MATH 2930 - Spring 2010
Worksheet 7
MATH 2930 - Differential Equations - Spring 2010
Worksheet 7 - March 11 & 12, 2010
Find the general solution to the following 2nd-order ODEs:
1. y 00 + 9y = cos(3t)
2. y 00 + y 0 + y = sin2 t
3. y 00 − 4y = cosh(t)
1
4. y 00 + 3y 0 + 2y = tet
Forced Oscillations & Resonance:
Newton’s II Law for a damped oscillator (with mass m, damping coefficient c, and spring
constant k) driven by a sinusoidal external force Fd (t) = F0 cos(ωt) is
my 00 + cy 0 + ky = F0 cos(ωt)
(1)
where y(t) is the displacement of the mass from its equilibrium position. Assume the oscillator is underdamped (c2 − 4mk < 0).
[Trick!] Dimensional reduction: Since m 6= 0, we may divide Eq. (1) by m and obtain
the ODE
F0
y 00 + γy 0 + ω02 y =
cos(ωt)
(2)
m
p
where ω0 ≡ k/m is the oscillator’s natural (angular) frequency (a simple fact you should
know by now), and γ ≡ c/m is its energy decay rate.2 In effect, the behavior of a driven
oscillator depends only on ω0 , γ, F0 /m (strength of the drive), and ω (drive frequency).
5. Verify that the solution to the homogeneous problem
yc00 + γyc0 + ω02 yc = 0
i.e., the complementary solution to Eq. (2), is
yc (t) = e−γt/2 [C1 cos(ω1 t) + C2 sin(ω1 t)]
where ω1 ≡
(3)
p
ω02 − (γ/2)2 .
ex + e−x
.
2
2
Recall that in the undriven case (F0 = 0), the oscillator’s amplitude y(t) decays exponentially as e−γt/2 .
You probably know from frosh physics that the oscillator’s mechanical energy E is proportional to the
amplitude squared, so E ∼ (e−γt/2 )2 = e−γt , whence the nomenclature.
1
Reminder: cosh x =
1
MATH 2930 - Spring 2010
Worksheet 7
6. Show that a particular solution to Eq. (2) is
F0
(ω02 − ω 2 )
γω
cos(ωt) + 2
sin(ωt)
yp (t) =
m (ω 2 − ω02 )2 + (γω)2
(ω − ω02 )2 + (γω)2
(4)
Rewrite yp (t) in the form R cos(ωt − φ), where R is the amplitude and φ is the phase
[Recall Wed 3/11 lecture]. Sketch the amplitude R vs. the drive frequency ω. What
is the (practical) resonant frequency? And (somewhat more challenging) how does the
phase φ vary with ω?
7. The complete solution to Eq. (2) is
y(t) = yc (t) + yp (t)
where yc and yp are given by Eqs. (3) and (4). Show that for sufficiently long times (t γ −1 ), yc (t) becomes negligible compared to yp (t), so the solution y(t) is ”essentially”
given by yp (t). This has a physical interpretation: physicists and engineers call yc (t)
, and yp (t) the
.
the
8. Applet time! Go to the following URLs (linkable from pdf document) and play
around with the applets.
• Simulate driven harmonic oscillation:
http://qbx6.ltu.edu/s_schneider/physlets/main/osc_damped_driven.shtml
• Plot amplitude of yp vs. drive frequency ω:
http://qbx6.ltu.edu/s_schneider/physlets/main/osc_damped_driven_amp.
shtml
Some suggestions for experimenting with the applets:
• Choose any m and k, and c sufficiently small. Can you estimate the practical
resonant frequency right away? Put your guess as the drive frequency ω, simulate,
and see if you get (near-)resonance.
• Next, change the drive frequency below (and, later, above) resonance. Does the
behavior match your expectation? Use the 2nd applet to plot amplitude vs. drive
frequency ω to get an overall picture.
• Now tune ω back to practical resonance. Change the damping coefficient c. How
does the (steady-state) amplitude vary with the amount of damping?
2
MATH 2930 - Spring 2010
Worksheet 7
Extra Credit3
9. Show that if the oscillator has zero damping (γ = 0) and is driven by a cosine force
F0 cos(ωt), then yp (t) in Eq. (4) has only the cosine term and no sine term. In other
words, the oscillator vibrates in phase with the driving force.4
10. Using dimensional analysis, show that the energy decay rate γ(≡ c/m) has the same
units as frequency (i.e., inverse time). [Hint: Do dimensional analysis on Eq. (2).]
11. An important concept in physics and engineering is the use of dimensionless variables
or parameters. They often tell us the intrinsic property of a system under study, or
give a criterion for the regime of validity in the system’s dynamics, etc. In damped
oscillators, the key dimensionless parameter is called the quality factor
ω0
Q≡
γ
An oscillator with higher Q will respond with a larger amplitude when on resonance.
To be more precise, use Eq. (4) to show that for long enough times, the amplitude R
reads
F0
1
s
R=
2
2
mω0
2
2
ω
ω
−2
−1 +Q
ω0
ω0
so
R=
Q
F
F
p
Q
≈
2
mω0 1 − (Q−2 /2)
mω02
on (practical) resonance
Thus for most purposes (Q ≥ 1), the resonant amplitude of a driven oscillator scales
linearly with Q.
12. (ECE people beware!) An L(inductor)-R(resistor)-C(capacitor) circuit in series powered by a sinusoidal voltage V0 cos(ωt) can be described by the following ODE:
1
LQ00 (t) + RQ0 (t) + Q(t) = V0 cos(ωt)
C
where Q(t) is the charge on the capacitor.5 This has an exact correspondence with
the mass-spring-dashpot problem. Exploit the correspondence to find the following
parameters associated with the circuit in terms of L, R, C, as needed:
•
•
•
•
the
the
the
the
natural frequency ω0
damping coefficient γ
quality factor Q
practical resonant frequency
3
Note: Some of these questions are drawn from Physics 2214 (Waves, Oscillations & Quantum Physics).
If you (have to) take this course in the future, these give a glimpse of the problems you’ll see during the first
2 weeks.
4
In a sense, the 90◦ -out-of-phase term (i.e., the sine term) appearing in the oscillator’s trajectory can
be attributed to the damping γ. In engineering parlance, the cosine term in Eq. (4) is called the reactive
response, while the sine term is called the dissipative or resistive response.
5
Note that the ODE is written down from Kirchhoff’s Law, and I(t) = Q0 (t) is the current of the circuit.
3