Section 3.7: Damped Free Vibrations

Here is the differential equation that models damped free vibrations:

L[y_,m_,γ_,k_]:=m D[y,{t,2}] + γ D[y,{t,1}] + k y ==0

Mathematica Difficulties--Why We Study Differential Equations

L[u[t],m,γ,k]

sol1=DSolve[L[u[t],m,γ,k],u[t],t]

Notice that Mathematica gets it WRONG--what if γ=2(Sqrt[k m])? The general solution should be y(t)=(Subscript[c, 1])(e^((Subscript[r, 1])t))+(Subscript[c, 2])t (e^((Subscript[r, 1])t)).

How significant is this? Let's invesitigate.We will include intial conditions u(0)=1 and u'(0)=0 (released from rest).

DSolve[{L[u[t],m,γ,k],u[0]==1,u'[0]==0},u[t],t];
Simplify[%]

This looks ok, until we realize the solution doesn't exist when γ=2(Sqrt[k m]). The problem is that what Mathematica found was not a general solution.

Here is the correct solution, which Mathematica does find if it knows γ=2(Sqrt[k m])to begin with.

DSolve[{L[u[t],1,2,1],u[0]==1,u'[0]==0},u[t],t]

The Three Types of Damping

Let's invesitigate the solutions for different values of γ.

(Subscript[m, 1])=1/2;
(Subscript[k, 1])=3/4;
(Subscript[γ, 1])=2(Sqrt[(Subscript[k, 1])(Subscript[m, 1])]);
sol=DSolve[{L[u[t],(Subscript[m, 1]),(Subscript[γ, 1]),(Subscript[k, 1])],u[0]==1,u'[0]==0},u[t],t];
usol[t_]=sol[[1,1,2]]
critical=Plot[usol[t],{t,0,12},PlotStyle->{Red,Thick},ImageSize->{500,Automatic},PlotRange->{All,All}]

(Subscript[m, 1])=1/2;
(Subscript[k, 1])=3/4;
(Subscript[γ, 1])=2(Sqrt[(Subscript[k, 1])(Subscript[m, 1])])+9/10;
sol=DSolve[{L[u[t],(Subscript[m, 1]),(Subscript[γ, 1]),(Subscript[k, 1])],u[0]==1,u'[0]==0},u[t],t];
usol[t_]=sol[[1,1,2]]
over=Plot[usol[t],{t,0,12},PlotStyle->{Blue,Thick},ImageSize->{500,Automatic},PlotRange->{All,All}]

(Subscript[m, 1])=1/2;
(Subscript[k, 1])=3/4;
(Subscript[γ, 1])=2(Sqrt[(Subscript[k, 1])(Subscript[m, 1])])-9/10;
sol=DSolve[{L[u[t],(Subscript[m, 1]),(Subscript[γ, 1]),(Subscript[k, 1])],u[0]==1,u'[0]==0},u[t],t];
usol[t_]=sol[[1,1,2]]
under=Plot[usol[t],{t,0,12},PlotStyle->{Green,Thick},ImageSize->{500,Automatic},PlotRange->{All,All}]

Show[critical,over,under,ImageSize->{500,Automatic},PlotRange->{All,All}]

Critical Damping: quickest return to equilibrium, at most one oscillation.
Over Damping: slower return to equilibrium than critical damping, no oscillations.
Under Damping: not enough damping to prevent oscillations during return to equilibrium.

Since the mass is most likely fixed, once can change out springs to affect how a system returns to equilibrium. For example, changing springs can affect how a car behaves when it hits a bump (different cars, different masses, different springs desired). What engineers want is to achieve critical damping, since it is the fastest return to equilibrium. You would also use this for doors that use springs to return them to equilibrium after being opened (think about swinging doors in a restaurant, for example).