Math 217: Differential Equations
Lecture Notes: Chapter 2 Apps
Mark Pedigo
1
Applications of Linear Second-Order DEs
Linear DEs often appear in mathematical models of mechanical systems and electric
circuits. We consider applications of the linear second-order DE
ax00 + bx0 + cx = F (x).
1.1
Vibrations of a mass
Free vibration of a mass
Example 1.1 (Free vibration of a mass). The system:
Force Diagram:
1
(1)
Fs : Force of spring on mass
• Assumed proportional to displacement x in the opposite direction (Hooke’s Law);
i.e., FS = −kx (“spring constant” k > 0)
• FS < 0 if x > 0 (spring stretched)
• FS > 0 if x < 0 (spring compressed)
FR : Force of dashpot
of the mass in opposite direction;
• Assumed proportional to velocity v = dx
dt
dx
i.e., FR = −cv = −c dt (“damping constant” c > 0)
• FR < 0 if v > 0 (motion of m to the right)
• FR > 0 if v < 0 (motion of m to the left)
If FR and FS are the only forces acting on m (that’s why it’s “free” motion) and its
, then
resulting acceleration is a = dv
dt
F = ma ⇒ mx00 = FS + FR
d2 x
dx
⇒ m 2 = −kx − c
dt
dt
dx
d2 x
⇒ m 2 + c + kx = 0,
dt
dt
which is a linear second-order homogeneous DE.
Forced vibration of a mass
Example 1.2 (Forced vibration of a mass). Similar setup to the last example, but now, in
addition to FS and FR , the mass is subject to an external force FE = F (t). Now,
F = ma ⇒ mx00 = FS + FR + FE
d2 x
dx
⇒ m 2 = −kx − c + F (t)
dt
dt
2
dx
dx
⇒ m 2 + c + kx = F (t),
dt
dt
which is a linear second-order nonhomogeneous DE.
1.2
Damped and Undamped Motion
If c = 0 (no dashpot), the motion is undamped. If c > 0 (there is a dashpot), the motion is
damped.
2
Free Undamped Motion
Suppose we have only a mass on a spring, with neither damping not external forces. Then
c = 0 and F (t) = 0 so
mx00 + cx0 + kx = F (t) ⇒ mx00 + kx = 0
k
⇒ x00 + x = 0.
m
Define ω0 =
q
k
m
(2)
and rewrite the (2) as
x00 + ω 2 x = 0.
To solve this equation, we note that characteristic equation is r2 + ω 2 = 0, with roots
r = −ω0 i. Thus, the position function is given by
x(t) = A cos(ω0 t) + B sin(ω0 t) .
Let C =
√
A2 + B 2 and let

−1

tan
α = π + tan−1


2π + tan−1
(3)
if A, B > 0 (Quadrant I)
if A < 0 (Quadrants II,III)
if A > 0, B < 0 (Quadrant IV)
Then
x(t) = A cos(ω0 t) + B sin(ω0 t)
A
B
=C
cos(ω0 t) + sin(ω0 t)
C
C
= C(cos α cos(ω0 t) + sin α sin(ω0 t)
= C(cos(ω0 t − α)
= C(cos(ω0 (t − δ))).
C: amplitude
ω0 : circular frequency
α: phase angle
δ := ωα0 : time lag
Period of the motion: time required for the system to complete one full oscillation: T = ω2π0
(sec) Frequency: number of complete cycles/second: f = T1 = ω2π0 (Hz)
This type of motion is known as simple harmonic motion. The mass oscillates to and fro
about its equilibrium position.
Example 1.3. See example 1, page 139.
3
Free damped Motion
Damped (so c > 0), but free (so no external force, i.e., F (t) = 0).
k
c 0
x + x=0
m
m
00
0
⇒ x + 2px + ω02 x = 0,
mx00 + cx0 + kx = 0 ⇒ x00 +
where ω0 :=
q
k
,
m
p :=
c
2m
> 0.
p
From the characteristic equation r2 + 2pr + ω02 , we see that r = −p ± p2 − ω02 . Thus, the
type and number of roots depends on whether p2 − ω02 is = 0 (one real root, with
multiplicity 2: “critically damped”), > 0 (2 real solutions: “overdamped”), or < 0 (2
complex solutions: “underdamped”).
1. Critically damped (p2 − ω02 = 0)
• Converges to 0 as fast as possible without oscillating
• Examples: Door closer seen on hinged doors in public buildings. Door closes
normally; the recoil mechanisms in most guns are critically damped so that they
return to their original positions, after the recoil, in the least possible time
• Critical roots: 1 real repeated root. Since the characteristic equation has a
double root −p the position function is
x(t) = (c1 + c2 t)e−pt
• See graph page 142.
2. Overdamped (p2 − ω02 > 0)
• Longer time to converge to 0 than the critically damped case
• Resistance of the dashpot damps out any oscillations
• Critical roots: 2 real, distinct, both negative. Since the characteristic equation
has two real roots r1 and r2 , the position function is
x(t) = c1 er1 t + er2 t
• See graph page 141.
3. Underdamped (p2 − ω02 < 0)
• The system will oscillate at the natural damped frequency
• Example: and underdamped swinging door would oscillate
4
• Criticalproots: 2 complex. The
p characteristic equation has two complex roots
2
2
−p ± i ω0 − p . Let ω1 = ω02 − p2 . Then
x(t) = e−pt (A cos(ω1 t) + B sin(ω1 t))
A
B
= Ce−pt ( cos(ω1 t) + sin(ω1 t))
C
C
= Ce−pt (cos α cos(ω1 t) + sin α sin(ω1 t))
= Ce−pt cos(ω1 t − α).
ω1 : “pseudofrequency”, or “circular frequence”
T1 := ω2π1 : “pseudoperiod” of oscillation
Ce−pt : time-varying amplitude.
• See graph page 142.
The action of the dashpot has two effects;
(a) It exponentially damps the oscillations according to the time-varying amplitude
(b) It slows the frequency of the motion.
Example 1.4. See Example 2, page 143
Recommended Homework
• §2.4: 1-4, 15-21.
1.3
The Simple Pendulum
A simple pendulum is an idealization of a real pendulum using the following assumptions:
• The rod/cord on which the bob swings is considered massless, inextensible, and
always remains taught.
• Motion occurs in two dimensions, so the bob always traces an arc.
• The motion does not lose energy to friction or air resistance.
Mathieu’s Equation: The DE which represents the motion of a simple pendulum is
d2 θ
g
+ sin θ = 0
2
dt
L
where g is the acceleration due to gravity, L is the length of the pendulum, and theta is the
angular displacement. (See Figure 2.4.3, page 137.)
5
Derivation. We are interested in the tangential acceleration (i.e., the component of the
acceleration tangent to the curve). Because we are only concerned with changes in speed,
and because the bob is forced to stay in a circular path, we apply Newton’s law to the
tangential axis only. So, since the force straight down is the bob’s weight mg, the
tangential acceleration is mg sin θ. By Newton’s law, F = ma = −mg sin θ (the negative
sign implies that a and θ point in opposite directions), so
a + g sin θ = 0.
(4)
= L dθ
and
Now, we look at a. Let s be the arc length. Then s = Lθ, so v = ds
dt
dt
2
2
2
d s
d θ
d θ
a = dt2 = L dt2 . Plugging a into (4) gives L dt2 + g sin θ = 0. Dividing out by (nonzero) L,
2
we get ddt2θ + Lg sin θ = 0.
When θ is “small” (less than 15 degrees), sin θ ≈ θ, so θ00 + kθ = 0, where k = Lg . Insert a
term cθ0 to account for frictional resistance of the surrounding medium (air):
θ00 + cθ0 + kθ = 0.
Recommended Homework
• §2.4: 5-8
6