Vibrational Motion Chapter 5 P. J. Grandinetti Chem. 4300 Sep. 13, 2017 P. J. Grandinetti (Chem. 4300) Vibrational Motion Sep. 13, 2017 1 / 20
Simple Harmonic Oscillator Simplest model for harmonic oscillator mass attached to one end of spring while other end is held fixed m -x 0 +x Mass at x = 0 corresponds to equilibrium position x is displacement from equilibrium. Assume no friction and that spring itself has no mass. P. J. Grandinetti (Chem. 4300) Vibrational Motion Sep. 13, 2017 2 / 20
Simple Harmonic Oscillator Pull the mass and let go. m -x 0 +x What happens? An oscillation. The time for one complete cycle is T. How do we get and solve the equation of motion for this system? P. J. Grandinetti (Chem. 4300) Vibrational Motion Sep. 13, 2017 3 / 20
Simple Harmonic Oscillator Hooke s law For small displacements from equilibrium restoring force is F = κ f x κ f is the force constant for the spring. Using Newton s 2nd law F = ma = κ f x we obtain differential equation of motion Propose the solution mẍ(t) + κ f x(t) = 0 x(t) = A cos(ωt + φ) Substituting into the differential equation gives ( κf mω 2) A cos(ωt + φ) = 0 P. J. Grandinetti (Chem. 4300) Vibrational Motion Sep. 13, 2017 4 / 20
Simple Harmonic Oscillator ( κf mω 2) A cos(ωt + φ) = 0 To make true for all values of t we set ω = ω 0 = κ f m. ω 0 is called the natural oscillation frequency Velocity of the mass is v(t) = ẋ(t) = ωa sin(ω 0 t + φ) Make equation satisfy initial conditions of x(t = 0) = A and ẋ(t = 0) = 0 by setting φ = 0 and A = x(0) to get final solution to equation of motion x(t) = x(0) cos ω 0 t P. J. Grandinetti (Chem. 4300) Vibrational Motion Sep. 13, 2017 5 / 20
Energy of simple harmonic oscillator P. J. Grandinetti (Chem. 4300) Vibrational Motion Sep. 13, 2017 6 / 20
Energy of simple harmonic oscillator Total energy of simple harmonic oscillator is sum of the kinetic and potential energy of mass and spring. Kinetic energy is given by where p is mass momentum, p = mv. K = 1 2 mv2, or K = p2 2m Potential energy is the energy stored in spring and is equal to work done in extending and compressing spring, x V(x) = F(x )dx x = κ f x dx = 1 2 κ fx 2 0 Expression above is work associated with extending spring. For work in compressing the spring just change integral limits to x to 0 (you get same result). P. J. Grandinetti (Chem. 4300) Vibrational Motion Sep. 13, 2017 7 / 20 0
Energy of simple harmonic oscillator Potential energy is given by V(x) = 1 2 κ fx 2 V(x) 0 x P. J. Grandinetti (Chem. 4300) Vibrational Motion Sep. 13, 2017 8 / 20
Energy of simple harmonic oscillator Although both K and V are time dependent during harmonic motion the total energy, E = K + V, for a simple harmonic oscillator remains time independent. E = 1 2 mv2 (t) + 1 2 κ fx 2 (t) or E = p2 (t) 2m + 1 2 κ fx 2 (t) Total Energy Potential Energy Energy Kinetic Energy P. J. Grandinetti (Chem. 4300) Vibrational Motion Sep. 13, 2017 9 / 20
Energy of simple harmonic oscillator Substitute equation of motion into energy expression E = 1 2 κ fx 2 (0) = 1 2 mω2 0 x2 (0) Solve for x(0) in terms of energy x(0) = 1 ω 0 2E m and rewrite oscillation as 2E x(t) = x(0) cos ω 0 t = ω 2 0 m cos ω 0 t P. J. Grandinetti (Chem. 4300) Vibrational Motion Sep. 13, 2017 10 / 20
Position probability distribution for harmonic oscillator P. J. Grandinetti (Chem. 4300) Vibrational Motion Sep. 13, 2017 11 / 20
Position probability distribution for harmonic oscillator Scale x(t) by initial amplitude x(0), to obtain a function, y(t), that oscillates between y = 1 and y = +1 y(t) = x(t) x(0) = cos ω 0 t Calculate normalized probability density, p(y), for finding the mass at any scaled position between y = ±1. Probability of finding mass in interval dy at given y is proportional to the time spent in dy interval, p(y) dy = b dt = b dt dy dt = b dy dy dy = b dy ẏ ẏ is the speed at a given y and b is proportionality constant. Derivative of y(t) is ẏ(t) = ω 0 sin ω 0 t so p(y) = b ẏ(t) = b ω 0 sin ω 0 t = b ω 0 (1 cos 2 ω 0 t) 1 2 = b ω 0 (1 y 2 ) 1 2 P. J. Grandinetti (Chem. 4300) Vibrational Motion Sep. 13, 2017 12 / 20
Position probability distribution for harmonic oscillator p(y) = b ω 0 (1 y 2 ) 1 2 After normalizing probability distribution becomes 10 p(y) = 1 π(1 y 2 ) 1 2 8 6 4 2 0-1.0-0.5 0.0 0.5 1.0 Mass spends majority of time at maximum excursions, that is, turning points where velocity is slowest and changes sign. P. J. Grandinetti (Chem. 4300) Vibrational Motion Sep. 13, 2017 13 / 20
Diatomic molecule vibration as Harmonic oscillator P. J. Grandinetti (Chem. 4300) Vibrational Motion Sep. 13, 2017 14 / 20
Diatomic molecule vibration as Harmonic oscillator Dashed line is harmonic oscillator potential. Solid line is Morse potential. V(r) reaches minimum at r e where restoring force is zero. V(r) causes repulsive force at r < r e and attractive force at r > r e. V(r) increases steeply at r < r e but levels out to constant at r > r e. At r there is no attractive force as V(r) has a slope of zero. P. J. Grandinetti (Chem. 4300) Vibrational Motion Sep. 13, 2017 15 / 20
Diatomic molecule vibration as Harmonic oscillator For small displacements from equilibrium bond length the interatomic potential can be approximated as a harmonic oscillator potential. Taylor series expansion of V(r) about equilibrium bond length, r = r e, gives dv(r V(r) V(r e )+ 0 e ) (r r dr e )+ 1 d 2 V(r e ) (r r 2! dr 2 e ) 2 + 1 3! d 3 V(r e ) dr 3 (r r e ) 3 + V(r e ) is the potential energy at equilibrium bond length. 1st-order term is zero since no restoring force, F = dv(r e ) dr, at r = r e Truncate expansion at the 3rd-order term and define 2 constants and get potential expansion κ f = d2 V(r e ) and γ dr 2 f = d3 V(r e ) dr 3 V(r) V(r e ) 1 2 κ f(r r e ) 2 + 1 6 γ f(r r e ) 3 + P. J. Grandinetti (Chem. 4300) Vibrational Motion Sep. 13, 2017 16 / 20
Diatomic molecule vibration as Harmonic oscillator V(r) V(r e ) 1 2 κ f(r r e ) 2 + 1 6 γ f(r r e ) 3 + For small displacements we drop 3rd-order term and see that potential energy has form of simple harmonic oscillator. For slightly larger displacements we could re-add 3rd-order term to potential energy to account for anharmonicity in bond vibration. P. J. Grandinetti (Chem. 4300) Vibrational Motion Sep. 13, 2017 17 / 20
Diatomic molecule vibration equations of motion Make harmonic oscillator approximation taking force on m 1 and m 2 as F 1 = κ f (r 1 r 2 + r e ) and F 2 = κ f (r 2 r 1 r e ) Equations of motion are 2 coupled differential equations: d 2 r m 1 1 dt = κ 2 f (r 2 r 1 r e ) and m d 2 r 2 2 dt = κ 2 f (r 2 r 1 r e ) Move into center of mass frame: M = m 1 + m 2 and R = 1 M (m 1r 1 + m 1 r 2 ) Obtain 2 uncoupled differential equations, M d2 R dt = 0, and μ d2 Δr = κ 2 dt 2 f Δr Δr = r 2 r 1 r e and μ is the reduced mass given by 1 μ = 1 m 1 + 1 m 2 P. J. Grandinetti (Chem. 4300) Vibrational Motion Sep. 13, 2017 18 / 20
Diatomic molecule vibration equations of motion Differential equation of motion describing the vibration μ dδr2 (t) dt 2 + κ f Δr(t) = 0 Same differential equation of motion as simple harmonic oscillator. Solutions takes the same form, Δr(t) = Δr(0) cos ω 0 t where ω 0 = κ f μ P. J. Grandinetti (Chem. 4300) Vibrational Motion Sep. 13, 2017 19 / 20
Vibration of Polyatomic Molecules Link: Normal modes of vibration P. J. Grandinetti (Chem. 4300) Vibrational Motion Sep. 13, 2017 20 / 20