me 392 chapter 7 single degree of freedom oscillator march 26 , 2012 week 11

ME 392Chapter 7

Single Degree of Freedom Oscillator

March 26, 2012week 11

Joseph Vignola

Assignments I would like to offer to everyone the extra help you might need to catch up.

Assignment 5 is due todayLab 3 is March 30 (next Friday)

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“ME_392_assignment_5_smith_johnson.doc”

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NameDate What it isLab partner

Single Degree of Freedom OscillatorThe single degree of freedom (SDoF) oscillator is a starting model for many problems

m

k b

F(t)

x(t)


A mass, m is free to move along one axes only. Here the x-axis

m

k b

F(t)

x(t)



What is the first thing you do with a problem like this?

m

k b

F(t)

x(t)




Draw a free body diagram

m

k b

F(t)

x(t)




m

k b

F(t)

x(t)

m

F(t)



A spring that pulls the mass back to its equilibrium position

m

k b

F(t)

x(t)

m

F(t)



A spring that pulls the mass back to its equilibrium position. The spring force is

m

k b

F(t)

x(t)

m

F(t)




A damper slows the mass by removing energy. Force is proportional to velocity

m

k b

F(t)

x(t)

m

F(t)




A damper slows the mass by removing energy. Force is proportional to velocity

A force drives the mass

m

k b

F(t)

x(t)

m

F(t)




A damper slows the mass by removing energy. Force is proportional to velocity A force drives the mass

m

k b

F(t)

x(t)

m

F(t)


m

k b

F(t)

x(t)


This equation can be written as

m

k b

F(t)

x(t)



Let’s solve the inhomogeneous problem

m

k b

F(t)

x(t)



Define two terms

m

k b

x(t)



Define two terms

m

k b

x(t)

is called the natural frequency and has units of radian/second



Define two terms

m

k b

x(t)


is the damping ratio and is dimensionless

Single Degree of Freedom OscillatorYou will determine the natural frequency and damping ratio of Lab 3

Define two terms

m

k b

x(t)


is the damping ratio and is dimensionless


The solution to this ODE with initial conditions

is…

m

k b

x(t)

The behavior of the system depends on



is

m

k b

x(t)



is

m

k b

x(t)

The period of the oscillation is



is

m

k b

x(t)



is

m

k b

x(t)

In this expression the time constantis related to other physical parameters by

The system response is sinusoidal

has natural frequency of

There's an exponential decay Where

So we can extract the damping ratio, ζ if we can measure

Summary of Free Ring-down

m

k b

x(t)

The system response is sinusoidal

has natural frequency of

There's an exponential decay Where

So we can extract the damping ratio, ζ if we can measure


m

k b

x(t)

The greater the damping the wider the resonance peak


m

k b

And plot response as a function of frequency

This leads to another way to estimate the damping ratio, ζ

we can drive the oscillator at a series for frequencies and measure the response amplitude

Single Degree of Freedom OscillatorAnd plot response as a function of frequency

We always assume that there is some error in our measurement.

Single Degree of Freedom Oscillator… so for a plot with perhaps 20 measurements

Single Degree of Freedom Oscillator… so for a plot with perhaps 20 measurements we can curve fit to extract the width of the resonance curve

Details of the Time FitLet’s assume we have noisy ring down data


We can generate the envelope using the magnitude of the Hilbert transform



On a log scale at least the beginning looks linear

This means that we can use polyfit.m



On a log scale at least the beginning looks linear

This means that we can use polyfit.m

Where is the time constant

Details of the Time Fitsf = 10000;N = 10000;si = 1/sf;k = 1e5;m = 2;x0 = 3;[f,t] = freqtime(si,N); omegac = sqrt(k/m);fc = omegac/(2*pi);zeta = .05;tau= 1 ./(omegac*zeta); env = x0*exp(-t*(1../tau));displacement = env.*(cos(omegac*t)*ones(size(tau))) + .075*randn(size(t));DISPLACEMENT = fft(displacement);[a,b] = max(abs(DISPLACEMENT));fc_data = f(b); env = abs(hilbert(displacement));lenv =log(env);fit_range = [.01 .2];[a,bin_range(1)] = min(abs(t-fit_range(1)));[a,bin_range(2)] = min(abs(t-fit_range(2)));p = polyfit(t(bin_range(1):bin_range(2)),lenv(bin_range(1):bin_range(2)),1); fit = polyval(p,t);tau_from_fit = -1/p(1);

Details of the Frequency Domain FitLet’s assume we have noisy FRF data


And we expect the FRF to be of the form

We need you find



We need you find

That best fit the data



We need you find

That best fit the data

Use fminsearch.m

Using the Lorentzian Fitfminsearch.m requires that you 1)make a fitting function 2)a guess or a starting point

My fitting program used three additional subroutines there are

Three_parameter_curve_fit_test.m (main program)lorentzian_fit_driver3.mlorentzian3.mlorentzian_fit3.m

These m-files can be found on the class webpage

me 392 chapter 7 single degree of freedom oscillator march 26 , 2012 week 11

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