Lecture D32: Damped free vibration Spring-Dashpot-Mass System C Spring force Fs=ka k>0 Dashpot Fd=-ca C>0 Newton's Second Law(mi=2F) m+ ca+k =0 (Define )Natural Frequency wn=vk/m,and Period T=2T/Wn (Define)Damping Factor s=c/(2mwn) Equation of motion +2(um+u2m=0
Lecture D32 : Damped Free Vibration Spring-Dashpot-Mass System Spring Force Fs = −kx, k > 0 Dashpot Fd = −cx˙, c > 0 Newton’s Second Law (mx¨ = P F) mx¨ + cx˙ + kx = 0 (Define) Natural Frequency ωn = q k/m, and Period τ = 2π/ωn (Define) Damping Factor ζ = c/(2mωn) Equation of motion x¨ + 2ζωnx˙ + ω 2 nx = 0 1
Solution a(t=Ae Characteristic polynomia 入2+2cn入+ Roots 1 (-c+ 1),入2 General Solution(superposition) 入1t t a2e 入)t (-c+√c2-1)unt t Ae 1)wnt
Solution Try x(t) = Aeλt Characteristic Polynomial λ 2 + 2ζωnλ + ω 2 n = 0 Roots λ1 = ωn(−ζ + q ζ 2 − 1), λ2 = ωn(−ζ − q ζ 2 − 1) General Solution (superposition) x = A1e λ1t + A2e λ2t = A1e (−ζ+ √ ζ 2−1)ωnt + A2e (−ζ− √ ζ 2−1)ωnt 2
Types of solutions ·(>1( Overdamped) 1>0→A12<0 .S=1(Critically Damped) 1 <0 (t)=(A1+Aten =5 =25 =1 c=25 One zero crossing at most !
Types of Solutions • ζ > 1 (Overdamped) q ζ 2 − 1 > 0 ⇒ λ1,2 < 0 • ζ = 1 (Critically Damped) λ1 = λ2 = −ωn < 0 x(t) = (A1 + A2t)e −ωnt 0 0.2 0.4 0.6 0.8 1 0 1 2 3 4 -0.2 One zero crossing at most !! 3
Types of Solutions(cont'd) s<1(Underdamped) Damped Natural Frequency(and Period) T d ad Solution r(t)=e-swnt(A cos wdt+Bsin wdt) or SUnt sin Wn= 6 =0.0833 t 0
Types of Solutions (cont’d) • ζ < 1 (Underdamped) Damped Natural Frequency (and Period) ωd = ωn q 1 − ζ 2 , τd = 2π ωd Solution x(t) = e −ζωnt (A cos ωd t + B sin ωd t) or, x(t) = Ce−ζωnt sin(ωd t + φ) -1 -0.5 0 0.5 1 1 2 3 4 4
Damping Factor Needs to be estimated experimentally Measure the ratio of two(or more)succes sive amplitudes 1 and 2, Ce 1 2 Ce-Swn(t1+Ta) Let 8=In (a1/x2). Since wd=wnV1-s V(2丌)2+6 For lightly damped systems 2丌
Damping Factor Needs to be estimated experimentally: Measure the ratio of two (or more) successive amplitudes x1 and x2, x1 x2 = Ce−ζωnt1 Ce−ζωn(t1+τd) = e ζωnτd Let δ = ln(x1/x2). Since ωd = ωn q 1 − ζ 2 , ζ = δ q (2π) 2 + δ 2 For lightly damped systems, ζ ≈ δ 2π 5
Energy Decay For Undamped harmonic oscillator: E=-mi2+=kx2=-ka2og(constant) For Underdamped oscillator Cma(t)=amaz(O)e-Sun Thus, we expect E(t)=E(Oe-2sant valid for lightly damped systems This can be used to estimate s if Energy (decay) can be measured
Energy Decay For Undamped harmonic oscillator: E = 1 2 mx˙ 2 + 1 2 kx2 = 1 2 kx2 max (constant) For Underdamped oscillator xmax(t) = xmax(0)e −ζωnt Thus, we expect E(t) = E(0)e −2ζωnt . . . valid for lightly damped systems This can be used to estimate ζ if Energy (decay) can be measured. 6