美国麻省理工大学：《动力工程学》（英文版）Lecture D33: Forced Vibration

Lecture d33. Forced vibration Fo sin wt Spring force Fs==k k>0 Dashpot Fd=-ca Forcing Feat= Fo sin wt Newton's Second Law(mi=2F) ma+ ca f ka Fo sin wt Wn=vk/m, S=c/(2mwn) Equation of motion F 十2n+nc 0 sin wt

Lecture D33 : Forced Vibration Spring Force Fs = −kx, k > 0 Dashpot Fd = −cx˙, c > 0 Forcing Fext = F0 sin ωt Newton’s Second Law (mx¨ = P F) mx¨ + cx˙ + kx = F0 sin ωt ωn = q k/m, ζ = c/(2mωn), Equation of motion x¨ + 2ζωnx˙ + ω 2 nx = F0 m sin ωt 1

Undamped forced vibration sin wt General solution c(t)=rc(t)+Ep(t) (t) is genera/ solution of ctni=0 have already seen p(t) is any solution of (1) Tryp(t)= X sin wt→ Fo/k (u/un)21-(u/un)2

Undamped Forced Vibration x¨ + ω 2 nx = F0 m sin ωt (1) General Solution x(t) = xc(t) + xp(t) • xc(t) is general solution of x¨ + ω 2 nx = 0 . . . have already seen • xp(t) is any solution of (1) Try xp(t) = X sin ωt ⇒ X = F0/k 1 − (ω/ωn) 2 = δst 1 − (ω/ωn) 2 2

Undamped forced vibration Particular solution t=X sin wt st sin wt (w/wn) W/G Cosine

Undamped Forced Vibration Particular Solution xp(t) = X sin ωt = δst 1 − (ω/ωn) 2 sin ωt -6 -4 -2 0 2 4 6 0 0.5 1 1.5 2 2.5 3 ω ωn 3

Damped forced vibration cWna+wnic sin wt (2) General solution c(t)=ac(t)+Ep(t) (t)is general solution of a+2CWn C+Wn=o have already seen Ep(t)is any solution of(2) Ep(t)=X1 CoS wt+ X2 sin wt or mp(t=X sin(wt-p)

Damped Forced Vibration x¨ + 2ζωnx˙ + ω 2 nx = F0 m sin ωt (2) General Solution x(t) = xc(t) + xp(t) • xc(t) is general solution of x¨ + 2ζωnx˙ + ω 2 nx = 0 . . . have already seen • xp(t) is any solution of (2) Try xp(t) = X1 cos ωt + X2 sin ωt or, xp(t) = X sin(ωt − ψ) 4

Damped forced vibration cct is general solution of +2Swn +wna=o have already seen (t) is any solution of(2) p(t=X1 cos wt+ X2 sin wt or Ep(t)= X sin(at-) Fo/k {[1-(u/u)2]2+[2cu/m]2 tan 2w/w

Damped Forced Vibration • xc(t) is general solution of x¨ + 2ζωnx˙ + ω 2 nx = 0 . . . have already seen • xp(t) is any solution of (2) Try xp(t) = X1 cos ωt + X2 sin ωt or, xp(t) = X sin(ωt − ψ) ⇒ . . . X = F0/k {[1 − (ω/ωn) 2] 2 + [2ζω/ωn] 2} 1/2 ψ = tan−1 " 2ζω/ωn 1 − (ω/ωn) 2 # 5

Damped forced vibration {[1-(u/un)2]2+[2u/wm2y2 c=01 =0.2 =0.5 0.5 15 2.5 /w

Damped Forced Vibration X = δst {[1 − (ω/ωn) 2] 2 + [2ζω/ωn] 2} 1/2 0 1 2 3 4 5 6 0 0.5 1 1.5 2 2.5 3 6

Damped forced vibration yb=tan 25w/wn (wn)2 =0 =0.2 C=0.5 0.1 0 Fo sint Xsin(wt-al Fo A wt t

Damped Forced Vibration ψ = tan−1 " 2ζω/ωn 1 − (ω/ωn) 2 # 0 1 2 3 7

Vibration isolation Fo sin wt How much of the applied force is transmitted to the wall Transmitted force Transmissibility= Applied Force Applied Force ect F sIn wt Transmitted Force(spring+ dashpot) Fw ka+ci = kX sin(at-b)+wX cos(wt-vD) XVk2+(cw )2 sin (t-o) 8

Vibration Isolation How much of the applied force is transmitted to the wall? Transmissibility = |Transmitted force | |Applied Force| Applied Force Fext = F0 sin ωt Transmitted Force (spring + dashpot) Fw = kx + cx˙ = kX sin(ωt − ψ) + cωX cos(ωt − ψ) = X q k 2 + (cω) 2 sin(ωt − σ) 8

Vibration isolation Transmissibility XVk2+(cw)2 0 1+(2cu/n) N(1-(u/un)2)2+(2(/n) =0 c=01 =0.2 C=0.5 For w/wn >v2 damping increases transmissi- bility ! For w/wn v2 having a spring increases trans- miscibility !

Vibration Isolation Transmissibility = X q k 2 + (cω) 2 F0 = vuut 1 + (2ζω/ωn) 2 (1 − (ω/ωn) 2) 2 + (2ζω/ωn) 2 0 1 2 0 1 2 3 Transmissibility For ω/ωn > √ 2 damping increases transmissi￾bility !! For ω/ωn < √ 2 having a spring increases trans￾missibility !! 9