2. Damped Oscillators SHO’s: Oscillations  “Free oscillations” because once begun, they will never stop! In real physical situations & for real physical systems, of course: –There are retarding forces: These will damp oscillations, which eventually die away to the stop motion. Usual approximation: Damping force F r  v = x In what fallows, take F r = -bv, b = a constant > 0, which depends on the system. Prototype oscillator: Mass m in 1d under combined linear restoring force - kx + retarding force -bv.

3. Prototype Damped Oscillator Newton’s 2 nd Law: F = ma = - bx - kx

4. Newton’s 2 nd Law Equation of Motion: F = ma = - bv - kx Or:m(d 2 x/dt 2 ) = - b(dx/dt) - kx Or: mx + bx + kx = 0 Definitions: –Damping Parameter: β  [b/(2m)] –Characteristic angular frequency: ω 0 2  (k/m) Equation of motion becomes: x + 2βx + (ω 0 ) 2 x = 0

5. Equation of motion: x + 2βx + ω 0 2 x = 0 A standard, homogeneous, 2 nd order differential equation. GENERAL SOLUTION (Appendix C!) has the form: x(t) = e -βt [A 1 e αt + A 2 e -αt ] where α  [β 2 - ω 0 2 ] ½ A 1, A 2 are determined by initial conditions: (x(0), v(0)). [A 1 e αt + A 2 e -αt ] can be oscillatory or exponential depending on the relative sizes of β 2 & ω 0 2.

6. x(t) = e -βt [A 1 e αt + A 2 e -αt ] with α  [β 2 - ω 0 2 ] ½ 3 cases of interest (Figure): Underdamping  ω 0 2 > β 2 Critical damping  ω 0 2 = β 2, Overdamping  ω 0 2 < β 2 x(t) (qualitative) in the 3 cases

7. Underdamped Case x(t) = e -βt [A 1 e αt + A 2 e -αt ] with α  [β 2 - ω 0 2 ] ½ Underdamping  ω 0 2 > β 2 Define: ω 1 2  ω 0 2 - β 2 > 0 ω 1  “Angular frequency” of the damped oscillator. –Strictly speaking, we can’t define a frequency when we have damping because the motion is NOT periodic! The oscillator never passes twice through a given point with the same velocity (it is slowing down!) –However we can define: ω 1 = (2π)/(2T 1 ), T 1 = time between two adjacent crossings of x = 0 –Note: ω 1 = [ω 0 2 - β 2 ] ½ < ω 0 –If the damping is small, ω 1  ω 0

8. x(t) = e -βt [A 1 e αt + A 2 e -αt ] with α  [β 2 - ω 0 2 ] ½ Underdamping  ω 0 2 > β 2  ω 1 2  ω 0 2 - β 2 > 0 –The solution is an oscillatory function multiplied by an exponential “envelope”. x(t) = e -βt [A 1 exp(iω 1 t) + A 2 exp(-iω 1 t)] Or, defining 2 new integration constants A & δ: x(t) = A e -βt cos(ω 1 t - δ) The oscillatory part looks just like the undamped solution with ω 0 2  ω 1 2  ω 0 2 - β 2 The maximum amplitude of motion decreases with time by a factor e -βt. This function forms an “envelope” function for the oscillatory part: x en =  A e -βt

9. x(t) = A e -βt cos(ω 1 t - δ), x en =  A e -βt Underdamping! Shown in fig. for δ = 0. Clearly the period is longer (the frequency is shorter) than for the undamped case!

10. x(t) = A e -βt cos(ω 1 t - δ) Underdamping! Ratio of the amplitudes at 2 successive maxima is of interest. –Define: T  Time at which a maximum occurs. τ 1  Time between 2 successive maxima. τ 1  (2π)/ω 1 –The ratio of the amplitudes at 2 successive maxima: D  [Aexp(-βT)]/[Aexp(-β{T+ τ 1 })] or D = exp(βτ 1 ) D  The “Decrement” of the motion ln(D) = βτ 1  “Logarithmic Decrement” of the motion

11. x(t) = A e -βt cos(ω 1 t - δ) Underdamping! Consider:v(t) = x(t) = (dx/dt) = - β[A e -βt cos(ω 1 t - δ) - ω 1 [A e -βt sin(ω 1 t – δ)] Consider the Mechanical Energy E: –Due to damping, E is not a constant in time! E IS NOT CONSERVED! –Energy is continually given up to the damping medium. Energy dissipation in terms of heat. E = T + U (U is due to the restoring force -kx only, not due to the retarding force -bv). U = U(t) = (½)k[x(t)] 2, T = T(t) = (½)m[v(t)] 2. E is clearly mess! Clearly, it decays in time as e -βt

12. Energy for underdamped case. See figure. (Also see Prob. 3-11)

13. Energy Loss Rate in the underdamped case. See figure. (dE/dt)  [v(t)] 2 (See Prob. 3-11) (dE/dt) = is a max when v(t) = max. (near equilibrium position). (dE/dt) = 0 when v(t) = 0

14. Example 3.2: Phase Diagram - Underdamped case. Use a computer to construct a phase diagram for the underdamped oscillator. A = 1 cm, ω 0 = 1 rad/s, β = 0.2 s -1, δ = (½)π. Define: u = ω 1 x, w = βx + v. Polar coordinates: ρ = [u 2 +w 2 ] ½ φ = ω 1 t. Get a logarithmic spiral for the phase trajectory. w vs. u:

15. Phase Diagram - Underdamped case. A spiral phase path. The continually decreasing magnitude of the radius vector is always an indication of damped oscillations. v vs. x:

16. x(t) & v(t) for the case just shown