Single Degree of Freedom System 1.003J/1.053J Dynamics and Control I, Spring 007 Professor Thomas Peacock 5//007 Lecture 0 Vibrations: Second Order Systems with One Degree of Freedom, Free Response Single Degree of Freedom System Figure 1: Cart attached to spring and dashpot. Figure by MIT OCW. System response? What is x(t)? Use 18.03 Background. mẍ + cẋ + kx = F (t) x(t) = Free Response + Response Due to Forcing }{{}}{{} Complementary Solution, when F (t)=0 Particular Solution This lecture will cover the Free Response.
Free Response Look at k 0 Figure : Cart with dashpot only. Figure by MIT OCW. mẍ + cẋ = 0 Assume conditions x(0) = x 0 and ẋ(0) = v 0. mẍ + cẋ = mv + cv = 0 v = v 0 e ( ct/m) already used ẋ(0) = v 0 Integrate v(t) once. Using x(0) = x 0, we obtain: x = x 0 + mv 0 ( 1 e c t ) m c Figure 3: Solution to differential equation. Solution attenuates to a steady state value. Figure by MIT OCW.
3 Figure 4: Velocity profile of solution. Velocity attenuates to zero. Figure by MIT OCW. No oscillations. Because k = 0, there was no restoring term. Look at m 0 or Therefore: cẋ + kx = 0 k ẋ = x c x(0) = x 0 x(t) = x c t 0 e k Figure 5: Solution to differential equation. Position decays to zero. Figure by MIT OCW.
4 Figure 6: Velocity profile of solution. Value attenuates to steady state value. Figure by MIT OCW. kx 0 k c t ẋ = e c No oscillations in this system. Dashpot force balances the spring force as x 0, spring force 0. Vibrations require a restoring force (e.g. spring) and inertia (e.g. mass). Full Free Response Problem So let us consider the full problem: mẍ + cẋ + kx = 0 (1) Note that cẋ(c > 0) is a damping term and is responsible for decay of oscillations. Examination of Energy d d ( ) 1 1 kx (T + V ) = mẋ + = mẋẍ+ kx ẋ = ẋ(mẍ+ kx) = ẋ( cẋ) = cẋ dt dt For c > 0: d (T + V ) < 0 dt Damping. Mechanical energy is dissipated. For c < 0:
5 d (T + V ) > 0 dt Energy input (Control system providing energy) Solution of the Equation with Engineering Quantities Rewrite as: mẍ + cẋ + kx = 0 ẍ + ζω n ẋ + ω n x = 0 () ω k n = m c ζ = mωn ω n : Natural Frequency ζ: Damping Ratio To solve, we assume a solution of the form x = Ae (λt) Substitute in Equation (): λ + ζω n λ + ω n = 0 When ζ > 1 and ζ < 1, the behavior is different. Assume c 0. (ζ 0) We have the following cases. Case 1: Overdamped Case : Critically Damped λ = ζω n ± ω n ζ 1 (3) ζ > 1 λ 1, λ = Real Negative Numbers x = A ± e ζωn± ζ 1 ζ = 1 λ 1, λ 0 as t = ω n x = (A 1 + A t)e ωnt 0 as t (4) Equation (4) is the fastest approach to the set point. That is why it is named critically damped.
6 Case 3: Underdamped 0 ζ < 1 λ 1, λ = ζω n ± iω d ω d = ω n 1 ζ Underdamped (Not enough damping to prevent oscillations). When ζ 0, ω n (Natural frequency). ω d x = [ A 1 e iω dt + A e iω dt ] e ζωnt Must have that A 1 and A are complex conjugates because x is real. x =[A 1 (cos ω d t + i sin ω d t) + A (cos ω d t i sin ω d t)]e ζωnt =[(A 1 + A )cos ω d t + i(a 1 A )sin ω d t]e ζωnt }{{}}{{} A 4 A 1 + A = i(a 1 A ) = A 4 A 4 x = cos ω d t + sin ω d t e ζω nt x = cos ω d t + tan φ sin ω d t e ζωnt x = cos ω d t cos φ + sin ω d t sin φ e ζωnt cos φ Note the trigonometric identity. e ζωnt : Decaying in time cos(ω d t φ): Oscillatory Behavior C and φ can be found from initial conditions. x(t) = Ce ζωnt cos(ω d t φ) (5) C = (6) cos φ A 4 φ = arctan (7)
7 Equations (6) and (7) relate C and φ to and A 4. But cos 1 φ = 1 + tan φ. 1 A 4 = 1 + cos φ A 3 1 A = 3 + A 4 C = cos φ + A 4 If 0 ζ < 1, the solution will show decaying oscillations. How do we determine (C and φ) or ( and A 4 )? Often easier to relate and A 4 to initial conditions. Initial Conditions: x(0) = x 0, ẋ(0) = v 0 At t = 0, x 0 = (using x(0) = x 0 ) x = [ cos ω d t + A 4 sin ω d t]e ζωnt At t = 0: ẋ =[ ω d sin ω d t + A 4 ω d cos ω d t]e ζωnt ζω n [ cos ω d t + A 4 sin ω d t]e ζωnt v 0 = A 4 ω d ζω n = A 4 ω d ζω n x 0 v 0 + ζω n x 0 A 4 = ω d ( ) v 0 + ζω n x 0 C = x 0 + (8) ω d Examine solution. v 0 + ζω n x 0 tan φ = (9) ω d x 0 x(t) = Ce ζωnt cos(ω d t φ)
8 e ζωnt : Decay cos(ω d t φ): Oscillating Figure 7: Solution both decays and oscillates given the presence of exponential solution and sinusoidal solution. Figure by MIT OCW. Calculate Amplitude. x(t) e ζωnt ζω = = e nnτ d x(t + nτ d ) e [ ζωn(t+nτ d)] x(t) ω n π ω n π π ln = nζω n τ d = nζ = nζ = nζ (10) x(t + nτ d ) ω d ω n 1 ζ 1 ζ For ζ << 1: Need ω n, ζ to define system. x(t) ln = πnζ (11) x(t + nτ d )
9 Example Experiment: Flexible Rod. Figure 8: Flexible rod. Figure by MIT OCW. Measure frequency of oscillation: ω d. x(t) Measure amplitude over several periods to obtain x(t+nτd ). This ratio is related to the damping ratio ζ by the equations (10) or (11) if ζ << 1. With ω d and ζ, one can calculate the natural frequency ω n.