## 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

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

Cheers

Greg Locock

## RE: harmonic forcing under natural frequency

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 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

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

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

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

## RE: harmonic forcing under natural frequency

## RE: harmonic forcing under natural frequency

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

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

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

## RE: harmonic forcing under natural frequency

Cheers

Greg Locock

## RE: harmonic forcing under natural frequency

## RE: harmonic forcing under natural frequency

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

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

Cheers

Greg Locock

## RE: harmonic forcing under natural frequency

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

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

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

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

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

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

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

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

## RE: harmonic forcing under natural frequency

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.