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(24), and (25), the nature of the solution changes as y passes through the value 2\fkm. This value is known as critical damping, while for larger values of y 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 y. 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.
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
u" + 0.125 u’ + 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 t such that lu(t)| < 0.1 for all t > t.
The solution of Eq. (29) is
u = e-t/16
' . V255 D . V255
A cos , , t + B sin , , t
To satisfy the initial conditions we must choose A = 2 and B = 2/V255; hence the solution of the initial value problem is
u = e-t/16 (^cos^2551 + 16
2 V255 \
sin ,, t V255 16
where tan S = 1/V255, so S = 0.06254. The displacement of the mass as afunctionof time is shown in Figure 3.8.7. For the purpose of comparison we also show the motion if the damping term is neglected.
FIGURE 3.8.7 Vibration with small damping (solid curve) and with no damping (dashed curve).
The quasi frequency is ^ = V255/16 = 0.998 and the quasi period is Td = 2n/^ = 6.295 sec. These values differ only slightly from the corresponding values (1 and 2n, 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
3.8 Mechanical and Electrical Vibrations
graph it appears that t is about 47.5 and by a more precise calculation we find that t = 47.5149 sec.
To find the time at which the mass first passes through its equilibrium position, we refer to Eq. (30) and set V255t/16 — S equal to n/2, the smallest positive zero of the cosine function. Then, by solving for t, we obtain
16 / n \
t = ( —+ S) = 1.637 sec.
V255 V2 /
u, 0.1 0.05 O II 3 c
u = AT e-tnecos (a t- 0.06254) V255 ( 16 )
- 0.05 - 0.1 - 0.15 40 45 1 50 55 60 1
FIGURE 3.8.8 Solution of Example 3; determination of t .
Electric Circuits. A second example of the occurrence of second order linear differential 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 = dQ/dt. (31)
Resistance R Capacitance C
?l Inductance L
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 closed
circuit, 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 IR.
The voltage drop across the capacitor is Q/ C.
The voltage drop across the inductor is LdI/dt.
Hence, by Kirchhoff’s law,
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
Thus we must know the charge on the capacitor and the current in the circuit at some initial time t0.
Alternatively, we can obtain a differential equation for the current I by differentiating Eq. (33) with respect to t, and then substituting for dQ/dt from Eq. (31). The result is
Hence I0 is also determined by the initial charge and current, which are physically 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.