2. Damped Oscillators, Continued The General Solution was: x(t) = e -βt [A 1 e αt + A 2 e -αt ] where α  [β 2 - ω 0 2 ] ½ 3 cases of interest are: –Underdamping  ω 0 2 > β 2 We just did this case in detail! –Critical damping  ω 0 2 = β 2 –Overdamping  ω 0 2 < β 2 We’ll now briefly do the last 2 cases!

3. Critical Damping x(t) = e -βt [A 1 e αt + A 2 e -αt ] where α  [β 2 - ω 0 2 ] ½ –Critical damping  ω 0 2 = β 2  α = 0. Clearly, there are no oscillations! Rather than use this general solution, we actually need to go back & solve the differential equation again, putting in the fact that ω 0 2 = β 2 beforehand. Get (Appendix C): x(t) = [A + B t]e -βt

4. x(t) = [A + B t]e -βt For given initial conditions, a critically damped oscillator will approach equilibrium at a rate more rapidly than either the underdamped or the overdamped oscillator. –This has important practical implications & applications! For example, in the design of various oscillator devices when a rapid return to equilibrium is required: Galvanometers Pneumatic screen door closer. …..

5. Damped Oscillators, Continued Back to the General Solution x(t) = e -βt [A 1 e αt + A 2 e -αt ] where α  [β 2 - ω 0 2 ] ½ 3 cases of interest are: –Underdamping  ω 0 2 > β 2 We did this case in detail! –Critical damping  ω 0 2 = β 2 We just discussed this! –Overdamping  ω 0 2 < β 2

6. Overdamping x(t) = e -βt [A 1 e αt + A 2 e -αt ] where α  [β 2 - ω 0 2 ] ½ Overdamping  ω 0 2 < β 2 Define ω 2  α  [β 2 - ω 0 2 ] ½ –Following the book, write the solution as (Appendix C): x(t)= e -βt [A 1 exp(ω 2 t) + A 2 exp(-ω 2 t)] ω 2  α is not an angular frequency & the motion is not oscillatory. x(t)  Equilibrium asymptotically as t gets large.

7. x(t)= e -βt [A 1 exp(ω 2 t) + A 2 exp(-ω 2 t)] In this case, it can have “strange” x – v phase diagram behavior! The figure shows 3 cases with x(0) = x 0 > 0

8. x(t)= e -βt [A 1 exp(ω 2 t) + A 2 exp(-ω 2 t)] Case I (from “strange” x – v phase diagram): x(0) = x 0 > 0, x(0) = v(0) = x 0 > 0: x(t) reaches a max at some time t > 0 before it  0. x(t)v(t)

9. x(t)= e -βt [A 1 exp(ω 2 t) + A 2 exp(-ω 2 t)] Case II (from “strange” x – v phase diagram): x(0) = x 0 > 0, x(0) = v(0) = x 0 < 0: x(t) & v(t) both monotonically  0 as t  0 x(t) v(t)

10. x(t)= e -βt [A 1 exp(ω 2 t) + A 2 exp(-ω 2 t)] Case III (from “strange” x – v phase diagram): x(0) = x 0 > 0, x(0) = v(0) = x 0 < 0: Numbers are so it is a special case in the phase diagram. x 0 is below the curve x = - (β + ω 2 )v. x(t) goes < 0 before it  0 & v(t) = x goes > 0 before it  0. (“Oscillatory”, but only one crossing of 0 for both x & v) x(t) v(t)

11. Example 3.3 Pendulum: Length. Mass m. Moves in oil with a small displacement angle θ. Oil gives a retarding force F r = K(dθ/dt). K  2m(g ) ½. Initial conditions: t = 0, θ = α; (dθ/dt) = 0. Find the angular displacement θ = θ(t) & velocity (dθ/dt) = θ(t). Sketch the phase diagram if (g/ ) ½ = 10s -1, α = 10 -2 rad.

12. N’s 2 nd Law eqtn of motion (rotations): I (d 2 θ/dt 2 ) = N (total torque)  m 2 θ = -mg sinθ - F r Using F r = 2m(g ) ½ (dθ/dt) & simplifying, eqtn of motion is:  θ + 2(g/ ) ½ θ + (g/ )θ = 0 Mathematically identical to the damped harmonic oscillator eqtn of motion x + 2βx + (ω 0 ) 2 x = 0  Identify the pendulum frequency: (ω 0 ) 2 = (g/ ) & the damping constant: β 2 = (g/ ). Since (ω 0 ) 2 = β 2, the pendulum is a critically damped oscillator!  Using the results for that case just discussed gives: θ(t) = [A + B t]e -βt

13. Critically damped pendulum θ(t) = [A + B t]e -βt with β = (g/ ) ½. Initial conditions: (t = 0) θ = α =10 -2 rad, (dθ/dt) = 0 β = (g/ ) ½ =10 s -1 Using these in θ(t), algebra gives constants A & B:  A = α = 10 -2 rad, B = βA = (g/ ) ½ A = 10 -1 s -1 So, θ(t) = α[1 + (g/ ) ½ t] exp[-(g/ ) ½ t] or θ(t) = [0.01 + 0.1t]e -10t rad

14. Phase Diagram, Critically Damped Pendulum θ(t) = [A + B t]e -βt = 10 -2 [1 + 10t] exp[-10t] rad θ(t) = (dθ/dt) = - α(g/ )t exp[-(g/ ) ½ t] = -t e -10t rad/s θ(t) vs. θ(t)