1	SmallOscillations Normal Mode analysis of Lagrange equations for systems undergoing small oscillations about equilibrium
2	Linear symmetrical triatomic molecule
3	Generalized coordinates
4	Potential energy: harmonic approximation
5	Matrix V: potential energy
6	Kinetic energy T and its corresponding matrix T
7	Equation for eigenvalues
8	Evaluation of determinant equation
9	Eigenvalue equation
10	Normal mode frequencies
11	Physical interpretation of frequencies 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	Physical interpretation … 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	Modes of oscillation
14	“Eigenvectors”
15
16	Recall structure of V, T
17	Equations for eigenvectors
18	Normalization condition
19	"Zero frequency solution:" Zero frequency solution: first and third equations then require that all three coefficients are equal
20	Second mode
21	Second mode (from normalization)
22	Third mode
23	Normal modes The general solution of equations of motion are
24	Initial conditions
25	Matrix form of initial conditions
26	Matrix form of initial conditions (continued)
27	Normal coordinates
28	Potential energy in normal coordinates
29	Kinetic energy in normal coordinates
30	Lagrange equations in normal coordinates
31	Forced undamped oscillator
32	General Solution of equation of motion
33	Particular solution for periodic force
34	General solution oscillates at two frequencies: natural and forcing frequencies 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	Forcing frequency equals natural frequency of oscillation
36	General solution when forcing frequency equals natural frequency 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	Forced damped oscillator If one adds a dissipation function to the Lagrangian (as before), then one obtains a equation for the damped, forced oscillator
38	General solution of damped, forced oscillator Choose coefficients beta and omega real; Alpha and Gamma are complex (phase differences) . Take real part of solution
39	Natural frequency of damped harmonic oscillator It is easy to show that the homogeneous solution is a solution of the damped harmonic oscillator, if
40	Particular solution of damped harmonic oscillator Try complex solution: take real part later
41	Particular solution Assume force is also of complex form:
42	Steady state response amplitude
43	AC Circuit Analogue In AC electrical circuits, the driving force is the applied voltage, V(t), and the response of interest is the current
44	Steady state current 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	Steady state response amplitude
46	Steady state rate of dissipation of energy In the steady state, energy is dissipated at the rate it is fed in by the driving force.
47	Time average 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	Intensity So all that survives in the time average is the contribution from the damping term:
49	Intensity (continued) Thus the explicit form of the intensity is
50	Maximum intensity The maximum value of the intensity occurs when Omega equals the natural frequency of the undamped, unforced oscillator
51	Square amplitude of response of damped harmonic oscillator:
52	Response of damped harmonic oscillator: Intensity
53	Chain of N coupled harmonic oscillators The potential energy V is
54	Periodic boundary conditions (atoms on a ring) If go around chain of atoms once, the N+1 atom is the same as atom 1, etc
55	Try plane wave solution (and take real part) Let x(t) be the real part of the following form:
56	PBC requires discrete values of wavenumber k From PBC
57	Is plane wave a solution of Lagrange equations? Substitute the plane wave into the equation of motion for x:
58	Dispersion relation This plane wave is a solution if and only if the frequency depends on k as
59	Actual displacements The actual solutions are the real or imaginary parts of the plane wave: Thus
60	Dispersion relation: first Brillouin zone
61	Group velocity: velocity with which the energy of wave packet travels If general displacement of atoms is made up of superposition of these waves, the energy travels at group velocity