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complex numbers, a procedure we shall introduce in the next chapter.
21-4 Initial conditions
NÑow let us consider what determines the constants A and ?Ö, or ø and A. Of
course these are determined by how we start the motion. If we start the motion
with just a small displacement, that is one type of oscillation; 1Ÿ we start with
an initial displacement and then push up when we let go, we get still a diferent
motion. The constants A and ?Ö, or a and A, or any other way of putting it, are
determined, of course, by the way the motion started, not by any other features
of the situation. Thhese are called the #miii@Ï conditions. We would like to connect
the initial conditions with the constants. Although this can be done using any
one of the forms (21.6), it turns out to be easiest if we use Eq. (21.6c). Suppose
that at ý —= 0 we have started with an initial displacement øo and a certain
velocity øọ. Thịs is the most general way we can sbart the motion. (We cannot
specify the acceleration with which it started, true, because that is determined
by the spring, once we speclfy #o.) Now let us calculate 4 and Ø. We start with
the equation for #,
z = Acosœg£ + Bsin œ0f.
Since we shall later need the velocity also, we differentiate z and obtain
= —ứg Äsin œg£ + œ0. cos 0g.
'These expressions are valid for all ¿, but we have special knowledge about z and 0
at £—=0. So 1ƒ we put ‡ = 0 into these equations, on the left we get #øo and 0o,
because that is what øz and 0 are at ý = 0; also, we know that the cosine oŸ zero
1s unity, and the sine of zero is zero. Therefore we get
#e=A-1+:0=A4A
0 — —œoA-0+œgB: 1 = g8.
So for this particular case we find that
A =zo, B = %o(ua.
trom these values of Á and Ö, we can get ø and A if we wish.
--- Trang 386 ---
'That is the end of our solution, but there is one physically Interesting thing
to check, and that is the conservation of energy. 5ince there are no frictional
losses, energy ought to be conserved. Let us use the formula
= acos (0g# + A);
0 = —ưgøasin (@g£ + A).
Now let us ñnd out what the kinetic energy 7' is, and what the potential energy
is. The potential energy at any moment is skz”, where # is the displacement
and & is the constant of the spring. If we substitute for +, using our expression
above, we get
U = šk#? = $kaŸ cos” (œạt + A).
Of course the potential energy is not constant; the potential never becomes
negative, naturally——there is always some energy in the spring, but the amount
of energy Ñuctuates with z. The kinetic energy, on the other hand, is sinu, and
by substituting for 0 we get
T= ÿmwŸ = š mua“ sinŸ (wọt + A).
Now the kinetic energy is zero when zø is at the maximum, because then there
1s no velocity; on the other hand, it is maximal when z is passing through zero,
because then it is moving fastest. This variation of the kinetic energy is just
the opposite of that of the potential energy. But the total energy ought to be a
constant. IÝ we note that k = mi, we see that
T+U= smưufa”[cos” (œạt + A) + sin” (œạt + A)] = 3mafdŸ.
The energy is dependent on the square of the amplitude; 1ƒ we have twice the
amplitude, we get an oscillation which has four times the energy. The øuerøge
potential energy is half the maximum and, therefore, half the total, and the
average kinetic energy is likewise half the total energy.
21-5 Forced oscillations
Next we shall discuss the ƒorced harmonic oscdllator, i.e., one in which there
is an external driving force acting. The equation then is the following:
md2+z/df? = —kaz + F(). (21.8)
--- Trang 387 ---
We would like to fnd out what happens in these cirecumstances. The external
driving force can have various kinds of functional dependence on the time; the
first one that we shall analyze is very simple—we shall suppose that the force is
oscillating:
†{) = Focos úf. (21.9)
Notice, however, that this œ is not necessarily œạ: we have œ under our control;
the forcing may be done at diferent frequencies. So we try to solve Eq. (21.8)
with the special force (21.9). What is the solution of (21.8)? One special solution,
(we shall discuss the more general cases laber) is
% = Ccosưf, (21.10)
where the constant is to be determined. In other words, we might suppose that
1f we kept pushing back and forth, the mass would follow back and forth in step
with the force. We can try it anyway. So we put (21.10) and (21.9) into (21.8),
and get
— nu cosÈ = —muf cos w‡ + Fù cos 0F. (21.11)
We have also put in k = múa, so that we will understand the equation better at
the end. Now because the cosine appears everywhere, we can divide it out, and
that shows that (21.10) is, in fact, a solution, provided we pick Πjust right. The
answer is that Πmust be
Œ = Fụ/m(wạ — œ2). (21.12)
'That Is, mm oscillates at the same frequency as the force, but with an amplitude
which depends on the frequency of the force, and also upon the frequency of the
natural motion of the oscillator. It means, frst, that if œ is very small compared
with œọ, then the displacement and the force are in the same direction. Ôn the
other hand, if we shake it back and forth very fast, then (21.12) tells us that Ở is
negative iŸ œ is above the natural frequenecy œọ oŸ the harmonic oscillator. (We
will call œọ the natural frequency of the harmonic oscillator, and œ the applied
frequency.) At very hiph requency the denominator may become very large, and
there is then not much amplitude.
Of course the solution we have found is the solution only 1ƒ things are started
Just right, for otherwise there is a part which usually dies out after a while. This
other part is called the #rønsient response to Ƒ(£), while (21.10) and (21.12) are
called the s£eadu-state response.
--- Trang 388 ---
According to our formula (21.12), a very remarkable thing should also occur:
1Ý œ is almost exactly the same as œ, then Œ should approach infinity. So If we
adjust the Írequenecy of the force to be “in time” with the natural frequenecy, then
we should get an enormous displacement. 'This is well known to anybody who
has pushed a child on a swing. It does not work very well to elose our eyes and
push at a certain speed at random. lf we happen to get the right timing, then
the swing goes very high, but if we have the wrong timing, then sometimes we