline
cases are the starting points for understanding these more complicated systems. We'll also address some of
these more complicated applications when we move on to systems of differential equations, in a few
weeks.
http://en.wikipedia.org/wiki/Mechanical_resonance (wikipedia page with links)
http://www.nset.org.np/nset/php/pubaware_shaketable.php (shake tables for earthquake modeling)
http://www.youtube.com/watch?v=M_x2jOKAhZM (an engineering class demo shake table)
http://www.youtube.com/watch?v=j-zczJXSxnw (Tacoma narrows bridge)
http://en.wikipedia.org/wiki/Electrical_resonance (wikipedia page with links)
http://en.wikipedia.org/wiki/Crystal_oscillator (crystal oscillators)
Damped forced oscillations c O 0 for x t :
m x##C c x#C k x = F0 cos w t
Undetermined coefficients for xP t :
k xP = A cos w t C B sin w t
C c xP #=KA w sin w t C B w cos w t
2
2
C m xP ##=KA w cos w t K B w sin w t
.
_________________________________________
2
L xP = cos w t
k ACc B wKm A w
2
C sin w t
k BKc A wKm B w
.
Collecting and equating coefficients yields the matrix system
kKm w
2
Kc w
cw
A
kKm w
B
2
F0
=
,
0
which has solution
A
B
=
kKm w
1
kKm w
2 2
C c2w
2
2
cw
Kc w
F0
kKm w
2
F0
=
0
kKm w
kKm w
2 2
C c2w
2
2
cw
In amplitude-phase form this reads
xP = A cos w t C B sin w t = C cos w t K a
with
F0
C=
. (Check!)
kKm w
cos a =
2 2
C c2w
2
kKm w
kKm w
sin a =
2
2 2
C c2w
2
cw
kKm w
2 2
.
C c2w
2
And the general solution x t = xP t C xH t is given by
x = xsp C xtr = C cos w t K a C eKp t C1 cos w1 t K a 1 .
,
underdamped:
,
critically-damped: x = xsp C xtr = C cos w t K a C eKp t c1 t C c2 .
, over-damped:
Important to note:
Kr t
x = xsp C xtr = C cos w t K a C c1 e
F0
1
Kr t
C c2 e
2
.
if w z w0 and c z 0 , because the
m
denominator will then be close to zero. This phenomenon is practical resonance.
, The phase angle a is always in the first or second quadrant.
,
The amplitude C in xsp can be quite large relative to
Exercise 0) ( a cool M.I.T. video.) Here is practical resonance in a mechanical mass-spring demo. Notice
that our math on the previous page exactly predicts when the steady periodic solution is in-phase and when
it is out of phase with the driving force, for small damping coefficient c! Namely, for c small, when
2
2
2
2
w !! w0 we have a near zero (in phase) for xsp ,because sin a z 0, cos a z 1; when w OO w0
we have a near p (out of phase), because sin a z 0, cos a zK1; for w z w0 , a is near
p
,
2
because sin a z 1, cos a z 0 .
http://www.youtube.com/watch?v=aZNnwQ8HJHU
Exercise 1) Solve the IVP for x t :
x##C 2 x#C 26 x = 82 cos 4 t
x 0 =6
x# 0 = 0 .
Solution:
x t =
41 cos 4 t K a C 10 eKt cos 5 t K b
a = arctan 0.8 , b = arctan K3 .
> with DEtools :
> dsolve x## t C 2$x# t C 26$x t = 82$cos 4$t , x 0 = 6, x# 0 = 0 ;
x t = K3 eKt sin 5 t C eKt cos 5 t C 5 cos 4 t C 4 sin 4 t
(1)
Practical resonance: The steady periodic amplitude C for damped forced oscillations (page 3) is
F0
C w =
.
kKm w
Notice that as w/0, C w /
F0
k
2 2
C c2w
2
and that as w/N, C w /0. The precise definition of practical
resonance occuring is that C w have a global maximum greater than
F0
k
, on the interval 0 ! w !N .
2
(Because the expression inside the square-root, in the denominator of C w is quadratic in w it will have
2
at most one minimum in the variable w , so C w will have at most one maximum for non-negative w. It
will either be at w = 0 or for w O 0, and the latter case is practical resonance.)
Exercise 2a) Compute C w for the damped forced oscillator equation related to the previous exercise,
except with varying damping coefficient c:
x##C c x#C 26 x = 82 cos w t .
2b) Investigate practical resonance graphically, for c = 2 and for some other values as well. Then use
Calculus to test verify practical resonance when c = 2.
> with plots :
82
> C d w, c /
2 2
:
2
26 K w
C c2 $w
> plot C w, 2 , w = 0 ..20, color = black, title = `practical resonance when c=2` ;
practical resonance when c=2
8
5
3
1
0
>
5
10
w
15
20
The mechanical-electrical analogy, continued: Practical resonance is usually bad in mechanical systems,
but good in electrical circuits when signal amplification is a goal....recall from earlier in the course:
http://cnx.org/content/m21475/latest/pic012.png
Kirchoff's Law: The sum of the voltage drops around any closed circuit loop equals the applied voltage
V t (volts).
1
For Q t :
L Q ## t C R Q # t C
Q t = V t = E0 sin w t
C
1
For I t :
L I## t C R I# t C
I t = V# t = E0 w cos w t .
C
Transcribe the work on steady periodic solutions from the preceding pages! The general solution for I t
is
I t = Isp t C Itr t .
Isp t = I0 cos w t K a = I0 sin w t K g , g = a K
F0
C w =
kKm w
2 2
C c2w
0 I0 w =
The denominator
1
KL w
Cw
2
2
E0 w
0 I0 w =
1
2
KL w
C
E0
1
KL w
Cw
p
.
2
2
C R2 w
2
.
2
C R2
C R2 of I0 w is called the impedance Z w of the circuit (because
the larger the impedance, the smaller the amplitude of the steady-periodic current that flows through the
circuit). Notice that for fixed resistance, the impedance is minimized and the steady periodic current
1
amplitude is maximized when
= L w , i.e.
Cw
1
C=
if L is fixed and C is adjustable (old radios).
2
Lw
1
L=
if C is fixed and L is adjustable
2
Cw
Both L and C are adjusted in this M.I.T. lab demonstration:
http://www.youtube.com/watch?v=ZYgFuUl9_Vs.
If you're an engineer concerned about resonance or practical resonance, you need to know how to deduce
the natural frequencies of undamped mechanical (or electrical) systems. Usually the best way is to use
conservation of energy in mechanical systems, which is an integrated and more generally applicable
version of Newton's second law for mechanical systems (when energy is actually conserved).
Conservation of potential energy around closed loops for undamped electrical circuits is Kirchoff's law.
Here are two examples, one old and one new:
,
We've carefully discussed the (linearized) pendulum model, which leads to w0 =
g
:
L
Exercise 3) Multicomponent systems are best understood using conservation of energy, when Newton's
law may not apply in any obvious way. For example, consider the following "rolling mass" configuration
(the spring constant of the massless spring is not shown, but as usual we call it k.)
Find the natural angular frequency for the configuration above. Do this by first computing
TE = KE C PE . Then express the fact that TE is constant once the system is set in motion, by computing
its time derivative and setting it equal to the zero function. Use:
(1) The potential energy from stretching/compressing a spring from equilbrium x = 0 is the work done in
moving it from equilbrium to displacement x, i.e.
x
PE = W =
k s ds =
0
1
k x2 .
2
(2) The KEtrans from the linear motion (translation) of the disk is
1
m x# t 2 , since x t is tracking the
2
center of mass' displacement from equilbrium
(3) The KErot from the rotation of the disk is given by
1
2
Iw ,
2
where w = q' t is the angular frequency of the rotation and I is the moment of inertia, which for a uniform
1
disk of mass m and radius a is given by I = ma2 . (Directly computing this rotational KE is a double
2
integral computation, that works out nicely using polar coordinates. You might even have done it in your
multivariable calculus class if and when you discussed moments of intertia....in general, moments of
intertia are used to compute rotational kinetic energy about centers of mass, and moments are used to
compute angular momentum, as well as centers of mass....this is why Calculus classes have units about
these topics.)
2
k
k
(The answer is w0 =
z .82
, which is slower than if the mass wasn't rolling.)
3
m
m
KErot =