1

 Normal Mode analysis of Lagrange equations for systems undergoing small
oscillations about equilibrium

2


3


4


5


6


7


8


9


10


11

 The zero frequency corresponds to a rigid translation of the molecule
along its axis. There is no
change in potential energy and the restoring force against this
translation is zero. There are
actually only two degrees of freedom; one of the three we started with
is a rigid body degree of freedom

12

 The second solution is the solution for a mass m undergoing oscillations
with a spring of force constant k.
Therefore we should find later that only the end atoms
participate in this mode of oscillation.
The center atom remains fixed.
 The mass M only oscillates in the third normal mode solution.

13


14


15


16


17


18


19

 Zero frequency solution: first
and third equations then require that all three coefficients are equal

20


21


22


23

 The general solution of equations of motion are

24


25


26


27


28


29


30


31


32


33


34

 The constants A and delta in the
natural oscillation depend on the initial conditions. The oscillation
associated with the periodic force depends on the driving frequency and
amplitude of the force.

35


36

 In this case the inhomogeneous term eventually overwhelms the
homogeneous term. This is
unphysical; every real system has some damping. Without damping the system has no way
to dissipate the energy given it by F(t)

37

 If one adds a dissipation function to the
 Lagrangian (as before), then one obtains a equation for the damped,
forced oscillator

38

 Choose coefficients beta and omega real;
 Alpha and Gamma are complex (phase differences) . Take real part of solution

39

 It is easy to show that the homogeneous solution is a solution of the
damped harmonic oscillator, if

40

 Try complex solution: take real part later

41

 Assume force is also of complex form:

42


43

 In AC electrical circuits, the driving force is the applied voltage,
V(t), and the response of interest is the current

44

 The homogeneous solution decays to zero due to the damping factor in the
exponential. Thus the steady
state value of the “current” comes from the time derivative of the
particular solution. So

45


46

 In the steady state, energy is dissipated at the rate it is fed in by
the driving force.

47

 There are two terms in the rate of dissipation that both average to
zero, as they are out of phase with each other (and hence are as often
positive as negative in one period).
These are

48

 So all that survives in the time average is
 the contribution from the damping term:

49

 Thus the explicit form of the intensity is

50

 The maximum value of the intensity occurs when Omega equals the natural
frequency of the undamped, unforced oscillator

51


52


53

 The potential energy V is

54

 If go around chain of atoms once, the
 N+1 atom is the same as atom 1, etc

55

 Let x(t) be the real part of the following form:

56


57

 Substitute the plane wave into the equation of motion for x:

58

 This plane wave is a solution if and only if the frequency depends on k
as

59

 The actual solutions are the real or imaginary parts of the plane
wave: Thus

60


61

 If general displacement of atoms is made up of superposition of these
waves, the energy travels at group velocity
