Free Harmonic Oscillations

Machines with rotating components commonly involve mass-spring systems or their equivalents in which the driving force is simple harmonic. The motion of a mass attached to a spring serves as a simple example of vibrations that occur in more complex mechanical systems.

From a teaching point of view it is suitable to consider a body of mass m attached to one end of a spring that resists compression as well as stretching. A rod attached to the mass carries a disk moving in an oil-filled cylinder (a dashpot). The other end of the spring could be attached to a fixed wall and vibrating horizontally. The resultant force on the body

is the sum of the restoring force - and the damping force -, is the force constant, is the distance of the body of mass from its equilibrium position, t is time and and r is the damping constant. We take >0 when the spring is stretched. The differential equation of motion is therefore:

> deq1:=m*diff(x(t),t$2)+r*diff(x(t),t)+k*x(t)=0;

If we set in deq1 the motion is undamped. Otherwise the solution of deq1 presents three distinct cases of damping according to whether is greater than, equal to, or less than zero.

> eq:=d=r^2-4*m*k:

> solm:=solve(eq,m):

Let us replace m with a new variable mv in deq1 .

> mv:=unapply(solm,d,r,k);

> deq2:=collect(4*k*subs(m=mv(d,r,k),deq1),diff(x(t),t$2));

With this differential equation, deq2 , the system is overdamped if d > 0 , critically damped if d = 0 and underdamped if

d < 0 . If we request Maple to solve deq1 for the undamped case and deq2 for each of the damped cases,subject to initial condition and = 0

> init:=x(0)=x[0],D(x)(0)=0: #initial conditions

we get:

UNDAMPED

With r = 0 we get:

> deq1a:=subs(r=0,deq1);

> dsolve({deq1a,init},x(t));

A typical graph of with and = 1 is shown i Figure 1. Figure 2 shows the animation of the undamped motion.

> xk:=unapply(rhs(%),m,k,x[0],t): # defines x = x(m,k,x[0],t)

OVERDAMPED

> interface(showassumed=0):

> assume(d>0):

> dsol:=dsolve({deq2,init},x(t));

> subs(eq,dsol);

> xo:=unapply(rhs(%),m,r,k,x[0],t): # defines x = x(m,r,k,x[0],t)

The solution consists of two exponential terms when r > . In all subsequent figures we take and
= 1 . Figure 3 shows some typical graphs of the position function for the overdamped case. Figure 4 shows the animation of the motion with the damping constant r = 4 .

CRITICALLY DAMPED

> lhs(dsol)=limit(rhs(dsol),d=0);

> subs(r=sqrt(4*m*k),%);

> xc:=unapply(rhs(%),m,k,x[0],t): #defines x = x(m,k,x[0],t)

The solution consists of one exponential term when r = . The graph in Figure 5 resemble those of the overdamped case in Figure 3. The animation of a critically damped motion with is shown in Figure 6.

UNDER DAMPED

> assume(d<0):

> dsolve({deq2,init},x(t));

> subs(eq,%);

> xu:=unapply(rhs(%),m,r,k,x[0],t): #defines x = x(m,r,k,x[0],t)

The solution represents exponentially damped oscillations of the mass-spring system about its equilibrium position as shown in both Figure 7 and Figure 8.

Undamped Motion

Overdamped Motion

Critically Damped Motion

Underdamped Motion