LECTURE 12: MECHANICAL VIBRATIONS
MINGFENG ZHAO
September 29, 2014
Recall in the last lecture:
I. Mass-Spring System:
Figure 1. Mass-Spring System
Let x(t) be the displacement of the mass, then
mx00 + cx0 + kx = F (t) .
II. RLC Circuit System:
Figure 2. RLC Circuit System
1
2
MINGFENG ZHAO
Let I(t) be the current in the circuit, then
LI 00 (t) + RI 0 (t) +
1
I(t) = E 0 (t) .
C
Example 1 (Pendulum Problem). Suppose we have a mass m (in kilograms) on a pendulum of length L (in meters).
Figure 3. Pendulum Problem
Let θ(t) be the angle between the vertical line and the pendulum at time t (in seconds), by Newton’s law, we have
mLθ00 = −mg sin(θ),
where Lθ00 denotes the angular acceleration. So we get
θ00 +
g
sin(θ) = 0.
L
When |θ| is very small, we can think sin(θ) ≈ θ, recall lim
x→0
θ00 +
sin(x)
= 1, then
x
g
θ=0.
L
Free undamped motion [Simple harmonic motion]
The free undamped motion (that is, c = 0 and F (t) ≡ 0):
mx00 + kx = 0.
r
Let ω0 =
k
, then our equation becomes:
m
x00 + ω02 x = 0.
LECTURE 12: MECHANICAL VIBRATIONS
The general solution to x00 + ω02 x = 0 is:
x(t) = A cos(ω0 t) + B sin(ω0 t) .
By a trigonometric identity, we have
x(t) = A cos(ω0 t) + B sin(ω0 t) = C cos(ω0 t − γ) ,
where
C=
p
A2 + B 2 ,
and γ = arctan
B
.
A
Figure 4. Free undamped motion
Here are some terminology:
√
• C = rA2 + B 2 is called the amplitude.
k
• ω0 =
is called the frequency.
m
B
• γ = arctan
is called the phase shift.
A
2π
• T =
is called the period of the motion.
ω0
Free damped motion
Let’s focus on damped motion:
mx00 + cx0 + kx = 0.
3
4
MINGFENG ZHAO
Rewrite the equation as:
x00 + 2px0 + ω02 x = 0,
where
c
p=
(damping term),
2m
r
and
ω0 =
k
(frequency).
m
The characteristic equation of x00 + 2px0 + ω02 x = 0 is:
r2 + 2pr + ω02 = 0.
Solve r2 + 2pr + ω02 = 0, we get
r1,2 = −p ±
q
p2 − ω02 .
Hence the solutions to x00 + 2px0 + ω02 x = 0 depend on p2 − ω02 =
c2 − 4km
:
4m2
I. Overdamping: c2 − 4km > 0, that is, p2 − ω02 > 0. Then we have two different real roots:
r1 = −p −
q
p2 − ω02 < 0,
and r2 = −p +
q
p2 − ω02 < 0.
Then the solution is:
y(t) = Aer1 t + Ber2 t .
Since r1 < 0 and r2 < 0, then
y(t) → 0,
as t → ∞.
Figure 5. Overdamped motion
LECTURE 12: MECHANICAL VIBRATIONS
5
Example 2. Let the mass be on a nonsmooth ground with friction constant c > 0, suppose the mass is release
from the displacement x0 which is away from the rest position, then
mx00 + cx0 + kx = 0,
x(0) = x0 ,
and x0 (0) = 0.
II. Critical damping: c2 − 4km = 0, that is, p2 − ω02 = 0. Then we have the same real roots:
r1 = r2 = −p < 0.
Then the solution is:
y(t) = Ae−pt + Bte−pt .
Since r1,2 = −p < 0, then
y(t) → 0,
as t → ∞.
Figure 6. Critical damped motion
III. Underdamping: c2 − 4km < 0, that is, p2 − ω02 < 0. Then we have two different real roots:
q
r1 = −p − i ω02 − p2 ,
q
and r2 = −p + i ω02 − p2 .
Then the solution is:
y(t) = Ae
−pt
cos
q
ω02
−
p2 t
+ Be
−pt
sin
q
ω02
−
p2 t
= Ce−pt cos (ω1 t − γ) ,
where
w1 =
q
ω02 − p2 < ω0 ,
and γ = arctan
B
A
6
MINGFENG ZHAO
Since p > 0, then
y(t) → 0,
as t → ∞.
Figure 7. Underdamped motion
In Figure 7, there are two envelop curves y1 (t) = Cy −pt and y2 (t) = −Ce−pt , the solution y(t) is oscillating
between these two envelop curves.
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca