Proof That The Following System Does Uniform Oscilations

  • Thread starter DorelXD
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  • #76
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But I have a question. The derivate at point ##y_0## shouldn't be:

defined as the limit of $$ \frac {R(y_0 + \Delta y) - R(y_0)} {\Delta y} $$ when ## \Delta y \to 0 ## ???

You wrote:

## \frac {R(y_0 + \Delta y) - R(y_)} {\Delta y} ##
Well, we found the tension when ## y = 0 ## :

[tex] R(0) = kc ( a\frac{1}{ \sqrt{ c^2 - b^2 } } -1 ) [/tex]
 
  • #77
6,054
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But I have a question. The derivate at point ##y_0## shouldn't be:

defined as the limit of $$ \frac {R(y_0 + \Delta y) - R(y_0)} {\Delta y} $$ when ## \Delta y \to 0 ## ???
Yes you right, I made a mistake.

[tex] R(0) = kc ( a\frac{1}{ \sqrt{ c^2 - b^2 } } -1 ) [/tex]
You are forgetting the force of weight in the resultant. But, again, think of this physically: ## y = 0 ## is the equilibrium point. What is the net (resultant) force on a body in equilibrium?
 
  • #78
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Ahhhhh well, it is obviously 0 . I'm sorry for not answering so quick, but personally I'm not in a hurry, and I hadn't enough time to use the laptop or my smartphone :(
 
  • #79
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Do you see then, that all that you need is to find ##R'(0)##?
 
  • #80
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So, [itex] R(0) = kc ( a\frac{1}{ \sqrt{ c^2 - b^2 } } -1 ) + mg [/itex]

And I need to derivate this function, right ?
 
  • #81
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Yes, you need to differentiate ##R(y)##.
 
  • #82
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But you said that I need to differentiate R(0), not R(y) .
 
  • #83
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You cannot differentiate ##R(0)##. It is a number, not a function. You differentiate the function with respect to its arguments, which gives you its derivative function, then you evaluate the derivative at 0.
 
  • #84
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But I've never differentiated something like that:

[tex] R(y) = k(y-c)( a \frac{1}{ \sqrt{ (c-y)^2 - b^2 } } - 1 ) + mg [/tex]

It looks difficult...
 
  • #85
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I am not sure you have to. If you have this problem while not having sufficient knowledge of calculus, it may be that you are supposed to approach it differently. I asked you earlier what you had learnt in your class before, that might give you a clue.
 
  • #86
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There is indeed a different approach. I have just talked to somebody I know, and he gave me a hint. Actually he remindend me of some basic geomtery. Well, I will shortly post the begining of the solution. You'll se that it's quite interesting, and that we both missed some things.

The solution I will post uses the aproximation of the trigonomteric functions when the angles are very small, and some basic euclidian geometry. It was fairly simple. I don't know how we both missed it...
 

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