Section 3. 7 Mass Spring Systems (Damped)

Key Terms/ Ideas:

• Dashpot

• Vibration isolators

• Damped Free Vibrations• Critical Damping• Over Damped• Under Damped

Vibrations, in the design of mechanical systems, are normally an undesirable occurrence, and engineers attempt to either reduce vibrations to a minimum, or eliminate them completely.

“Vibration Isolators” are commonly designed and used to minimize vibration of mechanical systems, for example:

VIPS series of vibration isolation platforms are fully assembled isolated platforms for microscopes, interferometers and other vibration sensitive equipment. Just set on a flat surface, connect one air-line, put your instrument on top and in under 5 minutes, you are isolated. Top surface has no holes.

http://www.vere.com/html/isolation_platforms.html

Adapted from work by Dr. Tai-Ran Hsu at SJSU.

More vibration isolators or damping products.

Heavy duty truck suspension.

A mundane item, a storm/screen door damper.

Inside the cylinder is a spring. If the spring was the only thing inside the cylinder you would get a door that slams shut and is annoying. So there is also an air cylinder. When you open the door, the cylinder fills with air. When the door is closing, the spring pushes a piston, which forces the air out of the cylinder through a small hole (which is adjustable to control the closing rate). The leaking air creates the hissing noise that you normally hear with these door closers.

http://usersites.horrorfind.com/home/halloween/wolfstone/HalloweenTech/pnucls_PneumaticCloser.html

Damped Free Vibrations

To the DE for Undamped Free Vibrations mu(t)'' + ku(t) = 0 we add a resistive force Fd

which is always opposite to the direction of motion. The force Fd is called a damping force.

This force may arise from several sources: resistance from the air or other medium in which the mass moves, internal energy dissipation due to the extension or compression of the spring, friction between the mass and the guides (if any) that constrain its motion to one dimension, or a mechanical device (dashpot) that imparts a resistive force to the mass. In any case, we assume that the resistive force is proportional to the speed |du/dt|of the mass; this is usually referred to as viscous damping.

We will take the equation of the damping force to be Fd = -γu'(t) where y (gamma) is a positive constant of proportionality known as the damping constant.

The IVP in this case is mu'' + γu' + ku = 0, u(0) = u0, u'(0) = v0 . We can view the DE in the following way:

[inertia] × u'' + [damping] × u' + [stiffness] × u = 0.

Mass is that quantity that is solely dependent upon the inertia of an object. The more inertia that an object has, the more mass that it has.

Recall mu'' = mg + Fs + Fd ; now it includes damping.

Here are some basic units in metric and English form. There are times when we need to convert units to their proper form based on the units for acceleration of gravity (denoted by g.

English g = 32 ft/sec2 Metric g = 9.8 m/sec2

The IVP for Damped Free Vibration mu'' + γu' + ku = 0, u(0) = u0, u'(0) = v0 has positive coefficients m, γ, and k so this a special class of second order linear IVPs.

In each of the three possible solutions exponentials are raised to a negative power, hence the solution u(t) in all cases converges to zero as t →∞.

Discriminant γ2 – 4km > 0 distinct real roots solution

γ2 – 4km = 0 repeated real roots solution

γ2 – 4km < 0 complex roots (-γ/2m ± µi) solution

r1 and r2 < 0

The roots of the characteristic equation are

u= (A + Bt)e-γt/(2m)

u= e-γt/(2m) (A cos(µt) + B sin(µt))

Since γ and m are positive we have negative exponents in these two cases.

Since m, y, and k are positive, y2 - 4km is always less than y2. Hence, if y2 - 4km > 0, then the values of distinct roots r1 and r2 are negative because the numerator is negative.

2-γ ± γ - 4km

1 2r t r tu = Ae +Be

The fact that the solution of the IVP for Damped Free Vibrationmu'' + γu' + ku = 0, u(0) = u0, u'(0) = v0 has u(t) → 0 as t →∞

confirms our intuitive expectation, namely, that damping gradually dissipates the energy initially imparted to the system, and consequently the motion dies out as time increases.

There are descriptive names associated with each of the cases. In addition, the graphical behavior of u(t) for each case is distinctive.

γ2 – 4km > 0 distinct real roots solution Over damped

γ2 – 4km = 0 repeated real roots solution Critically damped u= (A + Bt)e-γt/(2m)

The motion of the system in either of these cases crosses the equilibrium point either once or never, depending upon initial conditions. Thus u(t) “creeps” back to the equilibrium solution u(t) = 0.

Ref: Farlow

1 2r t r tu = Ae +Be

Example: Suppose that the motion of a spring-mass system is governed by the initial value problem

u'' +5u' +4u = 0, u(0) = 2,u'(0) = 1

Determine the solution of the IVP and find the time at which the solution is largest.

The characteristic equation is r2 + 5r + 4 = 0, so the roots are r = -1 and r = -4. The system is over damped.

The general solution of the DE is -t -4tu(t) = Ae +Be

Applying the initial conditions we have, u(0) = 2 2 = A +B

u'(0) = 1 1= -A- 4B

Solving for A and B we find that A = 3 and B = -1, thus the solution of the IVP is

-t -4tu(t) = 3e e

0 0.5 1 1.5 2 2.5 30

0.5

1

1.5

2

2.53*exp(-t)-exp(-4*t)

The max occurs at a horizontal tangent; set u'(t)= 0 and solve for t.

-t -4tu'(t) = -3e + 4e = 0

-4t -t -3t4e = 3e e = 3 / 4

ln(3 / 4)t = 0.09589sec.

-3

Over damped.

-t -4tu'(t) = -Ae - 4Be

γ2 – 4km < 0 complex roots (-γ/2m ± µi) solution

Under damped

u= e-γt/(2m) (A cos(µt) + B sin(µt))

In this case the damping coefficient γ is considered rather weak in comparison to the spring constant k, and we expect oscillatory motion.

Although the motion is not periodic, the parameter µdetermines the frequency with which the mass oscillates back and forth; consequently, µ is called the quasi frequency.

The time between successive maxima or successive minima is denoted by Td and is called the quasi period.

At http://phy.hk/wiki/englishhtm/Damped.htm is an applet that lets you select various values for the mass, the damping, and the spring constant.

Set the parameters m, γ, and k using the sliders.

The initial conditions are fixed values for this demo.

Because of Java restrictions this applet may not work on your computer.

Graphical demo; mp4 video of an applet that includes damping.

Example: Suppose that the motion of a spring-mass system is governed by the initial value problem

We have m = 1, γ = 0.125, and k = 1. So the roots of the corresponding characteristic polynomial are

So the solution of the IVP is

Using the techniques discussed previously we can express the sinusoid portion of the solution as a shifted cosine curve;

ω02 = k/m , A = 2 and B = 2/ and R = (A2 + B2)1/2255

The graph of this solution, along with solution to the corresponding undampedproblem, is given below.

It is evident from the graphs, that the damped and undamped solutions rise and fall almost together. The damping coefficient is small in this example, only one-sixteenth of the critical value, in fact. Nevertheless, the amplitude of the oscillation is reduced rather rapidly.

If the DE were u''+2u' + u = 0Then we would have a critically damped DE.

2/16 = 1/8 = .125

The quasi period is defined as Td = 2/ ≈ 6.295 sec.

We can determine further information about the solution

We have m = 1, γ = 0.125, and k = 1. The quasi frequency is given by

To find the time at which the mass first passes through its equilibrium position, we refer to the solution expression

and set ,, the smallest positive zero of the cosine function.

Then, by solving for t, we obtain

Here

δ = 0.06254

Discussion about damping.

Let’s assume that m, k, u0, and v0 are fixed and we get to vary the damping coefficient γ.

The roots of the characteristic polynomial are given by

Over damped distinct real roots γ2 – 4km > 0 γ2 > 4km 4mk/γ2 < 1

Critically damped repeated real roots γ2 – 4km = 0, 4mk/γ2 = 1

If we decrease γ slightly we can get the system to be

If we decrease γ a little more we can get the system to be

Under damped complex roots γ2 – 4km < 0, 4mk/γ2 > 1

Solution quickly becomes asymptotic to u = 0; maybe one local extrema.

Oscillatory with decreasing amplitude.

As the damping coefficient we have significant changes;• the quasi frequency goes to zero• the quasi period goes to infinity

Go to http://phy.hk/wiki/englishhtm/Damped.htm and experiment.

Here you can change the mass, the spring constant and the damping along with the amplitude and initial velocity.

http://phy.hk/wiki/englishhtm/Damped.htm