Download Section 3.8

Section 3.8, Mechanical and Electrical Vibrations
Mechanical Units in the U.S. Customary and Metric Systems
Unit
Distance
Mass
Time
Force
g (Earth)
U.S. Customary
feet ft
slugs
seconds sec
pounds lb
32 ft/sec2
MKS System
meters m
kilograms kg
seconds sec
Newtons N
9.81 m/sec2
CGS System
centimeters cm
grams g
seconds sec
dynes dyn
981 cm/sec2
Consider the spring of length l with a weight attached of mass m.
When a mass attached to the spring is allowed to hang under the influence of gravity
The spring becomes stretched by a distance L beyond its natural length. The length l + L is called the
equilibrium length of the spring-mass system.
There are two forces acting where the mass is attached to the spring.
The gravitational force, or weight of the mass, acts downward and has magnitude mg, where g is
the acceleration due to gravity.
Example 1 illustrations
The force Fs, the restorative force of
the spring, acts upward.
Example 1. Illustrate the lengths l and L and
the forces mg and Fs.
Hooke’s law (the linear spring assumption): When stretched, a spring exerts a restoring force opposite
to the direction of its elongation with a magnitude directly proportional to the amount of elongation.
Hooke’s law is a reasonable approximation until the spring is stretched to near its elastic limit.
Fs
kL , where k, the spring constant, is always > 0. The minus sign is due to the fact that the spring
force acts in a direction opposite the stretching direction (one direction is positive, the other negative).
Example 2. Find the spring constant (in the ft-lb system) if a 20-lb weight stretches a spring 6 in.
Sec. 3.8, Boyce & DiPrima, p. 1
Let u(t) be the displacement of a mass from its equilibrium position at time t.
According to Newton’s second law (when mass remains constant) mu' ' (t)
represents the sum of forces acting on the mass at time t.
ma
f (t) where f(t)
The Forces Acting on the Mass m
Gravity. The force of gravity, w, is a downward force with magnitude mg, where g is the acceleration
due to gravity. Hence, w = mg.
Spring (restoring) Force. The spring force is proportional to the total elongation
L + u of the spring and always acts to restore the spring to its natural position. Since this force always
acts in the direction opposite of u we have Fs = k(L + u).
Damping Force. This is a frictional or resistive force acting on the mass. For example, this force may
be air resistance or friction due to a shock absorber. In either case we assume (an assumption that may
be open to question, see p. 194 of text) that the damping force is proportional to the magnitude of the
velocity of the mass, but opposite in direction. That is, Fd = u (t), where > 0 is the damping
constant.
External Forces. Any external forces acting on the mass (for example, a magnetic force or the forces
exerted on a car by bumps in the pavement) will be denoted by F(t). Often an external force is periodic.
These forces considered as a single function are often referred to as the forcing function.
Mass-Spring-Damper System Equations
At the equilibrium position the only forces acting on the spring-mass system are w and Fs. Since there is
kL mg 0 . When
no acceleration there, the total force is zero. Thus at equilibrium ma Fs w
you recall that the spring was stretched an amount L under the influence of gravity, it makes sense that w
and Fs should cancel each other out at the equilibrium position.
For a spring not at equilibrium, that is, for a stretched or compressed spring, we can use Newton’s
second law to get:
mu' ' (t )
mg
Fs (t ) Fd (t ) F (t ) or mu' ' (t)
mg k(u(t) L)
u' (t) F(t) .
Rearranging this last equation we get mu' ' mg ku kL u' F(t) .
Recalling that kL mg 0 , we can simplify this equation to obtain mu' ' (t)
the mass-spring-damper system equation of motion
When = 0, the system is said to be undamped; otherwise it is damped.
When F(t) = 0, the motion is said to be free; otherwise the motion is forced.
Sec. 3.8, Boyce & DiPrima, p. 2
u' (t) ku(t) F(t) ,
I.
Undamped Free Vibrations (Simple Harmonic Motion)
In the case where = 0 and F(t) = 0 the system equation above reduces to
k
mu + ku = 0. If we divide by m and let 02
, we obtain u' ' 02 x 0 .
m
2
2
The characteristic equation, r
0 always has complex roots + 0i.
0
The general solution is of the form u(t ) A cos 0 t B sin 0 t .
It is customary to simplify the general solution by employing a trig identity for the cosine of a
difference* as follows:
Let A R cos and B R sin . Then u(t ) A cos 0 t B sin 0 t
= R cos cos 0t R sin sin 0t
u(t) = R cos( 0t
)
B
.
A
(We must be very careful to choose the correct quadrant for
cos and sin .)
A2
Note that R
B 2 and = arctan
according to the signs of
R, the maximum displacement of the mass from the equilibrium, is called the amplitude of the
motion.
, a dimensionless parameter that measures the displacement of the wave from its normal
position corresponding to = 0, is referred to as the phase angle.
) goes through one period when 0
Since cosine is a periodic function with period 2 , cos( 0t
2
< 0t
< 2 . If we solve for t, we obtain
which tells us the period is T =
t
0
2
=2
0
m
k
0
0
1/ 2
.
k
, measured in radians per unit time, is called the natural
m
frequency of the spring-mass system.
The circular frequency
*
0
cos(a b) cos a cosb sinasinb
Sec. 3.8, Boyce & DiPrima, p. 3
Example 3. A mass weighing 4 lb stretches a spring 3 in. after coming to rest at equilibrium.
Neglecting any damping or external forces that may be present, determine the general solution of the
equation of motion.
Now find specific equations for the motion of the mass along with its amplitude, period, and natural
frequency under the following initial conditions:
a) The mass is pulled 6 in. down below the equilibrium point and given a downward velocity of 2
ft/sec.
Sec. 3.8, Boyce & DiPrima, p. 4
b) The mass is lifted 6 in. above the equilibrium point (compressing the spring) and given a downward
velocity of 2 ft/sec.
c) The mass is lifted 6 in. above the equilibrium point (compressing the spring) and given an upward
velocity of 2 ft/sec.
Sec. 3.8, Boyce & DiPrima, p. 5
Maple sketches of Example 3 solutions
The phase shift,
, measures the horizontal distance between the two graphs.
0
Sec. 3.8, Boyce & DiPrima, p. 6
Damped Free Vibrations
Here we interested in the various effects of damping, while assuming that f(t) = 0. The motion of the
system is governed by
mu + u + ku = 0
1. Underdamped or Oscillatory Motion.
This is the case where the characteristic equation mr 2 r k 0 has a complex pair of roots,
i . (note that the real part of the root is always negative). The general solution is
u(t ) c1e t cos t c2e t sin t .
+
As with harmonic motion, the trig identity for the cosine of a difference can be employed to express
u(t) in an alternate form:
B
u(t ) Re t cos( t
) , where, again, R
.
A2 B 2 and = arctan
A
Since the cosine takes on values between 1 and 1, we can write
Re t Re t cos( t
) Re t or Re t u(t ) Re t
1 cos( t
) 1 or
In other words, the solution u(t) is ‘sandwiched’ between two exponential envelopes which both
approach zero.
Re t is called the damping factor of the solution. The system is called underdamped because the
damping factor is too small to prevent the system from oscillating.
2. Critically Damped Motion
This is the case where the characteristic equation mr 2 r k 0 has one repeated root (of
multiplicity 2). In this case the general solution is u (t ) c1e rt c2te rt . All solutions of critically
damped motion tend to zero as t increases and the graph of the solution will cross the t-axis exactly
once (Where?).
This special case is called critically damped motion because if (in mr 2
smaller the roots would be complex and oscillation would occur.
r
k
0 ) were any
3. Overdamped Motion
This is the case where the characteristic equation mr 2 r k 0 has two distinct real (negative)
roots, r1 and r2. The roots must be negative because is always positive. In this case the general
solution is u(t ) c1e r1t c2e r2 t . Using calculus we can show that u(t) has at most one local max or
min. Thus, the motion is nonoscillatory.
Sec. 3.8, Boyce & DiPrima, p. 7
Example 4. Assume that the motion of a spring-mass system with damping is governed by
u + u + 25u = 0; u(0) = 1, u (0) = 0.
Find the equation of motion and sketch its graph for the three cases where = 8 (Case 1), 10 (Case 2),
and 12 (Case 3).
Sec. 3.8, Boyce & DiPrima, p. 8
Maple sketches of Example 4 solutions
Case 1 Solution
Case 2 solution (solid line) compared with case 3 solution (point line).
Sec. 3.8, Boyce & DiPrima, p. 9
Circuits
The quantities used to describe the state of a simple series electrical circuit are:
The constants which are all positive and assumed to be known.
o The impressed voltage E, measured in volts
o The resistance R, measured in ohms
o The capacitance C, measured in farads
o The inductance L, measured in henrys.
The current I, a function of time t, measured in amperes
The total charge Q, measured in coulombs, on the capacitor at time t
dQ
The relation between Q and I is I
LCR series circuit
dt
Example 5. Sketch a picture of a simple
LRC series electrical circuit.
Using the fundamental laws first formulated in the mid-nineteenth century by the German physicist G.R.
Kirchhoff,* a second-order linear model of an LCR circuit can be obtained:
d 2Q
dQ 1
L
R
Q E (t ) .
dt
dt C
1
Q E (t ) with the mechanical model,
If we compare the circuit model LQ ' ' RQ'
C
mu + u + ku = F(t)
we can see the following correlations:
Correlations
Mechanical System
Mass, m
Damping Constant, γ
Spring Constant, k
Impressed Force, F(t)
Displacement, u(t)
Velocity, u′(t)
Electrical System
Inductance, L
Resistance, R
Reciprocal of Capacitance, 1/C
Impressed Voltage, E(t)
Charge, Q(t)
Current, I = Q′(t)
Thus, we can compare the behavior of charge on an LRC circuit with harmonic, free, damped, etc.
mechanical vibrations.
*
Kirchhoff’s current law states that the total current entering any point of a circuit equals the total current leaving it.
Kirchhoff’s voltage law states that the sum of the voltage changes around any loop in a circuit is zero.
Sec. 3.8, Boyce & DiPrima, p. 10
Example 6. Find the charge Q on the capacitor in an L-R-C series circuit when L = 0.25 henry, R = 10
ohms, C = 0.001 farad, E(t) = 0, with initial conditions Q(0) = Qo coulombs, and I(0) = 0. Discuss the
behavior of this charge as time increases. With what kind of mechanical vibration does this circuit
compare?
Sec. 3.8, Boyce & DiPrima, p. 11