LECTURE 12: MECHANICAL VIBRATIONS
MINGFENG ZHAO
October 05, 2015
RLC Circuit System
There is a resistor with a resistance of R ohms, an inductor with an inductance of L henries, and a capacitor with
capacitance of C farads. There is also an electric source (such as a battery) giving a voltage of E(t) volts at time t (in
seconds).
Figure 1. RLC Circuit System
Kirchhoff’s Voltage Law:
The sum of the electrical potential differences (voltage) around any closed network is zero.
Let I(t) be the current in the circuit, by Kirchhoff’s Voltage Law, we have
Vresistor (t) + Vinductor (t) + Vcapacitor (t) = Vbattery (t).
Then we know that
• For Vresistor , by Ohm’s Law:
Vresistor (t) = RI(t).
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MINGFENG ZHAO
• For Vinductor , by Faraday’s Law of Induction:
Vinductor (t) = LI 0 (t).
• For Vcapacitor , let Q(t) be the charge (in coulombs) on the capacitor, by Capacitor equation:
Vcapacitor (t) =
Q(t)
.
C
So we get
LI 0 (t) + RI(t) +
Q(t)
= E(t).
C
On the other hand, we know that Q0 (t) = I(t). So we have the following differential equation:
LI 00 (t) + RI 0 (t) +
1
I(t) = E 0 (t) .
C
Pendulum Problem
Suppose we have a mass m (in kilograms) on a pendulum of length L (in meters).
Figure 2. 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
LECTURE 12: MECHANICAL VIBRATIONS
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]
Recall the Mass-Spring System:
Figure 3. Mass-Spring System
Let x(t) be the displacement of the mass, then
mx00 + cx0 + kx = F (t) .
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.
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) ,
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MINGFENG ZHAO
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
Now let’s focus on damped motion:
mx00 + cx0 + kx = 0.
Rewrite the equation as:
x00 + 2px0 + ω02 x = 0,
where
r
c
(damping term),
p=
2m
and
ω0 =
Let x = ert be a solution to x00 + 2px0 + ω02 x = 0, then
x0 = rert ,
and x00 = r2 ert .
k
(frequency).
m
LECTURE 12: MECHANICAL VIBRATIONS
So we have
0
= x00 + 2px0 + ω02 x
= r2 ert + 2prert + ω02 ert
= ert [r2 + 2pr + ω02 ].
So r is a zero of the characteristic equation of x00 + 2px0 + ω02 x = 0, that 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
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MINGFENG ZHAO
Example 1. 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 complex roots:
q
r1 = −p − i ω02 − p2 ,
q
and r2 = −p + i ω02 − p2 .
Then the solution is:
y(t) = Ae−pt cos
q
q
ω02 − p2 t + Be−pt sin
ω02 − p2 t = Ce−pt cos (ω1 t − γ) ,
where
w1 =
q
ω02 − p2 < ω0 ,
and γ = arctan
B
A
LECTURE 12: MECHANICAL VIBRATIONS
7
Since p > 0, then
y(t) → 0,
as t → ∞.
Figure 7. Underdamped motion
In Figure 7, there are two envelop curves y1 (t) = Ce−pt and y2 (t) = −Ce−pt , the solution y(t) is oscillating
between these two envelop curves.
Problems you can do:
Lebl’s Book [2]: All exercises on Page 68 and Page 69.
Braun’s Book [1]: Read all materials in Section 2.6 and Section 2.7.
References
[1] Martin Braun. Differential Equations and Their Applications: An Introduction to Applied Mathematics. Springer, 1992.
[2] Jiri Lebl. Notes on Diffy Qs: Differential Equations for Engineers. Createspace, 2014.
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