×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!
  • Students Click Here

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here

Jobs

harmonic forcing under natural frequency

harmonic forcing under natural frequency

harmonic forcing under natural frequency

(OP)
for a second order differential equation,( mass-spring-damping) system, if we have harmonic force acting on the system, we can find the solution x(t), but how about if the frequency of harmonic force is the natural frequency?  there will be resonant. we know it won't go to infinity because of damping.  Are there any formula availabe for the response x(t) of the system under this kind of  force?  Thanks!

RE: harmonic forcing under natural frequency

Here's what I think:

at resonance the mass accereration and spring force terms cancel.

We are left with F=C d/dt(X)
Assume X has the form X0sin(w0*t) where w0=sqrt(k/m)

F=C sqrt(K/M)X0 cos (w0t)

The magnitude of peak displacement X0 will be
X0=F/[C*sqrt(K/M)] (where F is peak force).  

The phase of the displacement will be lagging F by 90 degrees.

RE: harmonic forcing under natural frequency

That's the fastest derivation of that I've seen, there is a slight niggle in that it doesn't quite predict the maximum amplitude, as your assumption for the resonant frequency (an admittedly nebulous term) is for the undamped system. The answer you get is close enough for all practical purposes, I suspect.

Cheers

Greg Locock

RE: harmonic forcing under natural frequency

(OP)
thanks for the reply!
what I want is the analytical solution to the harmonic forcing at the natural frequency, find x(t)=F(t,c,m,k), which can be drawn against the time t,and the amplitude of x will go from zero to a certain value, state state. is that easy, I haven't find any book have the discussion on this topic.

RE: harmonic forcing under natural frequency

I agree with Greg's comment that the solution is approximate but will be pretty close for lightly-damped systems (probably as close as our limited ability to estimate the actual damping of a real-world system warrants).

I think we have given you what you asked for. The functional form of the solution is a sinusoid. The magnitude and phase have been discussed.

If I get a chance I'll try to work on the more precise solution... which won't change the functional form.

RE: harmonic forcing under natural frequency

Here is an attempt at the more exact solution... with a little help from my computer (Maple). Hopefully it comes out legible.

Differential equation:

f(t)=m*diff(x(t),t,t)+c*diff(x(t),t)+m*x(t);
                       / 2      \
                       |d       |     /d      \
              f(t) = m |--- x(t)| + c |-- x(t)| + m x(t)
                       |  2     |     \dt     /
                       \dt      /

Transfer function H(w)=X(w)/F(w):
> H(w):=m*(I*w)^2+c*(I*w)+k;

                                  2
                      H(w) := -m w  + I w c + k

Magnitude of transfer function:
> Hmag:=sqrt((k-m*w^2)^2+(w*c)^2);

                        2          2    4  2    2  2 1/2
              Hmag := (k  - 2 k m w  + w  m  + w  c )

Derivative of magnitude of transfer function wrt w
> Hmagdiff:=diff(Hmag,w);

                                           3  2        2
                             -4 k m w + 4 w  m  + 2 w c
          Hmagdiff := 1/2 ----------------------------------
                            2          2    4  2    2  2 1/2
                          (k  - 2 k m w  + w  m  + w  c )

Find the resonance at point where Hmag is max (deriv=0):
> wr:=solve(Hmagdiff=0,w);

                                           2 1/2
                               (4 k m - 2 c )
                     wr := 1/2 -----------------
                                       m

> Hmagresonance:=subs(w=wr,Hmag);

                 /                       2                     2 2
                 | 2       k (4 k m - 2 c )        (4 k m - 2 c )
Hmagresonance := |k  - 1/2 ---------------- + 1/16 ---------------
                 |                m                       2
                 \                                       m

                       2   2\1/2
           (4 k m - 2 c ) c |
     + 1/4 -----------------|
                   2        |
                  m         /

> Hangle_resonance:=arctan(Im(H(w))/Re(H(w)));

                                            2
                                      Im(m w  - I w c - k)
           Hangle_resonance := arctan(--------------------)
                                            2
                                      Re(m w  - I w c - k)
The solution X will have a magnitude given by the magnitude of H at resonance times the magnitude of F.  The angle of X with respect to F is given by Hangle_resonance above.

RE: harmonic forcing under natural frequency

Ah, I think hechengli perhaps wants to see the transient response as the harmonic force signal is applied? Obviously in the steady state it will tend towards your answer.

That is, he may be interested in the response to the excitation F(t)=if(t<start,0,sin(w*t)), where start>0 and w=sqrt(k/m)

if so that is a lot more interesting. So hechengli - do you want the transient response or the steady state response, which pete's already done for you? incidentally, are we doing your homework for you?

Cheers

Greg Locock

RE: harmonic forcing under natural frequency

Those computer-equations didn't work. Let me use the other format:
Differential equation:

f(t)=m*diff(x(t),t,t)+c*diff(x(t),t)+m*x(t);f(t) = m*diff(diff(x(t),t),t)+c*diff(x(t),t)+m*x(t)

Transfer function H(w)=X(w)/F(w):
> H(w):=m*(I*w)^2+c*(I*w)+k;
H(w) := -m*w^2+I*w*c+k

Magnitude of transfer function:
> Hmag:=sqrt((k-m*w^2)^2+(w*c)^2);
Hmag := (k^2-2*k*m*w^2+w^4*m^2+w^2*c^2)^(1/2)

Derivative of magnitude of transfer function wrt w
> Hmagdiff:=diff(Hmag,w);
Hmagdiff := 1/2/(k^2-2*k*m*w^2+w^4*m^2+w^2*c^2)^(1/2)*(-4*k*m*w+4*w^3*m^2+2*w*c^2)

Find the resonance at point where Hmag is max (deriv=0):
> wr:=solve(Hmagdiff=0,w);
wr := 1/2*(4*k*m-2*c^2)^(1/2)/m

> Hmagresonance:=subs(w=wr,Hmag);
Hmagresonance := (k^2-1/2*k/m*(4*k*m-2*c^2)+1/16*(4*k*m-2*c^2)^2/m^2+1/4*(4*k*m-2*c^2)/m^2*c^2)^(1/2)


> Hangle_resonance:=arctan(Im(H(w))/Re(H(w)));
Hangle_resonance := arctan(-w*c/(m*w^2-k) (I think)

The solution X will have a magnitude given by the magnitude of H at resonance times the magnitude of F.  The angle of X with respect to F is given by Hangle_resonance above

RE: harmonic forcing under natural frequency

wr has the pleasing property that it reduces to sqrt(k/m) when c=0.

RE: harmonic forcing under natural frequency

.... but the unpleasant property that with c=0 Hangle=arctan(0)=0.... should be pi/2. You'll have to find the error yourself.

RE: harmonic forcing under natural frequency

(OP)
thanks again!
what I want to see is the transient response of the system under the harmonic force at its natural frequency. Not the steady state.  When I do experiment, I found the magnitude goes from zero to steady state, so I am kind of interested in find the exact solution for a damped system, maybe transient response as Greg Locock said.

RE: harmonic forcing under natural frequency

This problem has a very ugly analytical solution, the step function in the forcing makes it a tricky case. I have a strong suspicion that a Laplace transform will be the easiest analytical approach. Note that this forcing function is not really a harmonic excitation, you get a fair bit of high frequency stuff at the switch-on.

My Excel simulation in the time domain looks plausible, but seems to be giving the wrong answers. grr

Cheers

Greg Locock

RE: harmonic forcing under natural frequency

Ok, I found there error in my steady state solution. I typed in H(w):=m*(I*w)^2+c*(I*w)+k... when that is actually the inverse of H(w).  Fortunately I still found the correct wr when I set d/dw=0 I found the min of the inverse which is still the max of the transfer function.  All that remains for the steady state solution is to substitute my value of wr into the correct H(w)=1/[m*(I*w)^2+c*(I*w)+k].

The transient solution as Greg said would probably best be done by Laplace.  If I get a chance I'll let my computer try.  It would of course be easier if you have in mind specific values of m, c, k and the phase theta of the force F=F0cos(wrt+theta) at the moment it is "turned on".

RE: harmonic forcing under natural frequency

Every vibrations book ever published has the closed form solution to this problem derived,rederived and rerederived 8000 times over..go to your nearest techicial library and check it out.

RE: harmonic forcing under natural frequency

Perhaps it does (tho I don't remember seeing the transient behaviour in an analytical form). But this is fun, for a strange value of fun.

Cheers

Greg Locock

RE: harmonic forcing under natural frequency

Maple tackled the transient problem pretty easily with laplace transforms.  The solution as Greg said is somewhat ugly, but when plotted with sample values the parameters it looks reasonable. It's somewhat more cumbersome than what I posted above. I'll post it in same format if you want.  In meantime I'll try to figure out a way to create a word file that captures the equations without mangling them.

RE: harmonic forcing under natural frequency

I've spent some time playing with the transient solution and it is very uninteresting to me. Assuming I start with x(0)=0 and v(0)=0 and a lightly damped system, then the highest magnitude reaches is at steady state. Changing the angle of application has little overall effect.  As I increase the damping the steady state level decreases and also the time to reach the steady state level also decreases. I was a little surprised at this since I expected damping to "slow down" the time to reach steady state... but not so.

If you are only interested in the maximum value and the system meets conditions above (lightly damped and initial conditions are zero), then all you need to study is the steady state solution.  By the way the steady state solution magnitude matched that predicted in my first post.

I'm still working on getting my transient solution into word.

RE: harmonic forcing under natural frequency

Umm, I'll see your Laplace transform and raise. It should overshoot in the first couple of cycles, Shirley?

My Excel simulation does anyway (tho I'm not happy with it).

Cheers

Greg Locock

RE: harmonic forcing under natural frequency

Greg - you may be right but my simulation didn't show any overshoot whatsoever.  I'm guessing that it takes time to put energy into the system.  
One set of parameters I used (see attached) were k=1,m=1,c=0.1,theta=0.3.  By the way with Fmax=1 this gave max displacement of 10 as predicted above.  I tried several others that I thought were representative and saw similar behavior. But certainly could be my model is wrong or my parameter selection not good.

Here's my file:
http://geocities.com/pschimpf/mass_spring_transient_link.htm

If you are interested to email me your excel file then go the webpage listed on my profile and then near the bottom link to my electricpete@.... email.  (can't list it here at the risk of getting spam-bombed).  Also if you like I can post your excel file for others.

I agree it's fun. It is too few times during real work that I get to use those types of tools.

RE: harmonic forcing under natural frequency

Greg
I've been trying to think of physical examples which might shed some light on whether overshoot is expected for lightly damped system with resonant sinusoidal excitation applied suddenly.

I can think of one that seems somewhat close. If you hold one end of a heavy flexible yardstick you can get it to bend with only a slight movement of your wrist... at the right frequency. But when you first start even at the precise frequency there is no bending until you have given the amplitude a chance to build up.  Not sure if that's exactly right (I think my forcing function is displacement at my end of the stick, not force at my end) but it seems close.

The electrical circuits I am familiar with that generate an overshoot are first-order systems. For instance a voltage applied to a series inductor/resistor circuit can cause current to go twice as high as the steady state value due to the decarying dc offset transient compoent.

An 2nd-order LC circuit generating an overshoot?  No examples pop into my head offhand (but that doesn't mean there aren't any).

Can you think of any examples?

I for one have a hard time imagining an every-day life examples where resonance occurs and the forcing function is "force". Part of the problem is it's tough for me to imagine applying force to something that's moving.

If you are pushing a child on a swing (a pendulum but similar to a mass spring I think), what is the best part of the swing cycle to push forward?  Intuition tells me that it's when the child is all the way back. But if I apply resonance theory I would push when the swing is directly vertical (at it's lowest point in travel).   That would equate to a forward peak in force occuring at a point in time 90 degrees ahead of the forward peak in travel.  (force leads displacement by 90 degrees in resonance). But it's darned tough to push at the bottom because the kid is moving so fast. Intuition fails me.

Any comments.

RE: harmonic forcing under natural frequency

Please don't attack my kid-on-a-swing example. When I say "intuition fails me"... it meant that this example fails to act like a resonant system.  (I know it is easier to push at the top because physically you can apply the force longer while the child is moving slowly at that point.....unrelated to the resonance situation I am trying to model).

I have a modification of the yardstick example that I think makes it work.  You are not holding the yardstick in your hand... the yardstick is rigidly clamped sticking out of the the ceiling or wall (cantilever).  You apply a force (not displacement) very near the clamped end of the yardstick where it's not moving. If the yardstick is at rest and you start tapping it with constant force, you can get it to start swinging wildly... but not at first, only after repeated taps.  You can also try to apply sinusoidal force with same result.

RE: harmonic forcing under natural frequency

Greg - hope you don't mind me beating a dead horse on this one. I checked my calcs and I'm pretty sure you will come around to my way of thinking if you set up your differential equation the same as mine. And I'm also pretty sure you know a helluva lot more about resonant behavior than I do... but sometimes like all of us it takes awhile to get to the right answer.

I agree with you that on a dc step input there would be an overshoot of the final dc value. I can't exactly apply that to this suddenly applied ac input signal which I can't decompose into a sum of step dc and ac input signals.  I can only think about it intuitively that the step response oscillates around a dc operating point which will be much lower than the peak level of the steady state response.  I base this statement on the frequency response curve.... which will have the amplification much higher at the resonant frequency than the dc (zero frequency). So if there is any overshoot to the portion of our input that acts like a step-dc-like input... it is still much lower than the final ac value.  

What do you think? Am I close or out in left field?

RE: harmonic forcing under natural frequency

I learned something new to me when exploring various cases with my computer model.

There are two classes of 2nd-order underdamped (less than critical) systems to consider, with the dividing line being the 45 degree angle from the imaginary axis in the complex plane.  If a pole lies to-the-right of this line (closer to the imaginary axis=less  damped), then the frequency plot will show a resonant peak.  If a pole lies to-the-left of this line (closer to the real axis... more damped... closer to critical damping), then the frequency response will resemable a low-pass filter... with the only peak being at frequency=0 (no resonant peaks).

I stumbled into this by accident because my program would not work for the highly-damped cases.  That is because the formula for wr breaks down in the area between critical damping and the 45-degree line. There is no maximum in the frequency spectrum (other than w=0).

For the resonant systems we are discussing, the system should be less damped than those systems on the 45 degree line.

RE: harmonic forcing under natural frequency

My 45-degree terminology may not be familiar to everyone.  I know some others talk about "damping factors", but I am not familiar with that terminology.

An example system on the 45 degree line would be m=1, c=sqrt(2), k=1. The roots are -c/2m +/- I * sqrt(4ma-c^2)/2m = -sqrt(2)/2 +/- I * sqrt(4-2)/2 = -sqrt(2)/2 +/-I*sqrt(2)/2.  The real parts and imaginary parts are equal, creating the 45 degree angle from origin in the complex plane.

RE: harmonic forcing under natural frequency

OK I ran it through mathcad and I found that the example I gave does not overshoot, according to their solver, ie my Excel model is faulty. So you were right.

That's fine at resonance, but would you believe that at c=.2, m=k=1, w=2, the mathcad solution does overshoot? Admittedly the amplitude is less, but it is still greater than the ss response.

Having got a model that does behave like you were predicting, I then had a look at how the amplitude builds up at resonance for the undamped case... and got quite a surprise (which was a bit less surprising on further thought, but is even more surprising on further further thought). The amplitude ramps up linearly, well ok, each cycle we are adding energy, so the PE at max amplitude is getting bigger each time. PE is proportional to x squared, F is constant, x is increasing, so we are adding more energy with each cycle, by F.x *some function. Makes sense, except that we are not adding energy since F and x are in quadrature. I shall mull that one over tonight.

i'm going to debug that excel thing now that I've got some better numbers to compare it with.

Incidentally all this malarkey in the time domain is probably a bit unnecessary for the original poster, his best bet is to get the frequency domain transfer function of his system (trivial, in all the books), FFT his drive signal, munge the two together and IFFT the result.

Cheers

Greg Locock

RE: harmonic forcing under natural frequency

Good point Greg, and some good thoughts.

I believe that we are adding energy each phase when F leads X by 90 degrees.  Because V=d/dt(x) also leads X by 90 degrees and therefore will be in phase with F.

Power=F(t)v(t) will be always positive if F and v are in phase.  
something like P=F0*V0*cos^2(wt)=F0V0(.5+.5cos(2w0t)
where F0 and V0 are peak amplitudes of F(t) and V(t)

I am not too surprised that there is an oversheet for your parameters with w=2*wresonant.  In my mind the ballpark predictor of overshoot is the quantity H(w)/H(0) where w is the exciting frequency. That is based on previous discussion that the dc-step behavior produces overshoot and ac component does not.  If the DC response exceeds the sinusoidal response, we might expect overshoot. For your parameters the dc response H(0)=1 and the ac response H(2)=0.33

RE: harmonic forcing under natural frequency

.... and H(1)=5 = 1/c of course.

RE: harmonic forcing under natural frequency

Greg,
And since you have admitted that I was right, I had better return the favor and admit that you were right.

I was able to generate about 5% overshoot using the following parameters:
k=1,m=1,c=1.40,theta=2.8

That lies just to the right of the boundary of the resonant class of systems that I discussed above (boundary point is k=1,m=1,c=1.414=sqrt(2)). And it doesn't overshoot for any values of theta except very close to theta=2.8 (see my word file for definition of theta).

My explanation.. Moving toward the boundary (more damping)  increase the dc response and decrease ratio of H(wr)/H(0), pushing towards overshoot.

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members! Already a Member? Login


Resources

White Paper – Choosing the Right Spring Loaded Connector
In today’s cost-sensitive world, designers are often driven to specify the lowest cost solution for every aspect of their designs to ensure that their solution is competitively priced and their company remains profitable. However, specifying a low-cost, low-quality connector solution can result in premature failure, considerable re-work costs and damage to reputations. Download Now
eBook – Own the Lifecycle: Sustainable Business Transformation
Increasingly, product and services companies are seeking more information and control in the operational lifecycle of their products, including service and use. Better information about the operational lifecycle, and the ability to use that information, requires more than just unstructured data flowing back from products in the field. Download Now

Close Box

Join Eng-Tips® Today!

Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.

Here's Why Members Love Eng-Tips Forums:

Register now while it's still free!

Already a member? Close this window and log in.

Join Us             Close