Lecture 13: Damped Oscillations

(13) Dissipation and Damped Oscillators 1 Dissipation We’ve been trying to ignore it, but in the real world there is friction. Friction means that mechanical energy is converted to thermal energy, and we no longer have a ‘conservative’ system. But we can try. Rather than just start with a damped oscillation (as in eqn 25.1 in LL), I will motivate a modified Euler-Lagrange equation which includes dissipation, and then use this to arrive at damped oscillations. Imagine some fraction of kinetic energy is couple to thermal energy per unit time . Z 0 L (q, q̇) = (T + Tµ ) − U = L +  ∂L0 ∂q̇ Z  d ∂ (T ) dt dt ∂q̇ ∂ ∂L0 (T ) = ∂q ∂q̇ ∂L0 ∂L given uniform  such that = ∂q ∂q   d ∂L ∂L ∂ = − (T ) ⇒ dt ∂q̇ ∂q ∂q̇ d dt  T dt   d ∂L = + dt ∂q̇   d ∂L = + dt ∂q̇ Generalizing T to any velocity dependent function, Let D= 1X bjk q̇j q̇k 2 “dissipative function” j,k or, for one degree of freedom, simply 1 D = bq̇ 2 2 1 modified E-L ∂L ∂D d ∂L − = ∂q̇ ∂q dt ∂q̇   ∂L − bq̇ such that ṗ = ∂q where ∂L ∂q is the conservative force, and bq̇ is the dissipative force. Damped systems lose energy with time until they come to rest. The rate of energy loss is given by the dissipation function.   dE d ∂L = q̇ −L dt dt ∂q̇     d ∂L ∂L ∂L ∂L ∂L = q̇ + q̈ − + q̇ + q̈ dt ∂q̇ ∂q̇ ∂t ∂q ∂q̇     d ∂L ∂L = q̇ − normally zero, but... dt ∂q̇ ∂q dE ∂D = −q̇ = −2D dt ∂q̇ Note that the last line is just the rate of work done by friction as force × velocity. All of this is assuming no external driving force (i.e., ∂L ∂t = 0). 2 Damped Oscillator Let’s put this to work on our harmonic oscillator to make a more realistic damped oscillator. for and 1 1 L = mẋ2 − kx2 2 2 1 D = bẋ2 2 2 the equations of motion are mẍ = −kx − bẋ, ω02 = k m or ẍ + 2λẋ + ω02 x = 0, 2λ = b m This differential equation is best solved with complex exponentials, but the solution can be written in real form as “under damped” x (t) = ae−λt cos (ω1 t + φ) for λ < ω0 q where ω1 = ω02 − λ2 “over damped” �
x (t) = e−λt a1 eβt + a2 e−βt for λ < ω0 q where β = λ2 − ω02 , note β < λ ⇒ decay 3 “critically damped” x (t) = e−λt (a1 + a2 t) for λ = ω0 To complete the picture, we should add a driving force to our damped oscillator. Returning to the equation of motion... f ẍ + 2λẋ + ω02 x = Fm(t) = m cos (ωt) ⇒ x (t) = a1 e−λt cos (ω0 t + φ) + a2 cos (ωt + θ) a2 = f q 2 m (ω 2 − ω02 ) + 4λ2 ω 2 2λω tan θ = 2 ω − ω02 where a1 and φ come from the initial conditions Again, the driven solution has 2 parts, one that depends on the initial conditions and another which is the response to the drive. With damping, we see that the first of these decays with time, such that the motion at t  λ1 is essentially only the driven response. 4 x (t) ' a2 cos (ωt + θ) → for t  x (t) a2 = F (t) f Transfer Functions 5 1 λ (1) (2)