Oscillation Question: Mean Value of x with Constant Force P

[skaboy607]

Homework Statement

What would be the mean value of oscillation x if f(t)=P. Where P is a constant force.

Homework Equations

Not known.

The Attempt at a Solution

Absolutely no idea where to start, happy to just be pointed in the right direction.

Thanks

 

[Defennder]

?? What is the setup and the corresponding assumptions? You left out a lot of details. You could start off by writing out the DE first then after that we could help you.

 

[skaboy607]

Thanks for replying. Thats it I'm afraid, that is all the question gives so I can't give anymore information. Ok, how would I do that, t is variable right?

 

[dx]

There's no way that's the whole question. What is f?

 

[skaboy607]

I think it is. Here is a link to it, its question 1:

http://i423.photobucket.com/albums/pp315/skaboy607/Questions.png

Does it make more sense if its read in conjunction with question 1?

 

[dx]

Yes, it does make more sense if read in conjunction with Q1.

 

[skaboy607]

ok, I am still stuck what should be first steps to figuring out what it means and how to answer it?

 

[dx]

You'll have to show some kind of attempt. Even if you don't know how to do it, give me some of your thoughts on the problem. What is the question asking you? How do you think you could solve it?

 

[skaboy607]

ok. Well I've done question 1 by drawing free body diagrams, and then writing the equations of motion for that mass. I could replace f(t) with P in that equation. Thinking about that system, I know that if a graph of data was collected after the force P was applied, then I think it would plot a sin wave. I don't know what the mean value of oscillation is but at a guess, I would say the height of the wave but then I'm not sure as that would be amplitude?

 

[dx]

There's damping, so the oscillation will not be a sine wave.

Here's a huge hint: combine the constant force term C with kx like this:

kx - C = k(x - (C/k)).

 

[dx]

The mean value is just the center of oscillation. Do the simpler case first: what would the center of oscillation be if the equation was

m\frac{d^2 x}{dt^2} + c\frac{dx}{dt} + kx = 0

?

 

[skaboy607]

Oh ok so damping-no sin wave, no damping-sin wave.

Right definitely lost now. Is this what youve done.

mx(dbl dot)+c(dot)+kx=P
mx(dbl dot)+c(dot)+k(x-p/k)=0

hmmm I haven't got a clue?

Thanks

 

[dx]

Now, change variables to X = x - P/k. What do you get?

 

[skaboy607]

hmm

mx(dbl dot)+cx(dot)+kX=0

 

[dx]

Right. Now what is the relationship between (d²x/dt²) and (d²X/dt²) and between (dx/dt) and (dX/dt)?

 

[skaboy607]

-(omega)^2*sin(omega)t and (omega)*cos(omega)t

Is that right...?

 

[dx]

What's that? Just differentiate X = x - P/k.

 

[skaboy607]

I was clutching at straws. What is the big X suppose to represent?

 

[dx]

It's just a new variable that we're introducing so that the problem changes into another problem whose solution we already know.

The derivatives of X are clearly equal to those of x because they only differ by a constant. Is that clear? So the equation turns into the following equation:

m(d²X/dt²) + c(dX/dt) + kX = 0.

Now this is the equation without the driving force and you already know that the mean value of X is <X> = 0. But <X> = <x> - (C/k). Substitute <X> = 0 in this, and you will get <x> = C/k.

 

[skaboy607]

hmmmmmmmmm, sorry i don't really get it, know anywhere where I can read up about this?

 

[skaboy607]

I guess what I'm saying is if we know that the position of equilibrium is always 0 for a damped oscillator why is there an equation for it?

Thanks

 

[dx]

The equation is the equation of motion. If you start off the oscillator by pulling it a little and giving it some initial velocity, the equation will determine how it moves after that.

 

[skaboy607]

Sure but I've read and you said in your post yesterday that force/k is equal to the mean value of oscillation. Which is always 0 for a damped oscillator right? So I guess I don't really see the point of the question that I was asked 'what is the mean value of oscillation?'

 

[dx]

It is zero for a damped oscillator without the constant force term P. When there is a constant force term, the equilibrium position is P/k.

 

[skaboy607]

Ohh I see, so is the case of velocity and acceleration equal to 0 the same for a system with forced vibration?

Shameful as it may seem, I didnt undersand this that you said yesterday.

'The derivatives of X are clearly equal to those of x because they only differ by a constant. Is that clear? So the equation turns into the following equation:'

 

[dx]

We defined X as X = x - P/k, so

\frac{dX}{dt} = \frac{dx}{dt} - \frac{d}{dt}\frac{P}{k} = \frac{dx}{dt}.

i.e.,

\frac{dX}{dt} = \frac{dx}{dt}

Because P/k is a constant and ^d/_dt (P/k) = 0.

Similarly with the second derivative.

 

[skaboy607]

Ok so I am with you to here,

m(d²X/dt²) + c(dX/dt) + kX = 0. (same as eq with no force) where mean value of x=0.

But lost on the next bit. Also, is x representing our mean value of oscillation? I thought that was the distance say mass1 moved with respect to a datum (say the floor)

 

[dx]

So we've transformed the original equation to m(d²X/dt²) + c(dX/dt) + kX = 0.

But this is just the equation whose equilibrium position is X = 0. But X = x - P/k. So if X = 0, that means (x - P/k) = 0 i.e. x = P/k.

But lost on the next bit. Also, is x representing our mean value of oscillation? I thought that was the distance say mass1 moved with respect to a datum (say the floor)

x is just the position of our mass.

 

[skaboy607]

Right got it! Thank you very much for your help!

So does x always represent the mean value of oscillation as at the top of this thread, you said it was just another variable introduced.

Thanks again.

 

[dx]

X (capital x) was the variable we introduced. x (small x) is the variable in the original equation which represents the position of the mass. What we've shown is that when the position is x = P/k, the mass will be in equilibrium.

 

[skaboy607]

Right got it! Thank you very much for all your help. And patience.

 

[dx]

No problem.