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§22
Forced oscillations
61
SOLUTION. In this case the extension of the spring is (if
= cos
The kinetic energy is T = 1mr2o2, and the frequency is therefore w = V[F(++1)/mrl].
PROBLEM 5. Find the frequency of oscillations of the pendulum shown in Fig. 2 (§5),
whose point of support carries a mass M1 and is free to move horizontally.
SOLUTION. For < 1 the formula derived in $14, Problem 3 gives
T
Hence
PROBLEM 6. Determine the form of a curve such that the frequency of oscillations of a
particle on it under the force of gravity is independent of the amplitude.
SOLUTION. The curve satisfying the given condition is one for which the potential energy
of a particle moving on it is U = 1ks2, where s is the length of the arc from the position of
equilibrium. The kinetic energy T = 1ms2, where m is the mass of the particle, and the fre-
quency is then w = (k/m) whatever the initial value of S.
In a gravitational field U = mgy, where y is the vertical co-ordinate. Hence we have
1ks2 = mgy or y = w2s2/2g. But ds2 = dx2+dy², whence
dy = SV[(g/2w2y)-1] dy.
The integration is conveniently effected by means of the substitution y = g(1-cos E)/4ww,
which yields x = g(s+sin 5)/4w2. These two equations give, in parametric form, the equation
of the required curve, which is a cycloid.
$22. Forced oscillations
Let us now consider oscillations of a system on which a variable external
force acts. These are called forced oscillations, whereas those discussed in
§21 are free oscillations. Since the oscillations are again supposed small, it
is implied that the external field is weak, because otherwise it could cause the
displacement x to take too large values.
The system now has, besides the potential energy 1kx2, the additional
potential energy Ue(x, t) resulting from the external field. Expanding this
additional term as a series of powers of the small quantity x, we have
Ue(x, t) 12 x[dUe/dx]x_0. The first term is a function of time only,
and may therefore be omitted from the Lagrangian, as being the total time
derivative of another function of time. In the second term - [dUe/dx]x_0 is
the external "force" acting on the system in the equilibrium position, and
is a given function of time, which we denote by F(t). Thus the potential
energy involves a further term -xF(t), and the Lagrangian of the system
is
L
=
(22.1)
The corresponding equation of motion is m+kx = F(t) or
(22.2)
where we have again introduced the frequency w of the free oscillations.
The general solution of this inhomogeneous linear differential equation
with constant coefficients is x = xo+x1, where xo is the general solution of
62
Small Oscillations
§22
the corresponding homogeneous equation and X1 is a particular integral of
the inhomogeneous equation. In the present case xo represents the free
oscillations discussed in $21.
Let us consider a case of especial interest, where the external force is itself
a simple periodic function of time, of some frequency y:
F(t) = f cos(yt+)).
(22.3)
We seek a particular integral of equation (22.2) in the form X1 = b cos(yt+B),
with the same periodic factor. Substitution in that equation gives
b = f/m(w2-r2); adding the solution of the homogeneous equation, we
obtain the general integral in the form
(22.4)
The arbitrary constants a and a are found from the initial conditions.
Thus a system under the action of a periodic force executes a motion which
is a combination of two oscillations, one with the intrinsic frequency w of
the system and one with the frequency y of the force.
The solution (22.4) is not valid when resonance occurs, i.e. when the fre-
quency y of the external force is equal to the intrinsic frequency w of the
system. To find the general solution of the equation of motion in this case,
we rewrite (22.4) as
x
=
a
where a now has a different value. Asy->w, the second term is indetermin-
ate, of the form 0/0. Resolving the indeterminacy by L'Hospital's rule, we
have
x = acos(wt+a)+(f/2mw)tsin(wt+B). =
(22.5)
Thus the amplitude of oscillations in resonance increases linearly with the
time (until the oscillations are no longer small and the whole theory given
above becomes invalid).
Let us also ascertain the nature of small oscillations near resonance, when
y w+E with E a small quantity. We put the general solution in the com-
plex form
= A exp(iwt) exp[i(w+t)) = [A+B exp(iet)]exp(ist)
(22.6)
Since the quantity A+B exp(iet) varies only slightly over the period 2n/w
of the factor exp(iwt), the motion near resonance may be regarded as small
oscillations of variable amplitude.t Denoting this amplitude by C, we have
= A B exp(iet)|. Writing A and B in the form a exp(ix) and b exp(iB)
respectively, we obtain
(22.7)
t The "constant" term in the phase of the oscillation also varies.
§22
Forced oscillations
63
Thus the amplitude varies periodically with frequency E between the limits
|a-b a+b. This phenomenon is called beats.
The equation of motion (22.2) can be integrated in a general form for an
arbitrary external force F(t). This is easily done by rewriting the equation
as
or
=
(22.8)
where
s=xtiwx
(22.9)
is a complex quantity. Equation (22.8) is of the first order. Its solution when
the right-hand side is replaced by zero is $ = A exp(iwt) with constant A.
As before, we seek a solution of the inhomogeneous equation in the form
$ = A(t) exp(iwt), obtaining for the function A(t) the equation À(t)
= F(t) exp(-iwt)/m. Integration gives the solution of (22.9):
& = -
(22.10)
where the constant of integration so is the value of $ at the instant t = 0.
This is the required general solution; the function x(t) is given by the imagin-
ary part of (22.10), divided by w.t
The energy of a system executing forced oscillations is naturally not con-
served, since the system gains energy from the source of the external field.
Let us determine the total energy transmitted to the system during all time,
assuming its initial energy to be zero. According to formula (22.10), with
the lower limit of integration - 00 instead of zero and with ( - 00) = 0,
we have for t
00
exp(-iwt)dt|
The energy of the system is
E = 1m(x2+w2x2)= = 1ME2.
(22.11)
Substituting we obtain the energy transferred
(22.12)
t The force F(t) must, of course, be written in real form.
64
Small Oscillations
§22
it is determined by the squared modulus of the Fourier component of the
force F(t) whose frequency is the intrinsic frequency of the system.
In particular, if the external force acts only during a time short in com-
parison with 1/w, we can put exp(-iwt) Ill 1. Then
This result is obvious: it expresses the fact that a force of short duration
gives the system a momentum I F dt without bringing about a perceptible
displacement.
PROBLEMS
PROBLEM 1. Determine the forced oscillations of a system under a force F(t) of the follow-
ing forms, if at time t = 0 the system is at rest in equilibrium (x = x = 0): (a) F = Fo,
a constant, (b) F = at, (c) F = Fo exp(-at), (d) F = Fo exp(-at) cos Bt.
SOLUTION. (a) x = (Fo/mw2)(1-cos wt). The action of the constant force results in a dis-
placement of the position of equilibrium about which the oscillations take place.
(b) x = (a/mw3)(wt-sin wt).
(c) x = - cos wt +(a/w) sin wt].
(d) x = wt + sin wt +
+exp(-at)[(wpta2-B2) cos Bt-2aB sin
This last case is conveniently treated by writing the force in the complex form
F=Foexp(-ati)t].
PROBLEM 2. Determine the final amplitude for the oscillations of a system under a force
which is zero for t<0, Fot/T for 0 <t<<, and Fo for t > T (Fig. 24), if up to time
t = 0 the system is at rest in equilibrium.
F
Fo
,
T
FIG. 24
SOLUTION. During the interval 0<+<T the oscillations are determined by the initial
condition as x = (Fo/mTw3)(wt-sin wt). For t > T we seek a solution in the form
=c1w(t-T)+c2 sin w(t - T)+Fo/mw2
The continuity of x and x at t = T gives C1 = -(Fo/mTw3) sin wT, C2 = (Fo/mTw3 X
X (1 - cos wT). The amplitude is a = = (2Fo/mTw3) sin twT. This is the smaller,
the more slowly the force Fo is applied (i.e. the greater T).
PROBLEM 3. The same as Problem 2, but for a constant force Fo which acts for a finite
time T (Fig. 25).
§23
Oscillations of systems with more than one degree of freedom
65
SOLUTION. As in Problem 2, or more simply by using formula (22.10). For t > T we have
free oscillations about x =0, and
dt
FO
f
T
FIG. 25
The squared modulus of & gives the amplitude from the relation = The result is
a = (2Fo/mw2) sin twT.
PROBLEM 4. The same as Problem 2, but for a force Fot/T which acts between t = 0 and
t = T (Fig. 26).
F
FO
,
T
FIG. 26
SOLUTION. By the same method we obtain
a = (Fo/Tmw3)/[wT2-2wT sin wT+2(1-cos - wT)].
PROBLEM 5. The same as Problem 2, but for a force Fo sin wt which acts between t = 0
and t = T = 2n/w (Fig. 27).
F
T
FIG. 27
SOLUTION. Substituting in (22.10) F(t) = Fo sin wt = Fo[exp(iwt)-exp(-iwt)]/2i and
integrating from 0 to T, we obtain a = Fon/mw2.
$23. Oscillations of systems with more than one degree of freedom
The theory of free oscillations of systems with S degrees of freedom is
analogous to that given in §21 for the case S = 1.
3*
66
Small Oscillations
§23
Let the potential energy of the system U as a function of the generalised
co-ordinates qi (i = 1, 2, ..., s) have a minimum for qi = qio. Putting
Xi=qi-qio
(23.1)
for the small displacements from equilibrium and expanding U as a function
of the xi as far as the quadratic terms, we obtain the potential energy as a
positive definite quadratic form
(23.2)
where we again take the minimum value of the potential energy as zero.
Since the coefficients kik and kki in (23.2) multiply the same quantity XiXK,
it is clear that they may always be considered equal: kik = kki.
In the kinetic energy, which has the general form () (see (5.5)),
we put qi = qio in the coefficients aik and, denoting aik(90) by Mik, obtain
the kinetic energy as a positive definite quadratic form
Emission
(23.3)
The coefficients Mik also may always be regarded as symmetrical: Mik=Mki.
Thus the Lagrangian of a system executing small free oscillations is
(23.4)
i,k
Let us now derive the equations of motion. To determine the derivatives
involved, we write the total differential of the Lagrangian:
- kikxi dxk - kikxxdxi).
i,k
Since the value of the sum is obviously independent of the naming of the
suffixes, we can interchange i and k in the first and third terms in the paren-
theses. Using the symmetry of Mik and kik, we have
dL =
Hence
k
Lagrange's equations are therefore
(i=1,2,...,s);
(23.5)
they form a set of S linear homogeneous differential equations with constant
coefficients.
As usual, we seek the S unknown functions xx(t) in the form
xx = Ak explicut),
(23.6)
where Ak are some constants to be determined. Substituting (23.6) in the
§23
Oscillations of systems with more than one degree of freedom
67
equations (23.5) and cancelling exp(iwt), we obtain a set of linear homo-
geneous algebraic equations to be satisfied by the Ak:
(23.7)
If this system has non-zero solutions, the determinant of the coefficients
must vanish:
(23.8)
This is the characteristic equation and is of degree S in w2. In general, it has
S different real positive roots W&2 (a = 1,2,...,s); in particular cases, some of
these roots may coincide. The quantities Wa thus determined are the charac-
teristic frequencies or eigenfrequencies of the system.
It is evident from physical arguments that the roots of equation (23.8) are
real and positive. For the existence of an imaginary part of w would mean
the presence, in the time dependence of the co-ordinates XK (23.6), and SO
of the velocities XK, of an exponentially decreasing or increasing factor. Such
a factor is inadmissible, since it would lead to a time variation of the total
energy E = U+: T of the system, which would therefore not be conserved.
The same result may also be derived mathematically. Multiplying equation
(23.7) by Ai* and summing over i, we have = 0,
whence w2 = . The quadratic forms in the numerator
and denominator of this expression are real, since the coefficients kik and
Mik are real and symmetrical: (kA*Ak)* = kikAAk* = k
= kikAkAi*. They are also positive, and therefore w2 is positive.t
The frequencies Wa having been found, we substitute each of them in
equations (23.7) and find the corresponding coefficients Ak. If all the roots
Wa of the characteristic equation are different, the coefficients Ak are pro-
portional to the minors of the determinant (23.8) with w = Wa. Let these
minors be . A particular solution of the differential equations (23.5) is
therefore X1c = Ca exp(iwat), where Ca is an arbitrary complex constant.
The general solution is the sum of S particular solutions. Taking the real
part, we write
III
(23.9)
where
(23.10)
Thus the time variation of each co-ordinate of the system is a super-
position of S simple periodic oscillations O1, O2, ..., Os with arbitrary ampli-
tudes and phases but definite frequencies.
t The fact that a quadratic form with the coefficients kik is positive definite is seen from
their definition (23.2) for real values of the variables. If the complex quantities Ak are written
explicitly as ak +ibk, we have, again using the symmetry of kik, kikAi* Ak = kik(ai-ibi)
X
= kikaiak kikbibk, which is the sum of two positive definite forms.
68
Small Oscillations
§23
The question naturally arises whether the generalised co-ordinates can be
chosen in such a way that each of them executes only one simple oscillation.
The form of the general integral (23.9) points to the answer. For, regarding
the S equations (23.9) as a set of equations for S unknowns Oa, as we can
express O1, O2, ..., Os in terms of the co-ordinates X1, X2, ..., Xs. The
quantities Oa may therefore be regarded as new generalised co-ordinates,
called normal co-ordinates, and they execute simple periodic oscillations,
called normal oscillations of the system.
The normal co-ordinates Oa are seen from their definition to satisfy the
equations
Oatwaia = 0.
(23.11)
This means that in normal co-ordinates the equations of motion become S
independent equations. The acceleration in each normal co-ordinate depends
only on the value of that co-ordinate, and its time dependence is entirely
determined by the initial values of the co-ordinate and of the corresponding
velocity. In other words, the normal oscillations of the system are completely
independent.
It is evident that the Lagrangian expressed in terms of normal co-ordinates
is a sum of expressions each of which corresponds to oscillation in one dimen-
sion with one of the frequencies was i.e. it is of the form
(23.12)
where the Ma are positive constants. Mathematically, this means that the
transformation (23.9) simultaneously puts both quadratic forms-the kinetic
energy (23.3) and the potential energy (23.2)-in diagonal form.
The normal co-ordinates are usually chosen so as to make the coefficients
of the squared velocities in the Lagrangian equal to one-half. This can be
achieved by simply defining new normal co-ordinates Qx by
Qa = VMaOa.
(23.13)
Then
The above discussion needs little alteration when some roots of the charac-
teristic equation coincide. The general form (23.9), (23.10) of the integral of
the equations of motion remains unchanged, with the same number S of
terms, and the only difference is that the coefficients corresponding to
multiple roots are not the minors of the determinant, which in this case
vanish.
t The impossibility of terms in the general integral which contain powers of the time as
well as the exponential factors is seen from the same argument as that which shows that the
frequencies are real: such terms would violate the law of conservation of energy
§23
Oscillations of systems with more than one degree of freedom
69
Each multiple (or, as we say, degenerate) frequency corresponds to a number
of normal co-ordinates equal to its multiplicity, but the choice of these co-
ordinates is not unique. The normal co-ordinates with equal Wa enter the
kinetic and potential energies as sums Q and Qa2 which are transformed
in the same way, and they can be linearly transformed in any manner which
does not alter these sums of squares.
The normal co-ordinates are very easily found for three-dimensional oscil-
lations of a single particle in a constant external field. Taking the origin of
Cartesian co-ordinates at the point where the potential energy U(x,y,2) is
a minimum, we obtain this energy as a quadratic form in the variables x, y, Z,
and the kinetic energy T = m(x2+yj++2) (where m is the mass of the
particle) does not depend on the orientation of the co-ordinate axes. We
therefore have only to reduce the potential energy to diagonal form by an
appropriate choice of axes. Then
L =
(23.14)
and the normal oscillations take place in the x,y and 2 directions with fre-
quencies = (k1/m), w2=1/(k2/m), w3=1/(k3/m). In the particular
case of a central field (k1 =k2=kg III three frequencies
are equal (see Problem 3).
The use of normal co-ordinates makes possible the reduction of a problem
of forced oscillations of a system with more than one degree of freedom to a
series of problems of forced oscillation in one dimension. The Lagrangian of
the system, including the variable external forces, is
(23.15)
where L is the Lagrangian for free oscillations. Replacing the co-ordinates
X1c by normal co-ordinates, we have
(23.16)
where we have put
The corresponding equations of motion
(23.17)
each involve only one unknown function Qa(t).
PROBLEMS
PROBLEM 1. Determine the oscillations of a system with two degrees of freedom whose
Lagrangian is L = (two identical one-dimensional systems of
eigenfrequency wo coupled by an interaction - axy).
70
Small Oscillations
§24
SOLUTION. The equations of motion are x+ wo2x = ay, j + wo2y = ax. The substitution
(23.6) gives
Ax(wo2-w2) = aAy,
(1)
The characteristic equation is (wo2-w2)2= a2, whence w12 = wo2-a, w22 = wo2-+x. For
w = W1, the equations (1) give Ax = Ay, and for w = w2, Ax = -Ay. Hence x =
(Q1+Q2)/V2, y = (Q1-Q2)/V2, the coefficients 1/V2 resulting from the normalisation
of the normal co-ordinates as in equation (23.13).
For a < wo2 (weak coupling) we have W1 all wo-1x, W2 ill wotla. The variation of x
and y is in this case a superposition of two oscillations with almost equal frequencies, i.e.
beats of frequency W2-W1 = a (see $22). The amplitude of y is a minimum when that of x
is a maximum, and vice versa.
PROBLEM 2. Determine the small oscillations of a coplanar double pendulum (Fig. 1, $5).
SOLUTION. For small oscillations (01 < 1, 02 < 1), the Lagrangian derived in §5, Problem
1, becomes
L =
The equations of motion are
= 0, lio +1202+802
=
0.
Substitution of (23.6) gives
41(m1+m2)(g-h1w2)-A2w2m2l2 0, = 0.
The roots of the characteristic equation are
((ma3
As m1
8 the frequencies tend to the values (g/l1) and /(g/l2), corresponding to indepen-
dent oscillations of the two pendulums.
PROBLEM 3. Find the path of a particle in a central field U = 1kr2 (called a space oscillator).
SOLUTION. As in any central field, the path lies in a plane, which we take as the xy-plane.
The variation of each co-ordinate x,y is a simple oscillation with the same frequency
= v(k/m): x = a cos(wt+a), y=b cos(wt+), or x = a cos , y = b cos(+8)
= b cos 8 cos -b sin 8 sin , where = wt +a, 8 = B-a. Solving for cos o and sin o and
equating the sum of their squares to unity, we find the equation of the path:
This is an ellipse with its centre at the origin.t When 8 = 0 or IT, the path degenerates to a
segment of a straight line.
$24. Vibrations of molecules
If we have a system of interacting particles not in an external field, not all
of its degrees of freedom relate to oscillations. A typical example is that of
molecules. Besides motions in which the atoms oscillate about their positions
of equilibrium in the molecule, the whole molecule can execute translational
and rotational motions.
Three degrees of freedom correspond to translational motion, and in general
the same number to rotation, so that, of the 3n degrees of freedom of a mole-
cule containing n atoms, 3n-6 - correspond to vibration. An exception is formed
t The fact that the path in a field with potential energy U = 1kr2 is a closed curve has
already been mentioned in $14.