振动学系列(Vibration)3._Viscous_Damping

3. Viscous DampingIn the spring-mass systems we have studied so far, the response will continue to infinity; that is, it will never die out. Real systems, of course, do die out eventually, and it is important that we account for this in our mathematical model.To create a response that dies out over time, we typically add damping to the system.The damping force is usually taken to be proportional to the velocity; that is:fc=cxAs velocity increases, so does the damping force.Damping is usually the result of one type of friction or another.The equation of motion for the damped case ismx+cx+kx=0This is also a second-order, homogeneous differential equation. The general solution to this type of equation isxt=aetxt=aetxt=2aetSubstitute this into the equation of motionm2aet+caet+kaet=0m2+c+kaet=0Here we have the same situation as earlier. Sinceaet0we havem2+c+k=0Using the quadratic formula, we find1,2=-cc2-4km2mThis is called the “characteristic equation” of the system. Observing this equation, we see that there are three possible solution types:1. c2-4km01,2 are real2. c2-4km 1 real2. 1In the case where 1, the are real1=-n-n2-12=-n+n2-1 Since we have two distinct roots, the solution to the equation of motion is a linear combinationxt=a1e1t+a2e2twhere we can solve for a1 and a2 using initial conditionsx0=a1+a2=x0v0=1a1+2a2=v0Solving both of these equations simultaneously givesa1=-v0+x022-1a2=v0-x012-1Substituting the expressions for 1,2 given above results ina1=-v0+-+2-1nx02n2-1a2=v0+2-1nx02n2-1This type of response is not oscillatory. The response decays back to its initial state exponentially.Since 1 and 2 are negative, the response decays with time. This case is called the “Overdamped” case.Case 2: 1Recall 1,2=-nn2-1If 1, then 2 1 0, and we are left with a complex solution. Let us rearrange2-1=-11-2=j1-2then1=-n-jn1-22=-n+jn1-2Let d=n1-2 = “damped natural frequency”. Then 1=-n-jd2=-n+jdThe general solution is stillxt=a1e1t+a2e2tbut the values are complex. We can again solve for a1 and a2 using initial conditionsx0=a1+a2=x0v0=1a1+2a2=v0Solving both of these equations simultaneously givesa1=-v0+x022-1=-v0+x0-n+jd2jda2=v0-x012-1=v0+x0n+jd2jdSubstituting this into the general solution givesxt=a1e-n-jdt+a2e-n+jdt=e-nta1e-jdt+a2ejdtWe next make use of Eulers identityej=cos+jsinSubstituting this into the general solution givesx(t)=e-nta1cosdt-jsindt+a2cosdt+jsindt=e-nta1+a2cosdt+ja2-a1sindt=e-nt2jdx02jdcosdt+j2v0+2nx02jdsindt=e-ntx0cosdt+v0+nx0dsindtIn Homework 1, we found that we could express Bcosdt+Csindt=Xsindt+Thusxt=Xe-ntsindt+wheretan=x0dv0+nx0X=x02+1d2v02-nx0What does the solution look like?The response oscillates as it decays exponentially.Case 3: = 1In the critically damped case, we find that1,2=-1nn1-11,2=-nThe solution takes the formxt=a1+a2te-ntAnd from the initial conditionsa1=x0a2=v0+nx0As before, this takes the form of a decaying exponential solution. The solution does not oscillate.What do each of these responses mean physically? Again, think of the car driving off the curb:the response oscillates about an equilibrium 1does not oscillate, returns to equilibrium slowly does not oscillate, returns to equilibrium quickly = 1 Underdamped response is desirable in guitar strings.Critically damped response is desirable in automotive suspensions.Overdamped response is desirable in a door closing mechanism.ExampleA car with mass 1000kg is to be designed such that its suspension is critically damped. The suspension deflects 10cm under the cars weight. Find k and c for this suspension.SolutionThe problem statement gives the static ride height deflection as 10cm. From the Baja car design example we havex0=mgkRearranging givesk=mgx0=1000kg9.81m/s20.1m=98.1kN/mThe formula for damping ratio is=c2kmTo be critically damped, of course, we must have = 1. Thusc=2km=298,100kN/m1000kg=19,800kg/sIf we add mass to the car by adding a 100kg passenger, what happens?The damping ratio will change slightly, since we have