Download MATH 415, SPRING 2008

MATH 415, SPRING 2008
MECHANICAL AND ELECTRICAL OSCILLATIONS
Spring-mass systems
Let u = u(t) denote the position, at any time t, of an object attached to a spring
and allowed to move along a line forming the “internal axis” of the spring. Thus, u is
represented by a number, and we adopt the convention that the conditions u = 0, u > 0,
u < 0 represent the cases where the spring is in equilibrium (at its “natural length”); or,
is compressed; or, respectively, is stretched.
The object is affected by the following forces, all of which are parallel to the axis of the
spring, and hence may be expressed as real numbers. First, there is an external (applied)
force F (t), allowed to vary with time, which may be of an arbitrary nature and due to any
possible agent; then, the spring-elasticity force, trying to restore the spring to its natural
length; finally, the damping force, due to viscosity of the medium the object is moving in
(if any), which opposes the object’s motion and tries to slow it down. The last two forces
thus are always opposite in sign to u and, respectively, the velocity v = u0 .
As a function of the time t, the position u satisfies the following second-order linear
ordinary differential equation with constant coefficients:
(1)
mu00 + γu0 + ku = F (t) ,
where m, γ, k are the object’s mass, the damping constant and the spring constant.
Formula (1) follows from Newton’s second law of dynamics (along with some other
principles, such as Hooke’s law). Namely, according to the second law, mu00 (mass times
acceleration) equals the total force Ftotal applied to the object. Now
(2)
Ftotal = F (t) − γu0 − ku ,
as explained above. (About the minus signs, see the opposite-sign remark two paragraphs
ago.) The fact that the expression (2) equals mu00 now gives (1).
The quantities involved represent specific physical units, rather than pure “dimensionless” numbers. In SI units, m is expressed in kilograms, γ in newton-seconds per meter
(i.e., newtons per meter-per-second), and k in newtons per meter. The SI unit of force,
the newton, is the kilogram times meter per second-squared, i.e., the force that imparts to
a mass of one kilogram the acceleration of one meter-per-second per second.
Simple electrical circuits
Let us now consider a single-loop RCL circuit, consisting of a coil (inductor) of inductance L, a resistor of resistance R, and a capacitor of capacitance C. The circuit is
connected to an electric power source, which provides some impressed voltage (electromotive force) E(t) that, again, is allowed to vary with time.
We denote by I = I(t) the current in the circuit, which in general varies with time,
and may assume values of both signs, as well as zero. The usual sign convention is that,
after selecting one of the two possible “orientations” of the circuit loop (i.e., the chronology
in which it is to be traversed), we treat I as positive (or, negative) of the electron flow
is opposite to (or, in agreement with) this chosen orientation. Thinking of the electron
flow as the opposite flow of some fictional positive charges, we see that the sign of I then
matches the way such positive charges move.
The SI units for L, R, C are the henry, the ohm, and the farad, while the current I
and voltage E(t) are expressed in amperes and volts. Note that the most basic SI units are
the meter, kilogram, second and ampere (plus a few others, such as the kelvin, which need
not concern us here). Other units are derived from the basic ones in such a way that there is
a perfect agreement of units in formula (4) below. Namely, the coulomb (SI unit of electric
charge) equals ampere times second, and is the amount of charge transferred by a steady
direct current of one ampere during one second. The volt expresses potential differences,
that is, amounts of work (energy) needed to transfer a coulomb of electric charge between
the two given locations, and so the volt is the joule divided by the coulomb. (The energy
unit, joule, equals newton times meter, i.e., describes the amount of work done by a force
of one newton displacing an object by the distance of one meter.) Next, the farad equals
coulomb per volt; the ohm is volt divided by ampere; and the henry equals volt times
second per ampere. (On the other hand, the unit of power output is the watt, defined to
be joule per second, and it is at the same time equal to volt times ampere.)
Let Q be the amount of excess positive charge on the plate of the capacitor which is
reached first when the circuit is traversed as indicated by the fixed orientation. (Thus, if
Q < 0, there is an excess of negative charge on that plate.) The other plate then will have
an excess positive charge of −Q, and Q may also vary with time. Now
(3)
LI 0 + RI +
1
Q = E(t) .
C
In fact, by Lenz’s law, a current I varying with time in an inductor of inductance L generates an additional amount of electromotive force equal to −LI 0 (with the negative sign
indicating that this extra electromotive force tries to oppose the change in current which
caused it). Similarly, the capacitor leads to an additional contribution to the electromotive
force, equal to −Q/C (the minus sign indicating, again, that the excess of positive charge
on the first-reached plate tries to slow down a further accumulation of positive charge
on it, due to electrostatic repulsion). The total effective electromotive force thus equals
E(t) − LI 0 − Q/C which, by Ohm’s law, must be the same as RI, and (3) follows.
Applying d/dt to both sides of (3), we further obtain
(4)
LI 00 + RI 0 +
1
I = E 0 (t) .
C
Solving this second-order linear equation with constant coefficients we can find precisely
how I varies with time, provided that the initial values of I and I 0 are given.