Forced Driven oscillator

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  • #1
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Homework Statement



http://fatcat.ftj.agh.edu.pl/~i7zebrow/rysunek.jpg
tring constant is k,
object mass is [tex]M_{1}[/tex]
Damping friction is b
and we wiggle the top end of spring in the above diagram with amount Asin(wt)
(Where A is a amplitude and w is a frequency).

Homework Equations


Spring Force:
[tex]Fs=-kx[/tex]
Damping force is:
[tex]Fb=-b\frac{dx}{dt}[/tex]

The Attempt at a Solution


I don't need full solution.I just looking for differential equation for this problem.My attempt is:
[tex]M_{1}\frac{d^{2}x}{dt^{2}}+b\frac{dx}{dt}+kx=Asin(\omega t)[/tex]
I am sure about left side of the eguation but the right side is propably wrong I hope that someone can tell me where I made an mistake.
 
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Answers and Replies

  • #2
671
2
Look at the dimensions, you are comparing force with distance. That is dimensionally inconsistent!

Consider the extension of the spring as a function of time, how does the oscillation of the end-point affect it? How does this translate into the force the spring exerts on the mass?
 
  • #3
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Do you mean that the force of spring should looks like this:

[tex]Fs=-k(x-Asin(wt))[/tex]
and now the differential equation will be:
[tex]
M_{1}\frac{d^{2}x}{dt^{2}}+b\frac{dx}{dt}+kx=k*Asin( \omega t)
[/tex]
Now the differential is dimensionally consistent.But, is that correct??
 
Last edited:
  • #4
671
2
That's how I would approach the problem, you might want to wait for confirmation from someone else that that's the way to go.

As for actually solving the differential equation, assume that the response, [tex]x(t)[/tex] is also a sinusoid. (Think about it, you're driving it at a certain frequency, why would it do anything but oscillate at that frequency?)

That is to say, assume a general solution of the form [tex]x(t)=B_0\sin{(\omega t)}[/tex] and then find [tex]B_0[/tex] and the conditions under which that is indeed a solution.
 
  • #5
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Ok thanks a lot.

ps.
I'm not sure about solution. there mus be transitory solution too(otherwise the stationary solution that is sinusoid).
 
  • #6
671
2
Ok thanks a lot.

ps.
I'm not sure about solution. there mus be transitory solution too(otherwise the stationary solution that is sinusoid).
To find the transient solution, all you need to do is add 0 to both sides! Remember that the sum of solutions is also a solution for such linear differential equations!
Do you remember doing this to find transients?

If not, this is how it's done (Very long and interesting lecture on the subject!):
http://ocw.mit.edu/OcwWeb/Physics/8-03Fall-2004/VideoLectures/detail/embed04.htm [Broken]
 
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  • #7
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Thank you for the link:) it's very helpful.
 

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