Just a quick one, Have I got my Mass/Spring/Damper model correct?

[Jowin86]

Homework Statement

Model the mass/spring/damper sdof system.

Homework Equations

Below left is a diagram of the system, and the attempt at the solution is obvious the right hand side one.

The Attempt at a Solution

Please see attached pdf.

Thanks for any help!

 

[Jowin86]

Here's an image of it instead of the PDF, hopefully get a reply now:

 

[Jowin86]

Although I think y=YoSinwt ?

 

[rude man]

That looks right if the function of the 'scotch yoke' and motor is to impart a linear excitation y(t). I never heard of a 'scotch yoke' before.

Are the 'bushes' bushings?

 

[Jowin86]

Yeah the scotch yoke converts the radial motion of the motor into linear motion. And the bushes are just like dry bearings on the rod, which connects the yoke to the spring, that restrict the system to one degree of freedom.

Thanks for your help :)

 

[rude man]

Looks OK except why did you reverse the positions of the mass and the damper?

What are you planning to do with your model, if anything?

 

[Jowin86]

Just because it was easier to draw. our lecturer said that the way it's hooked up it acts as if it were connected to ground. So I drew it like that.

It's for my assignment, we took measurements of time and acceleration from the free vibration response of the system and then used the electric motor to take forced response measurements. There's an accelerometer on the mass and the top of the spring so we can compare the input and output measurements. If that makes sense??

Basically we have to predict the frequency response of the system and compare our predictions to the actual test results. I am not 100% sure how to though :( I've got some of the way there, got the damping rato, natural and damped frequency of the system, and I think I have the parameter values too, these i got from the free vibration test results.

Got to make equations for magnification factor vs frequency ratio and then phase lag vs frequency ratio to show the frequency response now.. I think I have found them from my notes and all I have to do is change the values of " r " for frequency ratio and plot the answers ? That sound about right?

 

[rude man]

Just because it was easier to draw. our lecturer said that the way it's hooked up it acts as if it were connected to ground. So I drew it like that.

It's for my assignment, we took measurements of time and acceleration from the free vibration response of the system and then used the electric motor to take forced response measurements. There's an accelerometer on the mass and the top of the spring so we can compare the input and output measurements. If that makes sense??

Basically we have to predict the frequency response of the system and compare our predictions to the actual test results. I am not 100% sure how to though :( I've got some of the way there, got the damping rato, natural and damped frequency of the system, and I think I have the parameter values too, these i got from the free vibration test results.

Got to make equations for magnification factor vs frequency ratio and then phase lag vs frequency ratio to show the frequency response now.. I think I have found them from my notes and all I have to do is change the values of " r " for frequency ratio and plot the answers ? That sound about right?

Are you just looking for acceleration; what about displacement and velocity? Displacement you can possibly get from motor position (angle position indicator). For velocity, I don't know. I wouldn't want to try to time-integrate the accel output. I kind of gather you're just looking for acceleration of the mass.

The theoretical part is easier. You'll have to write the (1st order linear, constant-coefficient) differential equation for the system, with the forced excitation as the forcing function. That's very straight-forward if you've had diff. eq.'s. If not, find someone who did!

By all means use those notes. Otherwise, all the info you need is in the solution of the diff. eq. but it might be burdensome to plot.

 

[Jowin86]

Yeah we just took the voltages from the accelerometers, then I used them in the logarithmic decay method to find damping ratio etc for the first part..So I don't think we need velocity and displacement?

What I've got to do is use theory to predict the amplitude ratio and phase lag of the mass as a function of the forcing frequency ratio. Then compre the results with the test data. It also says for this test the system response is to be the ratio of, the output displacement of the mass, and the input displacement to the spring, and the phase lag will be considered too. I can't quite get my head around what that means though.

(btw is forcing frequency ratio ω, or is that just forcing frequency?)


Also to write a "1st order linear, constant-coefficient differential equation for the system, with the forced excitation as the forcing function", how is the forced excitation, or anything else for that matter, made the forcing function, or function of the equation? Is it that the equation will be something like x="loads of things multiplied and divided by the thing you want the function to be"?

I'm very new to this so I may not be making much sense or being vague so I do apologise! I appreciate the help :) Any further reading you could recommend to help with this would be good!

 

[rude man]

ω

Yeah we just took the voltages from the accelerometers, then I used them in the logarithmic decay method to find damping ratio etc for the first part..So I don't think we need velocity and displacement?

What I've got to do is use theory to predict the amplitude ratio and phase lag of the mass as a function of the forcing frequency ratio. Then compre the results with the test data. It also says for this test the system response is to be the ratio of, the output displacement of the mass, and the input displacement to the spring, and the phase lag will be considered too. I can't quite get my head around what that means though.

(btw is forcing frequency ratio ω, or is that just forcing frequency?)


Also to write a "1st order linear, constant-coefficient differential equation for the system, with the forced excitation as the forcing function", how is the forced excitation, or anything else for that matter, made the forcing function, or function of the equation? Is it that the equation will be something like x="loads of things multiplied and divided by the thing you want the function to be"?

I'm very new to this so I may not be making much sense or being vague so I do apologise! I appreciate the help :) Any further reading you could recommend to help with this would be good!

I noticed you said the system rsponse is considered to be the displacement of the mass compared to the displacement of the forcing function (scotch yoke). I was wondering how you propose to determine the instantaneous position of the mass, which is the output they apparently want. The accel will only give you the 2nd time derivative of displacement. This should be your first concern here.

You mentioned frequency and phase response graphs before. With those graphs and expressions you don't need to know how to solve the differential equation, it's already done for you. All you do is to compare your data with the graphs and expressions for amplitude and phase.

Of course, what will happen is that as your forcing function frequency ω gets close to the natural frequency ω0 of your system, the response will peak and the phase will be close to zero. Your charts should indicate that, as should your expressions. BTW ω is frequency, actually radian frequency which is frequency times 2∏, or ω = 2∏. (For example, your 115V outlets put out f = 60 Hz which is 377 rad/s. The frequency ratio you'll be using as your independnt variable is ω/ω0.

 

[Jowin86]

I think because we had an accelerometer on the input and the output, i.e the yoke and the mass, the voltage output of these is the same ratio as if we had actual displacement values. So the comparison between the two can still me made.

I have plotted my frequency response graphs, as expected the experimental results differ slightly from the theory. out of 4 tests i would say two of them are close enough to seem like the loss in the system due to friction etc could account for the difference. The other two however, both being around 0.9 frequency ratio (close to resonance) are massively different. The theory MF values are 3.9 and 3.66, and the experiment values are 1.3, and 1.4 for the same frequency ratio values.

Would you say that this is normal? My initial thoughts are either something is wrong here or there is something else happening. I watched a lecture online that talked about Beat response, am i right in thinking this could be the reason? It seems possible but i need to read up on it more and see if it applies to the situation i have here with the results..

 

[rude man]

As long as you stick to theoretical vs. empirical (experimental) displacement RATIOS AT A GIVEN FREQUENCY, you're right, accel ratio is same as diplacement ratio.

However, you seem to have graphed displacement vs. frequency, and that assumption is not correct.

Diplacement x = x0*sin(wt), acceleration d2x/dt2 = -w2*x0*sin(wt) = -w2*x.

'Beats' do not exist. The accel output always puts out the forcing frequency, and only that frequency.

I need to know (1) exactly what you graphed and compared with, and (2) what frequencies you ran. For eaxmple, the accel output: it's a sine wave, did you accurately determine the peak-to peak voltage, did you use a voltmeter or even a lock-in amplifier, or demodulator?

Your accel data should have been way more than displacement, relatively speaking, at the higher frequencies vs. at the lower.

 

[Jowin86]

1)

For the theory plot if you take a look at this pdf i found:

http://www.colorado.edu/engineering/CAS/courses.d/Structures.d/ASEN3112.Lect21.d/ASEN3112.Lect21.pdf

...on page 21-5, take a look at the equations in the box labeled (21.7). I have used the those to plot the theory by adding different values of "r" from 0.1 - 2.0. Accept i have in my notes that for the phase it is inverse tan so i did that.

For comparing these theory plots, i used the experimental data. This is 4 sets of data, each a set of two times and 4 voltages for the input peak and the output peak that follows it, (the four voltages are peak to peak) and also we had another time for the next input peak. From these i worked out the time period, lagging time of the output, amplitude input and output. then with that data i worked out the MF, phase lag and forcing frequency as follows:

MF=Amplitude output / amplitude input

phase angle=( lagging time of output to the input / forcing time period ) * 360

forcing frequency = (1 / forcing time period) * 2pie 2.)

We used an oscilloscope and recorded a small amount of the sine waves, then tracked along the display to the peaks and recorded values for the input and output sine waves as above.

the frequencies we ran at are roughly 28rads, 29rads, 44rads, 63rads. I don't know why the lecturer did 28 and 29 rads so close together though ?
Edit:

Just for reference, my graphs look pretty similar to figure 21.2 on that pdf for the lines of damping ratio=0.1... my damping ratio is 0.125

 

[rude man]

What sort of device is attached to the motor or whatever to measure displacement?

Was your 'scope input set to dc or ac? Scopes will sharply attenuate frequencies below a certain value if the input is ac-coupled, which could help explain your data.

What frequency gave you your highest MF?

Finally, give me your input and output peak-to-peak voltage readings for the lowest and the highest excitation frequencies you used.

 

[Jowin86]

Theres nothing attached to the motor to measure displacement, just an accelerometer to measure acceleration.

Im not sure if the scope input was ac or dc? Is 'scope input' a setting on the oscilloscope? I will ask my lecturer but I'm not sure what scope input is so not sure what to ask.

In the theory i got my biggest MF at 0.98 frequency ratio, i thought that was supposed to be at frequency ratio of 1?

the peak to peak input and output for lowest and highest are as follows:

Lowest excitation was 29rads:
Input peak to peak = 97.74mV
output peak to peak = 147.88mV

Highest excitation was 63rads:
Input peak to peak = 448.15mV
Output peak to peak = 95.1mV

the two lowest excitations are the ones that are significantly lower that theory states they should be.

 

[rude man]

Ah, so now I know. You're measuring input acceleration wheras I thought the sensor measuring input was measuring displacement. I see you had mentioned this already, sorry I didn't pick up on it sooner.

So now everything's fine. Your ratio is output accel/input accel which will be the same as output displacement/input displacement. So you can compare your results with the charrts you have.

You still didn't give me the frequency at which MF peaked, just the ratio.

The fact that you got a lower ratio than you expected at the lowest frequencies seems to corroborate my suspicion that your 'scope was ac-coupled. Yes, ac vs. dc coupling is a selectable input option on every 'scope I've ever seen, though some real low-end ones might not have that option. Check with your lab instructor. Check with me as to what to dowith your data if this turns out to have been the problem if you want.

The ratio of 0.98 is probably right-on. When damping is added to a system like yours, the frequency of max MF is lower than the natural frequency, which is √(k/m).

See http://en.wikipedia.org/wiki/Harmonic_oscillator for a good discussion.

 

[Jowin86]

Oh yeah sorry the frequency of the max MF was 31.17rads

Ok I'll send him an email, I should hear back tomorrow he's usually pretty good with replying. I haven't got time to read the wiki article till tomorrow but thank you very much!

By the way, is it really ok to discuss coursework on here? If it came up in the plagiarism checks they do I'm assuming it's just the same as discussing the problem with anyone in person or getting the answer from a textbook so it shouldn't be a problem?

 

[rude man]

Oh yeah sorry the frequency of the max MF was 31.17rads

By the way, is it really ok to discuss coursework on here? If it came up in the plagiarism checks they do I'm assuming it's just the same as discussing the problem with anyone in person or getting the answer from a textbook so it shouldn't be a problem?

Obviously, that depends on the view taken by your school.

This site is supposed to give students a hint, never an off-the-top solution. I myself don't even bother replying to anyone who hasn't evinced significant effort to solve the problem beforehand.

I believe this site is well known so if your school has a problem with it I would think they would have warned you. Obviously, you would not want to cheat on an in-class exam with your android using it!

 

[Jowin86]

Hi,

I have finally sorted the lower values out! Turns out if i use different method for working them out they come out about right.

I have however got two values on the phase lag vs frequency response graph that are odd. Two of the results are close to the theory plot at around r=0.8 to r=0.9, the other two are about 10 degrees lower than the theory. So the phase lag is less than predicted for the two tests where the system was ran at higher speeds.

Would this be something expected or is something wrong here? My thoughts are that the friction in the system is preventing the phase change making a full 180 degree swap. Am I on the right track there?

Also, I am so close to getting the equation of motion right for the system. Could you hint me in the right direction?

 

[rude man]

I forgot - is the mass dangling against gravity or on a frictionless horizontal plane? Since you don't have g in your equation I'll assume it's on a horizontal table or similar.

Now, as to your equation - I don't get the same thing, so could you go thru the steps you took to arrive at it?

In parti cular, where did the cwy0cos(wt) term come from?

 

[Jowin86]

the mass is dangling against gravity

 

[rude man]

Then where's your g term?

Like I said, need to see your derivation of your whole equation, starting from F = ma.

Confession: whether or not you include g isn't going to affect your data anyway. It will just bias x(t) by a constant value = Mg/k.

 

[Jowin86]

sorry my e-mail notification only showed me the first line of your reply so didn't see you assigning about the cwy0cos(wt) term...

here's my working, i haven't included gravity because like you say it won't affect the data. My thinking with this was, the force is going through the spring which absorbs or dissipates some of that force, then into mass which is damped my the damper. Something my lecturer said makes me think the spring part goes on the massxaccel side of the equation..

 

[rude man]

OK -all looks good except two things:

re: your 2nd equation:
1. why the cy' term?
2. double-ckeck the k(x-y) term?

 

[Jowin86]

Ok I'll have another look at that. I put c(x-y) because I had an example like that in the notes for base excitation, but the force was going through the damper on that one so maybe it's not needed for this model?

I though pt the -k(x-y) term was ok, guess I'll have to look,

P.s.

What do you think about the two values on the graph that are below the blue line (theory plot)

 

[rude man]

Ok I'll have another look at that. I put c(x-y) because I had an example like that in the notes for base excitation, but the force was going through the damper on that one so maybe it's not needed for this model?

Look at it this way - what if y(t) goes all over the place but x stays fixed - what would the damper do then?

I though pt the -k(x-y) term was ok, guess I'll have to look,

Hint: It's close. check the sign.

P.s.

What do you think about the two values on the graph that are below the blue line (theory plot)

I'll look at that later. That graph is your lab data, not a simulation based on your equation, right?

 

[Jowin86]

The graph is both lab data and simulation. The blue line is a plot of the calculated values (using theory) and the 4 symbols are the lab data plotted on there to compare to the calculated values

For -k(x-y) then, If the force compresses the spring then it dissipates some energy before passing it into the system, but the spring then has potential energy according to the amount of displacement which it passes into the system... so maybe:

K(x-y)

= Kx-kYoSinWt

So that would be

mx+kx-kYoSinWt = not figured this side out yet lol

Unless the force y(t) isn't dissipated it's just passed on plus the potential energy of the spring, in which case that would be k(x+y).

 

[Jowin86]

Oh, if y(t) went all over and x stayed still the damper would do nothing.. So on the other side of the = I just have -cx(dot) right?

 

[rude man]

Oh, if y(t) went all over and x stayed still the damper would do nothing.. So on the other side of the = I just have -cx(dot) right?

Absolutely right!

 

[rude man]

The graph is both lab data and simulation. The blue line is a plot of the calculated values (using theory) and the 4 symbols are the lab data plotted on there to compare to the calculated values

For -k(x-y) then, If the force compresses the spring then it dissipates some energy before passing it into the system, but the spring then has potential energy according to the amount of displacement which it passes into the system... so maybe:

K(x-y)

= Kx-kYoSinWt

So that would be

mx+kx-kYoSinWt = not figured this side out yet lol

Unless the force y(t) isn't dissipated it's just passed on plus the potential energy of the spring, in which case that would be k(x+y).

Don't be concerned with potential energy. We're dealing strictly with F = Ma.

Now - remebering that downward is +, and that we're writing an equation describing the motion of M, what force is applied to M if I give a + displacement to y, keeping x constant, and also vice-versa? Remember my hint, you were almost there.

If the blue line is theory then it will be wrong if it's based on your previous equation, so first order of business is to get that equation right. If you're using system simulation software like SIMULINK we'll have to think some more.

 

[Jowin86]

Yesssssss!

am i right about the other side? In which case the eom would be:

mx+kx-kYoSinWt=-cx (with the appropriate dots above the x's)

 

[Jowin86]

ignore the previous post I hadn't read your new post.


erm, if down is + then i'd say +. so it'd be:

mx+kx+kYosinWt=-cx

 

[rude man]

Yesssssss!

am i right about the other side? In which case the eom would be:

mx+kx-kYoSinWt=-cx (with the appropriate dots above the x's)

The sign of Ky0sin(wt) affects the phasing of x''/y''. +y(t) imparts a force to the spring which in turn tends to make M go positive. So +Ky is a positive force on M and that's the equation we're concerned with: sum of forces on M = M times M's acceleration.

 

[Jowin86]

mx+kx+kYosin(Wt)=-cx is the eom then?

It seems to make sense that way. Our lecturer didn't really teach us how to do this, he just did about 2 example and never said how he got the eom. :-/ guess that's uni for you hey.

Thank you for your guidance :)

Now i just have to sort the two odd values for phase lagg :'(

 

[rude man]

mx+kx+kYosin(Wt)=-cx is the eom then?

It seems to make sense that way. Our lecturer didn't really teach us how to do this, he just did about 2 example and never said how he got the eom. :-/ guess that's uni for you hey.

Thank you for your guidance :)

Now i just have to sort the two odd values for phase lagg :'(

No, the sign on Ky0sin(wt) is - if you put it on the left side of the eom.

Write the equation as mx'' = Ky0sin(wt) - cx' - Kx so it looks like ma = F. Then try to understand why each term's sign is + or -. Makes all the difference! For example, if either the cx' or Kx term were poitive per the above eom you'd get an exploding system! Those terms would represent what engineers call positive feedback which means the bigger the input, the bigger the output, feeding on itself until boom!