5.6 Series Solutions near a Regular Singular Point, Part I
Chapter 3. Second Order Linear Equations
3.8 Mechanical and Electrical Vibrations
One of the reasons that second order linear equations with constant coefficients are worth studying is that they serve as mathematical models of some important physical ODE
processes. Two important areas of application are in the fields of mechanical and electrical oscillations. For example, the motion of a mass on a vibrating spring, the torsional oscillations of a shaft with a flywheel, the flow of electric current in a simple series circuit, and many other physical problems are all described by the solution of an initial value problem of the form ay + by + cy = g(t),y(0) = y ,y(0) = y .
(1)
0
0
This illustrates a fundamental relationship between mathematics and physics: Manyphysical problems may have the same mathematical model. Thus, once we know how to solve the initial value problem (1), it is only necessary to make appropriate interpretations of the constants a, b, and c, and the functions y and g, to obtain solutions of different physical problems.
We will study the motion of a mass on a spring in detail because an understanding of the behavior of this simple system is the first step in the investigation of more complex vibrating systems. Further, the principles involved are common to many problems.
Consider a mass m hanging on the end of a vertical spring of original length l, as shown in Figure 3.8.1. The mass causes an elongation L of the spring in the downward (positive) direction. There are two forces acting at the point where the mass is attached to the spring; see Figure 3.8.2. The gravitational force, or weight of the mass, acts downward and has magnitude mg, where g is the acceleration due to gravity. There is also a force F , due to the spring, that acts upward. If we assume that the elongation L
s
of the spring is small, the spring force is very nearly proportional to L; this is known
l
l + L + u
L
m
m
u
FIGURE 3.8.1 A spring–mass system.
Fs = –kLw = mgFIGURE 3.8.2 Force diagram for a spring–mass system.
3.8Mechanical and Electrical Vibrations as Hooke’s8 law. Thus we write F = −k L, where the constant of proportionality k is
s
called the spring constant, and the minus sign is due to the fact that the spring force acts in the upward (negative) direction. Since the mass is in equilibrium, the two forces balance each other, which means that mg − k L = 0.
(2)
For a given weight w = mg, one can measure L and then use Eq. (2) to determine k.
Note that k has the units of force/length.
In the corresponding dynamic problem we are interested in studying the motion of the mass when it is acted on by an external force or is initially displaced. Let u(t), measured positive downward, denote the displacement of the mass from its equilibrium position at time t; see Figure 3.8.1. Then u(t) is related to the forces acting on the mass through Newton’s law of motion, mu(t) = f (t), (3)
where u is the acceleration of the mass and f is the net force acting on the mass.
Observe that both u and f are functions of time. In determining f there are four separate forces that must be considered:
1.
The weight w = mg of the mass always acts downward.
2.
The spring force F is assumed to be proportional to the total elongation L + u of
s the spring and always acts to restore the spring to its natural position. If L + u > 0, then the spring is extended, and the spring force is directed upward. In this case F = −k(L + u).
(4)
s
On the other hand, if L + u < 0, then the spring is compressed a distance |L + u|, and the spring force, which is now directed downward, is given by F = k|L + u|.
s
However, when L + u < 0, it follows that |L + u| = −(L + u), so F is again
s given by Eq. (4). Thus, regardless of the position of the mass, the force exerted by the spring is always expressed by Eq. (4).
3.
The damping or resistive force F always acts in the direction opposite to the direc d
tion of motion of the mass. This force may arise from several sources: resistance from the air or other medium in which the mass moves, internal energy dissipation due to the extension or compression of the spring, friction between the mass and the guides (if any) that constrain its motion to one dimension, or a mechanical device (dashpot) that imparts a resistive force to the mass. In any case, we assume that the resistive force is proportional to the speed |du/dt| of the mass; this is usually referred to as viscous damping. If du/dt > 0, then u is increasing, so the mass is moving downward. Then F is directed upward and is given by
d
F (t) = −γ u(t), (5)
d where γ is a positive constant of proportionality known as the damping constant.
On the other hand, if du/dt < 0, then u is decreasing, the mass is moving upward, and F is directed downward. In this case, F = γ |u(t)|; since |u(t)| = −u(t),
d
d 8Robert Hooke (1635 –1703) was an English scientist with wide-ranging interests. His most important book, Micrographia, was published in 1665 and described a variety of microscopical observations. Hooke first published his law of elastic behavior in 1676 as an anagram: ceiiinosssttuv; in 1678 he gave the solution ut tensio sic vis, which means, roughly, “as the force so is the displacement.”
Chapter 3. Second Order Linear Equations it follows that F (t) is again given by Eq. (5). Thus, regardless of the direction of
d
motion of the mass, the damping force is always expressed by Eq. (5).
The damping force may be rather complicated and the assumption that it is modeled adequately by Eq. (5) may be open to question. Some dashpots do behave as Eq. (5) states, and if the other sources of dissipation are small, it may be possible to neglect them altogether, or to adjust the damping constant γ to approximate them. An important benefit of the assumption (5) is that it leads to a linear (rather than a nonlinear) differential equation. In turn, this means that a thorough analysis of the system is straightforward, as we will show in this section and the next.
4.
An applied external force F (t) is directed downward or upward as F(t) is positive or negative. This could be a force due to the motion of the mount to which the spring is attached, or it could be a force applied directly to the mass. Often the external force is periodic.
Taking account of these forces, we can now rewrite Newton’s law (3) as mu(t) = mg + F (t) + F (t) + F(t)
s
d
= mg − k[L + u(t)] − γ u(t) + F(t).
(6)
Since mg − k L = 0 by Eq. (2), it follows that the equation of motion of the mass is mu(t) + γ u(t) + ku(t) = F(t), (7)
where the constants m, γ , and k are positive. Note that Eq. (7) has the same form as Eq. (1).
It is important to understand that Eq. (7) is only an approximate equation for the displacement u(t). In particular, both Eqs. (4) and (5) should be viewed as approximations for the spring force and the damping force, respectively. In our derivation we have also neglected the mass of the spring in comparison with the mass of the attached body.
The complete formulation of the vibration problem requires that we specify two initial conditions, namely, the initial position u and the initial velocity v of the mass:
0
0
u(0) = u ,u(0) = v .
(8)
0
0
It follows from Theorem 3.2.1 that these conditions give a mathematical problem that has a unique solution. This is consistent with our physical intuition that if the mass is set in motion with a given initial displacement and velocity, then its position will be determined uniquely at all future times. The position of the mass is given (approximately) by the solution of Eq. (7) subject to the prescribed initial conditions (8).
A mass weighing 4 lb stretches a spring 2 in. Suppose that the mass is displaced an E X A M P L E additional 6 in. in the positive direction and then released. The mass is in a medium that
1
exerts a viscous resistance of 6 lb when the mass has a velocity of 3 ft/sec. Under the assumptions discussed in this section, formulate the initial value problem that governs the motion of the mass.
The required initial value problem consists of the differential equation (7) and initial conditions (8), so our task is to determine the various constants that appear in these equations. The first step is to choose the units of measurement. Based on the statement of the problem, it is natural to use the English rather than the metric system of units. The only time unit mentioned is the second, so we will measure t in seconds. On the other hand, both the foot and inch appear in the statement as units of length. It is immaterial which one we use, but having made a choice, it is important to be consistent. To be definite, let us measure the displacement u in feet.
Since nothing is said in the statement of the problem about an external force, we assume that F (t) = 0. To determine m note that
w
lb-sec2 m =
=
4 lb
= 1 .
g
32 ft/ sec2
8
ft
The damping coefficient γ is determined from the statement that γ u is equal to 6 lb when u is 3 ft/sec. Therefore γ = 6 lb = lb-sec
2 .
3 ft/sec
ft
The spring constant k is found from the statement that the mass stretches the spring by 2 in., or by 1/6 ft. Thus
lb
k = 4 lb = 24 .
1/6 ft
ft Consequently, Eq. (7) becomes 1 u + 2u + 24u = 0, 8
or u + 16u + 192u = 0.
(9)
The initial conditions are u(0) = 1 ,u(0) = 0.
(10)
2
The second initial condition is implied by the word “released” in the statement of the problem, which we interpret to mean that the mass is set in motion with no initial velocity.
Undamped Free Vibrations.
If there is no external force, then F (t) = 0 in Eq. (7). Let us also suppose that there is no damping, so that γ = 0; this is an idealized configuration of the system, seldom (if ever) completely attainable in practice. However, if the actual damping is very small, the assumption of no damping may yield satisfactory results over short to moderate time intervals. Then the equation of motion (7) reduces to mu + ku = 0.
(11)
The general solution of Eq. (11) is u = A cos ω t + B sin ω t,
(12)
0
0
where ω2 = k/m.
(13)
0
The arbitrary constants A and B can be determined if initial conditions of the are given.
Chapter 3. Second Order Linear Equations
In discussing the solution of Eq. (11) it is convenient to rewrite Eq. (12) in the form u = R cos(ω t − δ),
(14)
0
or u = R cos δ cos ω t + R sin δ sin ω t.
(15)
0
0
By comparing Eq. (15) with Eq. (12), we find that A, B, R, and δ are related by the equations A = R cos δ,B = R sin δ.
(16)
Thus
R = A2 + B2, tan δ = B/A.
(17)
In calculating δ care must be taken to choose the correct quadrant; this can be done by checking the signs of cos δ and sin δ in Eqs. (16).
The graph of Eq. (14), or the equivalent Eq. (12), for a typical set of initial conditions is shown in Figure 3.8.3. The graph is a displaced cosine wave that describes a periodic, or simple harmonic, motion of the mass. The the motion is m 1/2
T = 2π = .
ω
2π
(18)
k
0
√
The circular frequency ω = k/m, measured in radians per unit time, is called the
0
the vibration. The maximum displacement R of the mass from equilibrium is of the motion. The dimensionless parameter δ is called the or phase angle, and measures the displacement of the wave from its normal position corresponding to δ = 0.
Note that the motion described by Eq. (14) has a constant amplitude that does not diminish with time. This reflects the fact that in the absence of damping there is no way for the system to dissipate the energy imparted to it by the initial displacement and velocity. Further, for a given mass m and spring constant k, the system always vibrates at the same frequency ω , regardless of the initial conditions. However, the
0
initial conditions do help to determine the amplitude of the motion. Finally, observe from Eq. (18) that T increases as m increases, so larger masses vibrate more slowly.
On the other hand, T decreases as k increases, which means that stiffer springs cause the system to vibrate more rapidly.
u
R
R cos δ
δ δ+ π δ + 2 π
ω0t
–R
FIGURE 3.8.3
Simple harmonic motion; u = R cos(ω t − δ).
0
3.8Mechanical and Electrical Vibrations
Suppose that a mass weighing 10 lb stretches a spring 2 in. If the mass is displaced an E X A M P L E additional 2 in. and is then set in motion with an initial upward velocity of 1 ft/sec, de 2
termine the position of the mass at any later time. Also determine the period, amplitude, and phase of the motion.
The spring constant is k = 10 lb/2 in. = 60 lb/ft and the mass is m = w/g = 10/32 lb-sec2/ft. Hence the equation of motion reduces to u + 192u = 0, (19)
and the general solution is
√
√
u = A cos(8 3t) + B sin(8 3t).
The solution satisfying the initial conditions u(0) = 1/6 ft and u(0) = −1 ft/sec is
√
√
u = 1 cos(8 3t) − 1 √ sin(8 3t).
(20)
6
8 3
√
The natural frequency is ω = 192 ∼ = 13.856 rad/sec, so the period is T = 2π/w ∼
=
0
0
0.45345 sec. The amplitude R and phase δ are found from Eqs. (17). We have R2 = 1 + 1 = 19 , so
R ∼ = 0.18162 ft.
36
192
576
√
The second of Eqs. (17) yields tan δ = − 3/4. There are two solutions of this equation, one in the second quadrant and one in the fourth. In the present problem cos δ > 0 and sin δ < 0, so δ is in the fourth quadrant, namely, √
δ = − arctan( 3/4) ∼ = −0.40864 rad.
The graph of the solution (20) is shown in Figure 3.8.4.
u
R = 0.182
~
u = 0.182 cos(8√3 t + 0.409)
0.2
0.5
1
1.5
t
– 0.2
T = 0.453
~
FIGURE 3.8.4
An undamped free vibration; u + 192u = 0, u(0) = 1/6, u(0) = −1.
Damped Free Vibrations.
If we include the effect of damping, the differential equation governing the motion of the mass is mu + γ u + ku = 0.
(21)
Chapter 3. Second Order Linear Equations
We are especially interested in examining the effect of variations in the damping coefficient γ for given values of the mass m and spring constant k. The roots of the corresponding characteristic equation are
−γ ± γ 2 − 4km
γ
r , r =
=
−1 ± 1 − 4km .
(22)
1
2
2m
2m
γ 2
Depending on the sign of γ 2 − 4km, the solution u has one of the following forms: γ 2 − 4km > 0, u = Aer t
t
1
+ Ber2 ;
(23)
γ 2 − 4km = 0, u = (A + Bt)e−γt/2m;
(24)
(
γ 4km − γ 2)1/2 2 − 4km < 0, u = e−γt/2m(A cos µt + B sin µt), µ = > 0.
2m
(25)
Since m, γ , and k are positive, γ 2 − 4km is always less than γ 2. Hence, if γ 2 − 4km ≥ 0, then the values of r and r given by Eq. (22) are negative. If γ 2 − 4km < 0, then
1
2
the values of r and r are complex, but with negative real part. Thus, in all cases,
1
2
the solution u tends to zero as t → ∞; this occurs regardless of the values of the arbitrary constants A and B, that is, regardless of the initial conditions. This confirms our intuitive expectation, namely, that damping gradually dissipates the energy initially imparted to the system, and consequently the motion dies out with increasing time.
The most important case is the third one, which occurs when the damping is small.
If we let A = R cos δ and B = R sin δ in Eq. (25), then we obtain u = Re−γ t/2m cos(µt − δ).
(26)
The displacement u lies between the curves u = ±Re−γ t/2m; hence it resembles a cosine wave whose amplitude decreases as t increases. A typical example is sketched in Figure 3.8.5. The motion is called a damped oscillation, or a damped vibration. The amplitude factor R depends on m, γ , k, and the initial conditions.
Although the motion is not periodic, the parameter µ determines the frequency with which the mass oscillates back and forth; consequently, µ is called the quasi
frequency. By comparing µ with the frequency ω of undamped motion, we find that
0
µ
(
γ
1/2
2
γ 2 = 4km − γ 2)1/2/2m
√
=
∼
=
.
ω 1 −
1 −
(27)
k/m
4km
8km
0
u
Re– γt/2m
R cos δ
δ
δ π
+
+ 2 δ π + 3 δ π
µt
−Re– γt/2m
FIGURE 3.8.5 Damped vibration; u = Re−γ t/2m cos(µt − δ).
3.8Mechanical and Electrical Vibrations
The last approximation is valid when γ 2/4km is small; we refer to this situation as “small damping.” Thus, the effect of small damping is to reduce slightly the frequency of the oscillation. By analogy with Eq. (18) the quantity T = 2π/µ is called the quasid
period. It is the time between successive maxima or successive minima of the position of the mass, or between successive passages of the mass through its equilibrium position while going in the same direction. The relation between T and T is given by
d
−
1/2
T
ω
γ 2
γ 2 d = 0 = 1 −
∼
= 1 + ,
(28) T
µ
4km
8km
where again the last approximation is valid when γ 2/4km is small. Thus, small damping increases the quasi period.
Equations (27) and (28) reinforce the significance of the dimensionless ratio γ 2/4km.
It is not the magnitude of γ alone that determines whether damping is large or small, but the magnitude of γ 2 compared to 4km. When γ 2/4km is small, then we can neglect the effect of damping in calculating the quasi frequency and quasi period of the motion.
On the other hand, if we want to study the detailed motion of the mass for all time, then we can never neglect the damping force, no matter how small.
As γ 2/4km increases, the quasi frequency µ decreases and the quasi period Td
√
increases. In fact, µ → 0 and T → ∞ as γ → 2 km. As indicated by Eqs. (23),
d
√
(24), and (25), the nature of the solution changes as γ passes through the value 2 km.
This value is known as critical damping, while for larger values of γ the motion is said to be overdamped. In these cases, given by Eqs. (24) and (23), respectively, the mass creeps back to its equilibrium position, but does not oscillate about it, as for small γ .
Two typical examples of critically damped motion are shown in Figure 3.8.6, and the situation is discussed further in Problems 21 and 22.
u
2
u(0) = , 1 u'(0) = 7
2
4
u = + 2
1
t e–t/2
(2 )
1
2
4
6
8
10
t
u(0) = , 1 u'(0) = – 7
2
4
u =
1 – 3 t e–t/2
–1
(
)
2
2
FIGURE 3.8.6
Critically damped motions: u + u + 0.25u = 0; u = (A + Bt)e−t/2.
The motion of a certain spring–mass system is governed by the differential equation E X A M P L E
3
u + 0.125u + u = 0, (29)
where u is measured in feet and t in seconds. If u(0) = 2 and u(0) = 0, determine the position of the mass at any time. Find the quasi frequency and the quasi period, as well Chapter 3. Second Order Linear Equations as the time at which the mass first passes through its equilibrium position. Also find the time τ such that |u(t)| < 0.1 for all t > τ .
The solution of Eq. (29) is
√
√
255
255
u = e−t/16 A cos t + B sin t .
16
16
√
To satisfy the initial conditions we must choose A = 2 and B = 2/ 255; hence the solution of the initial value problem is
√
√
255
255
u = e−t/16 2 cos t +
2
√
sin
t
16
255
16
√
= 32 √
255
e−t/16 cos t − δ ,
(30)
255
16
√
where tan δ = 1/ 255, so δ ∼ = 0.06254. The displacement of the mass as a function of time is shown in Figure 3.8.7. For the purpose of comparison we also show the motion if the damping term is neglected.
u
u" + u = 0 u(0) = 2, u' (0) = 0 u" + 0.125 u' + u = 0
2
1
10
20
30
40
50
t
–1
–2
FIGURE 3.8.7
Vibration with small damping (solid curve) and with no damping (dashed curve).
√
The quasi frequency is µ =
255/16 ∼ = 0.998 and the quasi period is T = 2π/µ ∼
=
d
6.295 sec. These values differ only slightly from the corresponding values (1 and 2π, respectively) for the undamped oscillation. This is evident also from the graphs in Figure 3.8.7, which rise and fall almost together. The damping coefficient is small in this example, only one-sixteenth of the critical value, in fact. Nevertheless, the amplitude of the oscillation is reduced rather rapidly. Figure 3.8.8 shows the graph of the solution for 40 ≤ t ≤ 60, together with the graphs of u = ±0.1. From the graph it appears that τ is about 47.5 and by a more precise calculation we find that τ ∼ = 47.5149 sec.
To find the time at which the mass first passes through its equilibrium position, we
√
refer to Eq. (30) and set 255t/16 − δ equal to π/2, the smallest positive zero of the cosine function. Then, by solving for t, we obtain
π
t =
16
√
+ δ ∼ = 1.637 sec .
255
2
u
u = 0.1
0.1
√ u =
32
e–t/16 cos 255 t – 0.06254
(
)
√255
16
0.05
τ
40
45
50
55
60
t
– 0.05 u = – 0.1 – 0.1 – 0.15 FIGURE 3.8.8
Solution of Example 3; determination of τ .
Electric Circuits.
A second example of the occurrence of second order linear differ ential equations with constant coefficients is as a model of the flow of electric current in the simple series circuit shown in Figure 3.8.9. The current I , measured in amperes, is a function of time t. The resistance R (ohms), the capacitance C (farads), and the inductance L (henrys) are all positive and are assumed to be known constants. The impressed voltage E (volts) is a given function of time. Another physical quantity that enters the discussion is the total charge Q (coulombs) on the capacitor at time t. The relation between charge Q and current I is I = d Q/dt.
(31)
Resistance R
Capacitance C
Inductance L
I Impressed voltage E(t) FIGURE 3.8.9
A simple electric circuit.
Chapter 3. Second Order Linear Equations
The flow of current in the circuit is governed by Kirchhoff’s9 second law: In a closedcircuit, the impressed voltage is equal to the sum of the voltage drops in the rest of the
circuit.
According to the elementary laws of electricity, we know that:
The voltage drop across the resistor is I R.
The voltage drop across the capacitor is Q/C.
The voltage drop across the inductor is Ld I /dt.
Hence, by Kirchhoff’s law, d I
L + RI + 1 Q = E(t).
(32)
dt
C
The units have been chosen so that 1 volt = 1 ohm · 1 ampere = 1 coulomb/1 farad = 1 henry · 1 ampere/1 second.
Substituting for I from Eq. (31), we obtain the differential equation L Q + R Q + 1 Q = E(t) (33)
C for the charge Q. The initial conditions are Q(t ) = Q ,Q(t ) = I (t ) = I .
(34)
0
0
0
0
0
Thus we must know the charge on the capacitor and the current in the circuit at some initial time t .
0
Alternatively, we can obtain a differential equation for the current I by differentiating Eq. (33) with respect to t, and then substituting for d Q/dt from Eq. (31). The result is L I + R I + 1 I = E(t), (35)
C
with the initial conditions I (t ) = I ,I (t ) = I .
(36)
0
0
0
0
From Eq. (32) it follows that E(t ) − R I − (I /C)Q
I =
0
0
0 .
(37)
0
L
Hence I is also determined by the initial charge and current, which are physically
0
measurable quantities.
The most important conclusion from this discussion is that the flow of current in the circuit is described by an initial value problem of precisely the same form as the one that describes the motion of a spring–mass system. This is a good example of the unifying role of mathematics: Once you know how to solve second order linear equations with constant coefficients, you can interpret the results either in terms of mechanical vibrations, electric circuits, or any other physical situation that leads to the same problem.
9Gustav Kirchhoff (1824 –1887), professor at Breslau, Heidelberg, and Berlin, was one of the leading physicists of the nineteenth century. He discovered the basic laws of electric circuits about 1845 while still a student at
Ko¨nigsberg. He is also famous for fundamental work in electromagnetic absorption and emission and was one of the founders of spectroscopy.
3.8Mechanical and Electrical Vibrations
PROBLEMS
In each of Problems 1 through 4 determine ω , R, and δ so as to write the given expression in
0
the form u = R cos(ω t − δ).
0
√
䉴 5. A mass weighing 2 lb stretches a spring 6 in. If the mass is pulled down an additional 3 in.
6. A mass of 100 g stretches a spring 5 cm. If the mass is set in motion from its equilibrium
7. A mass weighing 3 lb stretches a spring 3 in. If the mass is pushed upward, contracting
䉴 9. A mass of 20 g stretches a spring 5 cm. Suppose that the mass is also attached to a 䉴 10. A mass weighing 16 lb stretches a spring 3 in. The mass is attached to a viscous damper
11. A spring is stretched 10 cm by a force of 3 newtons. A mass of 2 kg is hung from the spring 12. A series circuit has a capacitor of 10−5 farad, a resistor of 3 × 102 ohms, and an inductor
14. Show that the period of motion of an undamped vibration of a mass hanging from a vertical
√
15. Show that the solution of the initial value problem
0
0
0
0
0
0
0
0
0
0
Chapter 3. Second Order Linear Equations
0
0
0
0
0
17. A mass weighing 8 lb stretches a spring 1.5 in. The mass is also attached to a damper with
0
0
0
0
0
the mass passes through its equilibrium position after it is released.
21.
Logarithmic Decrement.
(a) For the damped oscillation described by Eq. (26), show
d
d
d
24. The position of a certain spring–mass system satisfies the initial value problem
2
䉴 25. Consider the initial value problem
0
0
3.8Mechanical and Electrical Vibrations (e) Another way to proceed is to write the solution of the initial value problem in the 26. Consider the initial value problem
0
0
(a) Solve the initial value problem.
0
0
other parameters.
0
0
䉴 28. The position of a certain undamped spring–mass system satisfies the initial value problem (a) Find the solution of this initial value problem.
䉴 29. The position of a certain spring–mass system satisfies the initial value problem
4
(a) Find the solution of this initial value problem.
30. In the absence of damping the motion of a spring–mass system satisfies the initial value
problem
Your result should confirm the principle of conservation of energy for this system.
Chapter 3. Second Order Linear Equations
k
m
FIGURE 3.8.10 A spring–mass system.
䉴 32. In the spring–mass system of Problem 31, suppose that the spring force is not given by Hooke’s law but instead satisfies the relation
s
differential equation
Suppose that the initial conditions are
3.9 Forced Vibrations
Consider now the case in which a periodic external force, say F cos ωt with ω > 0, is
0
applied to a spring–mass system. Then the equation of motion is
ODE
mu + γ u + ku = F cos ωt.
(1)
0
First suppose that there is no damping; then Eq. (1) reduces to mu + ku = F cos ωt.
(2)
0
√
If ω = k/m = ω, then the general solution of Eq. (2) is
0
Fu = c cos ω t + c sin ω t +
0
cos ωt.
(3)
1
0
2
0
m(ω2 − ω2)
0
The constants c and c are determined by the initial conditions. The resulting motion
1
2
is, in general, the sum of two periodic motions of different frequencies (ω and ω) and
0
amplitudes. There are two particularly interesting cases.
3.9
Forced VibrationsBeats.
Suppose that the mass is initially at rest, so that u(0) = 0 and u(0) = 0. Then it turns out that the constants c and c in Eq. (3) are given by
1
2
F
c = −
0
,c = 0, (4)
1
m(ω2 − ω2)
2
0
and the solution of Eq. (2) is
F
u =
0
(cos ωt − cos ω t).
(5)
m(ω2 − ω2)
0
0
This is the sum of two periodic functions of different periods but the same amplitude.
Making use of the trigonometric identities for cos(A ± B) with A = (ω + ω)t/2 and
0
B = (ω − ω)t/2, we can write Eq. (5) in the form
0
2F
(ω − ω)t(ω + ω)tu =
0
sin
0
sin
0 .
(6)
m(ω2 − ω2)
2
2
0
If |ω − ω| is small, then ω + ω is much greater than |ω − ω|. Consequently,
0
0
0
sin(ω + ω)t/2 is a rapidly oscillating function compared to sin(ω − ω)t/2. Thus
0
0
the motion is a rapid oscillation with frequency (ω + ω)/2, but with a slowly varying
0
sinusoidal amplitude
2F
(ω − ω)t
0
sin
0 .
m(ω2 − ω2)
2
0
This type of motion, possessing a periodic variation of amplitude, exhibits what is called
Such a phenomenon occurs in acoustics when two tuning forks of nearly equal frequency are sounded simultaneously. In this case the periodic variation of amplitude is quite apparent to the unaided ear. In electronics the variation of the amplitude with
u
u = 2.77778 sin 0.1t
3
u = 2.77778 sin 0.1 t sin 0.9 t
2
1
t
10
20
30
40
50
60
–1
–2
u = – 2.77778 sin 0.1t
–3
FIGURE 3.9.1
A beat; solution of u + u = 0.5 cos 0.8t, u(0) = 0, u(0) = 0; u = 2.77778 sin 0.1t sin 0.9t.
Chapter 3. Second Order Linear Equations time is called amplitude modulation. The graph of u as given by Eq. (6) in a typical case is shown in Figure 3.9.1.
Resonance.
As a second example, consider the case ω = ω ; that is, the frequency
0
of the forcing function is the same as the natural frequency of the system. Then the nonhomogeneous term F cos ωt is a solution of the homogeneous equation. In this
0
case the solution of Eq. (2) is
Fu = c cos ω t + c sin ω t +
0
t sin ω t.
(7)
1
0
2
0
2mω
0
0
Because of the term t sin ω t, the solution (7) predicts that the motion will become
0
unbounded as t → ∞ regardless of the values of c and c ; see Figure 3.9.2 for a
1
2
typical example. Of course, in reality unbounded oscillations do not occur. As soon as u becomes large, the mathematical model on which Eq. (1) is based is no longer valid, since the assumption that the spring force depends linearly on the displacement requires that u be small. If damping is included in the model, the predicted motion remains bounded; however, the response to the input function F cos ωt may be quite
0
large if the damping is small and ω is close to ω . This phenomenon is known as
0
Resonance can be either good or bad depending on the circumstances. It must be taken very seriously in the design of structures, such as buildings or bridges, where it can produce instabilities possibly leading to the catastrophic failure of the structure.
For example, soldiers traditionally break step when crossing a bridge to eliminate the periodic force of their marching that could resonate with a natural frequency of the bridge. Another example occurred in the design of the high-pressure fuel turbopump for the space shuttle main engine. The turbopump was unstable and could not be operated over 20,000 rpm as compared to the design speed of 39,000 rpm. This difficulty led to a shutdown of the space shuttle program for 6 months at an estimated cost of u
10
u = 0.25 t sin t
5
u = 0.25 t
10
20
30
40
t
–5
u = – 0.25 t
–10
FIGURE 3.9.2
Resonance; solution of u + u = 0.5 cos t, u(0) = 0, u(0) = 0; u = 0.25t sin t.
3.9
Forced Vibrations $500,000/day.10 On the other hand, resonance can be put to good use in the design of instruments, such as seismographs, intended to detect weak periodic incoming signals.
Forced Vibrations with Damping.
The motion of the spring–mass system with damp ing and the forcing function F cos ωt can be determined in a straightforward manner,
0
although the computations are rather lengthy. The solution of Eq. (1) is u = c er t
t
1
+ c er2 + R cos(ωt − δ), r = r
(8)
1
2
1
2
where
F
− ω2)
γ ωR = 0 ,,
,
cos δ = m(ω20 sin δ =
(9) and
= m2(ω2 − ω2)2 + γ 2ω2.
(10)
0
In Eq. (8) r and r are the roots of the characteristic equation associated with Eq. (1);
1
2
they may be either real and negative or complex conjugates with negative real part. In either case, both exp(r t) and exp(r t) approach zero as t → ∞. Hence, as t → ∞, 1
2
Fu → U (t) =
0
cos(ωt − δ).
(11)
m2(ω2 − ω2)2 + γ 2ω2
0
For this reason u (t) = c er t
t
1
+ c er2 is called the U(t), which c
1
2
represents a steady oscillation with the same frequency as the external force, is called the steady-state solution or the forced response. The transient solution enables us to satisfy whatever initial conditions are imposed; with increasing time the energy put into the system by the initial displacement and velocity is dissipated through the damping force, and the motion then becomes the response of the system to the external force.
Without damping, the effect of the initial conditions would persist for all time.
It is interesting to investigate how the amplitude R of the steady-state oscillation depends on the frequency ω of the external force. For low-frequency excitation, that is, as ω → 0, it follows from Eqs. (9) and (10) that R → F /k. At the other extreme, for
0
very high-frequency excitation, Eqs. (9) and (10) imply that R → 0 as ω → ∞. At an intermediate value of ω the amplitude may have a maximum. To find this maximum point, we can differentiate R with respect to ω and set the result equal to zero. In this way we find that the maximum amplitude occurs when ω = ω , where
max
γ 2
γ 2 ω2 = ω2 − = ω2 1 − .
(12)
max
0
0
2m2
2km
Note that ω< ω and that ω is close to ω when γ is small. The maximum max
0
max
0
value of R is
F
Fγ 2
R
=
0
∼
=
0
1 + ,
(13)
max
γ ω 1 − (γ 2/4mk)γ ω
8mk
0
0
10 F. Ehrich and D. Childs, “Self-Excited Vibration in High-Performance Turbomachinery,” Mechanical Engineering (May 1984), p. 66.
Chapter 3. Second Order Linear Equations where the last expression is an approximation for small γ . If γ 2/2km > 1, then ω as
max given by Eq. (12) is imaginary; in this case the maximum value of R occurs for ω = 0 and R is a monotone decreasing function of ω. For small γ it follows from that R
∼
= F /γ ω . Thus, for small γ , the maximum response is much larger than the max
0
0
amplitude F of the external force, and the smaller the value of γ , the larger the ratio
0
R
/F . Figure 3.9.3 contains some representative graphs of Rk/F versus ω/ω for
max
0
0
0
several values of γ .
The phase angle δ also depends in an interesting way on ω. For ω near zero, it follows from Eqs. (9) and (10) that cos δ ∼ = 1 and sin δ ∼ = 0. Thus δ ∼ = 0, and the response is nearly in phase with the excitation, meaning that they rise and fall together, and in particular, assume their respective maxima nearly together and their respective minima nearly together. For ω = ω , we find that cos δ = 0 and sin δ = 1, so δ = π/2. In this
0
case the response lags behind the excitation by π/2; that is, the peaks of the response occur π/2 later than the peaks of the excitation, and similarly for the valleys. Finally, for ω very large, we have cos δ ∼ = −1 and sin δ ∼ = 0. Thus δ ∼ = π, so that the response is nearly out of phase with the excitation; this means that the response is minimum when the excitation is maximum, and vice versa.
In Figure 3.9.4 we show the graph of the solution of the initial value problem u + 0.125u + u = 3 cos 2t,u(0) = 2,u(0) = 0.
The graph of the forcing function is also shown for comparison. (The unforced motion of this system is shown in Figure 3.8.7.) Observe that the initial transient motion decays as t increases, that the amplitude of the steady forced response is approximately 1, and that the phase difference between the excitation and response is approximately π.
Rk/F0
3
= 0
Γ
2
Γ = 0.5
Γ = 1.0
Γ = 2.0
Γ = 3.0
1
1
2
ω/
ω0
FIGURE 3.9.3
Forced vibration with damping: amplitude of steady-state response versus frequency of driving force; = γ 2/m2ω2.
0
3.9
Forced Vibrations
u
Solution of
Forcing function 3 cos 2tu" + 0.125u' + u = 3 cos 2t
u(0) = 2, u'(0) = 0
3
2
1
10
20
30
40
50
t
–1
–2
–3
FIGURE 3.9.4
A forced vibration with damping; solution of u + 0.125u + u = 3 cos 2t, u(0) = 2, u(0) = 0.
√
More precisely, we find that = 145/4 ∼ = 3.0104, so R = F / ∼ = 0.9965. Further,
0
cos δ = −3/ ∼ = −0.9965 and sin δ = 1/4 ∼ = 0.08305, so that δ ∼ = 3.0585. Thus the calculated values of R and δ are close to the values estimated from the graph.
PROBLEMS
In each of Problems 1 through 4 write the given expression as a product of two trigonometric functions of different frequencies.
5. A mass weighing 4 lb stretches a spring 1.5 in. The mass is displaced 2 in. in the positive
6. A mass of 5 kg stretches a spring 10 cm. The mass is acted on by an external force of 䉴 7. (a) Find the solution of Problem 5.
(b) Plot the graph of the solution.
䉴 8. (a) Find the solution of the initial value problem in Problem 6.
(b) Identify the transient and steady-state parts of the solution.
(c) Plot the graph of the steady-state solution.
Chapter 3. Second Order Linear Equations 9. If an undamped spring–mass system with a mass that weighs 6 lb and a spring constant 10. A mass that weighs 8 lb stretches a spring 6 in. The system is acted on by an external force
11. A spring is stretched 6 in. by a mass that weighs 8 lb. The mass is attached to a dashpot (a) Determine the steady-state response of this system.
12. A spring–mass system has a spring constant of 3 N/m. A mass of 2 kg is attached to the 13. Furnish the details in determining when the steady-state response given by Eq. (11) is maximum; that is, show that ω2 and R are given by Eqs. (12) and (13), respectively.
max
max
䉴 14. (a) Show that the phase of the forced response of Eq. (1) satisfies tan δ = γ ω/m(ω2 − ω2).
0
(b) Plot the phase δ as a function of the forcing frequency ω for the forced response of u + 0.125u + u = 3 cos ωt.
15. Find the solution of the initial value problem
where
0
0
䉴 17. Consider a vibrating system described by the initial value problem
4
(a) Determine the steady-state part of the solution of this problem.
䉴 18. Consider the forced but undamped system described by the initial value problem
0
3.9
Forced Vibrations 䉴 19. Consider the vibrating system described by the initial value problem
䉴 20. For the initial value problem in Problem 18 plot u versus u for ω = 0.7, ω = 0.8, and ω = 0.9; that is, draw the phase plot of the solution for these values of ω. Use a t interval that is long enough so the phase plot appears as a closed curve. Mark your curve with arrows to show the direction in which it is traversed as t increases.
Problems 21 through 23 deal with the initial value problem u + 0.125u + u = F(t),u(0) = 2,u(0) = 0.
In each of these problems:
(a) Plot the given forcing function F(t) versus t and also plot the solution u(t) versus t on the same set of axes. Use a t interval that is long enough so the initial transients are substantially eliminated. Observe the relation between the amplitude and phase of the
√
forcing term and the amplitude and phase of the response. Note that ω = k/m = 1.
0
(b) Draw the phase plot of the solution, that is, plot u versus u.
䉴 21. F(t) = 3 cos(0.3t)
䉴 䉴 23. F(t) = 3 cos 3t
䉴 24. A spring–mass system with a hardening spring (Problem 32 of Section 3.8) is acted on by
5
䉴 25. Suppose that the system of Problem 24 is modified to include a damping term and that the resulting initial value problem is u + 1 u + u + 1 u3 = cos ωt,u(0) = 0,u(0) = 0.
5
5
(a) Plot a computer-generated solution of the given problem for several values of ω between 1/2 and 2 and estimate the amplitude R of the steady response in each case.
(b) Using the data from part (a), plot the graph of R versus ω. For what frequency ω is the amplitude greatest?
(c) Compare the results of parts (a) and (b) with the corresponding results for the linear spring.
REFERENCES Coddington, E. A., An Introduction to Ordinary Differential Equations (Englewood Cliffs, NJ: Pren tice Hall, 1961; New York: Dover, 1989).
There are many books on mechanical vibrations and electric circuits. One that deals with both is: Close, C. M., and Frederick, D. K., Modeling and Analysis of Dynamic Systems (2nd ed.) (Boston: Houghton-Mifflin, 1993).
Chapter 3. Second Order Linear Equations
A classic book on mechanical vibrations is: Den Hartog, J. P., Mechanical Vibrations (4th ed.) (New York: McGraw-Hill, 1956; New York; Dover, 1985).
A more recent intermediate-level book is: Thomson, W. T., Theory of Vibrations with Applications (3rd ed.) (Englewood Cliffs, NJ: Prentice Hall, 1988).
An elementary book on electrical circuits is: Bobrow, L. S., Elementary Linear Circuit Analysis (New York: Oxford University Press, 1996).