181 lines
8.2 KiB
Markdown
181 lines
8.2 KiB
Markdown
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title: 24-vibrations-of-molecules
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---
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SOLUTION. The equations of motion are x+ wo2x = ay, j + wo2y = ax. The substitution
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(23.6) gives
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Ax(wo2-w2) = aAy,
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(1)
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The characteristic equation is (wo2-w2)2= a2, whence w12 = wo2-a, w22 = wo2-+x. For
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w = W1, the equations (1) give Ax = Ay, and for w = w2, Ax = -Ay. Hence x =
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(Q1+Q2)/V2, y = (Q1-Q2)/V2, the coefficients 1/V2 resulting from the normalisation
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of the normal co-ordinates as in equation (23.13).
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For a < wo2 (weak coupling) we have W1 all wo-1x, W2 ill wotla. The variation of x
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and y is in this case a superposition of two oscillations with almost equal frequencies, i.e.
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beats of frequency W2-W1 = a (see $22). The amplitude of y is a minimum when that of x
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is a maximum, and vice versa.
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PROBLEM 2. Determine the small oscillations of a coplanar double pendulum (Fig. 1, $5).
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SOLUTION. For small oscillations (01 < 1, 02 < 1), the Lagrangian derived in §5, Problem
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1, becomes
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L =
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The equations of motion are
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= 0, lio +1202+802
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=
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0.
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Substitution of (23.6) gives
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41(m1+m2)(g-h1w2)-A2w2m2l2 0, = 0.
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The roots of the characteristic equation are
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((ma3
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As m1
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8 the frequencies tend to the values (g/l1) and /(g/l2), corresponding to indepen-
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dent oscillations of the two pendulums.
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PROBLEM 3. Find the path of a particle in a central field U = 1kr2 (called a space oscillator).
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SOLUTION. As in any central field, the path lies in a plane, which we take as the xy-plane.
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The variation of each co-ordinate x,y is a simple oscillation with the same frequency
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= v(k/m): x = a cos(wt+a), y=b cos(wt+), or x = a cos , y = b cos(+8)
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= b cos 8 cos -b sin 8 sin , where = wt +a, 8 = B-a. Solving for cos o and sin o and
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equating the sum of their squares to unity, we find the equation of the path:
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This is an ellipse with its centre at the origin.t When 8 = 0 or IT, the path degenerates to a
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segment of a straight line.
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$24. Vibrations of molecules
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If we have a system of interacting particles not in an external field, not all
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of its degrees of freedom relate to oscillations. A typical example is that of
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molecules. Besides motions in which the atoms oscillate about their positions
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of equilibrium in the molecule, the whole molecule can execute translational
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and rotational motions.
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Three degrees of freedom correspond to translational motion, and in general
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the same number to rotation, so that, of the 3n degrees of freedom of a mole-
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cule containing n atoms, 3n-6 - correspond to vibration. An exception is formed
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t The fact that the path in a field with potential energy U = 1kr2 is a closed curve has
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already been mentioned in $14.
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§24
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Vibrations of molecules
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71
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by molecules in which the atoms are collinear, for which there are only two
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rotational degrees of freedom (since rotation about the line of atoms is of no
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significance), and therefore 3n-5 vibrational degrees of freedom.
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In solving a mechanical problem of molecular oscillations, it is convenient
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to eliminate immediately the translational and rotational degrees of freedom.
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The former can be removed by equating to zero the total momentum of the
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molecule. Since this condition implies that the centre of mass of the molecule
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is at rest, it can be expressed by saying that the three co-ordinates of the
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centre of mass are constant. Putting ra = rao+Ua, where ra0 is the radius
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vector of the equilibrium position of the ath atom, and Ua its deviation from
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this position, we have the condition = constant = or
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= 0.
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(24.1)
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To eliminate the rotation of the molecule, its total angular momentum
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must be equated to zero. Since the angular momentum is not the total time
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derivative of a function of the co-ordinates, the condition that it is zero can-
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not in general be expressed by saying that some such function is zero. For
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small oscillations, however, this can in fact be done. Putting again
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ra = rao+ua and neglecting small quantities of the second order in the
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displacements Ua, we can write the angular momentum of the molecule as
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= .
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The condition for this to be zero is therefore, in the same approximation,
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0,
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(24.2)
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in which the origin may be chosen arbitrarily.
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The normal vibrations of the molecule may be classified according to the
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corresponding motion of the atoms on the basis of a consideration of the sym-
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metry of the equilibrium positions of the atoms in the molecule. There is
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a general method of doing so, based on the use of group theory, which we
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discuss elsewhere. Here we shall consider only some elementary examples.
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If all n atoms in a molecule lie in one plane, we can distinguish normal
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vibrations in which the atoms remain in that plane from those where they
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do not. The number of each kind is readily determined. Since, for motion
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in a plane, there are 2n degrees of freedom, of which two are translational
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and one rotational, the number of normal vibrations which leave the atoms
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in the plane is 2n-3. The remaining (3n-6)-(2n-3) = n-3 vibrational
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degrees of freedom correspond to vibrations in which the atoms move out
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of the plane.
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For a linear molecule we can distinguish longitudinal vibrations, which
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maintain the linear form, from vibrations which bring the atoms out of line.
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Since a motion of n particles in a line corresponds to n degrees of freedom,
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of which one is translational, the number of vibrations which leave the atoms
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t See Quantum Mechanics, $100, Pergamon Press, Oxford 1965.
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72
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Small Oscillations
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§24
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in line is n - 1. Since the total number of vibrational degrees of freedom of a
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linear molecule is 3n - 5, there are 2n-4 which bring the atoms out of line.
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These 2n-4 vibrations, however, correspond to only n-2 different fre-
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quencies, since each such vibration can occur in two mutually perpendicular
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planes through the axis of the molecule. It is evident from symmetry that
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each such pair of normal vibrations have equal frequencies.
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PROBLEMS
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PROBLEM 1. Determine the frequencies of vibrations of a symmetrical linear triatomic
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molecule ABA (Fig. 28). It is assumed that the potential energy of the molecule depends
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only on the distances AB and BA and the angle ABA.
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3
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B
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A
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(o)
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(b)
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(c)
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FIG. 28
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SOLUTION. The longitudinal displacements X1, X2, X3 of the atoms are related, according
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to (24.1), by MA(X1+x3) +mBX2 = 0. Using this, we eliminate X2 from the Lagrangian of the
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longitudinal motion
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L =
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and use new co-ordinatesQa=x1tx,Qx1-x3.The result
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where u = 2mA+mB is the mass of the molecule. Hence we see that Qa and Qs are normal
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co-ordinates (not yet normalised). The co-ordinate Qa corresponds to a vibration anti-
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symmetrical about the centre of the molecule (x1 = x3; Fig. 28a), with frequency
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wa = (k1u/mAmB). The co-ordinate Q8 corresponds to a symmetrical vibration (x1 = -x3;
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Fig. 28b), with frequency
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The transverse displacements y1, y2, y3 of the atoms are, according to (24.1) and (24.2),
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related by mA(y1+y2) +mBy2 = 0, y1 = y3 (a symmetrical bending of the molecule; Fig. 28c).
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The potential energy of this vibration can be written as 1/22/282, where 8 is the deviation of the
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angle ABA from the value IT, given in terms of the displacements by 8 = [(y1-yy)+(ys-y2)]/L.
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Expressing y1,y2, y3 in terms of 8, we obtain the Lagrangian of the transverse motion:
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L =
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whence the frequency is = V(2k2u/mAmB).
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t Calculations of the vibrations of more complex molecules are given by M. V. VOL'KENSH-
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TEIN, M. A. EL'YASHEVICH and B. I. STEPANOV, Molecular Vibrations (Kolebaniya molekul),
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Moscow 1949; G. HERZBERG, Molecular Spectra and Molecular Structure: Infra-red and
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Raman Spectra of Polyatomic Molecules, Van Nostrand, New York 1945,
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§24
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Vibrations of molecules
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73
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PROBLEM 2. The same as Problem 1, but for a triangular molecule ABA (Fig. 29).
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y
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A
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A
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3
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2a
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I
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2
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B
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(a)
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(b)
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(c)
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FIG. 29
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SOLUTION. By (24.1) and (24.2) the x and y components of the displacements u of the
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atoms are related by
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0,
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0,
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(y1-y3) sin x-(x1+x3) cos a = 0.
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The changes 8l1 and Sl2 in the distances AB and BA are obtained by taking the components
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along these lines of the vectors U1-U2 and U3-U2:
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8l1 = (x1-x2) sin cos a,
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8l2 = -(x3-x2) sin at(y3-y2) cos a.
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The change in the angle ABA is obtained by taking the components of those vectors per-
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pendicular to AB and BA:
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sin sin
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a].
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The Lagrangian of the molecule is
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L
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We use the new co-ordinates Qa = 1+x3, q81 = X1-x3, q82 = y1+y3. The components
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of the vectors u are given in terms of these co-ordinates by X1 = 1(Qa+q x3 = 1(Qa-981),
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X2
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= -MAQa/MB, = -MAQ82/MB. The
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Lagrangian becomes
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L
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=
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qksici +
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sin
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a
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cos
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a.
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74
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Small Oscillations
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