# Forced oscillations

1. Dec 12, 2003

### StephenPrivitera

For forced oscillations we have
$$\frac {d^2x} {dt^2} = -\omega_N^2x+\frac {F_0} {m} cos\omega_Ft$$
The solution is
$$x(t)=\frac {F_0} {m(\omega_N^2-\omega_F^2)}cos\omega_Ft$$
This doesn't seem to reduce to oscillations where F0=0. Shouldn't it?

2. Dec 12, 2003

### Integral

Staff Emeritus
If a spring is not given an initial displacement or initial velocity it will not oscililate, this appears to be a driven, undamped spring. Why would you expect it to oscillate with no driving force. If you solve the Diff Eq with either an inital displacement or velocity, in addtion to the driving function, then you shold see some terms representing non driven oscillations.

3. Dec 12, 2003

### StephenPrivitera

If you solve the Diff Eq with either an inital displacement or velocity, in addtion to the driving function, then you shold see some terms representing non driven oscillations.

Ok, That's relieving. So it does reduce.

How would you write the equation of motion with an initial displacement?

If the initial displacement is zero how come x(0) isn't zero (above)?

Last edited: Dec 12, 2003
4. Dec 12, 2003

### Integral

Staff Emeritus
I would guess it is not zero at t=0 because you are driving with a cos function.

To include initial conditions you specify x(0)= X or x'(0)=V Then solve the Diff Eq. The inital condions determine the 2 constants of integration that will appear in the solution.

The above solution is for a systme that is being, and HAS BEEN, driven by the given function. It does not start at rest, it is in motion as determined by the driving function.

5. Dec 12, 2003

### StephenPrivitera

It is very hard for me to understand something if I cannot derive it. I'm going to have to dig through my calc book to do this one.
I think in any event my professor is more concerned with the qualitative features of a driving force.
This is very confusing.
Is F always acting to pull the body away from equilibrium?
Why would you need to drive a harmonic oscillator? If it's not damped then once you start it going it should just keep going. If you keep applying a force shouldn't the amplitude get larger and larger with time?

6. Dec 12, 2003

### Integral

Staff Emeritus
I do not have my texts at hand (posting from work ;) ) So everything I write is off the top of my head. I am not sure about the definiton of your $$\omega_N$$ this should be the natural frequency of the system, there also should be a more fundamental expression of it. The consist of the spring constant and perhaps another factor, which is not comming to me. I believe damping generally consists of a lossy term in the basic Diff Eq which is porportioal to the velocity. Are you not given any more information about the equation you are given?

Consider what happens as

$$\omega_N \rightarrow \omega_F$$

That is if the Natural frequency equals the Driving frequency.

7. Dec 12, 2003

### dhris

Hang on, that x(t) is not the general solution to that equation. That is only the particular solution. You have omitted the homogeneous part:
$$x(t)=A \sin \omega_N t + B \cos \omega_N t + \frac{F_0}{m(\omega_N^2-\omega_F^2)}\cos \omega_F t$$
A and B are determined from the initial conditions. Your equation is the solution for certain initial conditions but is not the solution in general. As you can see $$F_0=0$$ does lead to oscillatory solutions in the general case.

dhris

8. Dec 12, 2003

### StephenPrivitera

No, I'm just given that a driving force is acting on the system. In my book it talks about driving forces in the context of couteracting the effects of damping. In class (where the above equation came from), we talked about driving forces without damping. But without damping the amplitude should go larger with time.

The fact that the amplitude goes huge when the two frequencies are close was his main point.

9. Dec 12, 2003

### Integral

Staff Emeritus
This is correct. I have being assuming that the initial conditions had been such that A=B=0 leaving only driven solution. If there is a damping term, the Solution to the homogenous equation will die out in time so what remains is only particular solution. This is called the steady state solution.

10. Dec 12, 2003

### StephenPrivitera

I still don't see how a driving force without damping won't cause the amplitude to keep increasing. Are you not continually doing positive work on the system and thus adding energy to the system?

11. Dec 12, 2003

### dhris

Why do you think the driving force always has to work constructively? In some cases it does, as when you approach the natural frequency in your example. In other cases it won't. Imagine I'm pushing a child on a swing. I can give him a couple of pushes to get him going, then I can hold out my arms and stop him. That's as much a "driving" force as anything else.

dhris

12. Dec 12, 2003

### Integral

Staff Emeritus
The solution you have been given is for LARGE time, it does not apply at t=0 for a real system. This is the steady state solution, and is valid only have all transitory effects, that is those due to the initial conditions have DAMPED out. I emphasized damped because this is supposed to be the Undamped solution. According to my text this solution is valid for t>> 1/(damping constant) in your case the damping constant = 0 so that implies that this solution in only good for $$t >> \infty$$ So this is the equation of motion for a body that has be oscillating forever.

We have a single frequency, that of the driving force, and an amplitude which is a function of the natural frequency and the driving frequency. This means that the system will settle into a uniform frequency and amplitude motion, given enough time. The system will oscillate that the driving frequency. IF the driving frequency is near the natural frequency then the amplitude of the oscillations will increase becoming infinitely large when the driving frequency equals the driving frequency.

So the key to understanding this equation is that it is only valid for systems that have been in motion for a long time. Understand, that due the periodicity of cos you can get meaningful results for any value of t, your are free to define t=0 but this does not represent the beginning of the oscillations, the equation represents a body that has been in motion for an infinitely long time.

13. Dec 12, 2003

### turin

You need to rephrase the above from:

"The solution is..."

to

"A solution is..."

This was basically pointed out by dhris:

It is not that you're solution assumes the system has been functioning for a long time; this is one explanation. In general, you must apply initial conditions to the solution to get "the" solution for the particular physical situation, even if the system is being driven; there is still a nontrivial response. It just happens that, in disappative systems, after what engineers call steady state is achieved, the dynamics can be characterized by the input, as you have written.

Last edited: Dec 12, 2003