How to show that the motion graph of a linear oscillator is an ellipse.

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RockenNS42
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Homework Statement


The motion of a linear oscillator may be represented by means of a graph in which x is abscissa and dx/dt as ordinate. The histroy of the oscillator is then a curve
a)show that for an undamped oscillator this curve is an ellipse
b) show (at least qualitatively) that if a damping curve is introduced on gets a curve spiraling into origin.

Homework Equations


The Attempt at a Solution


a) I got that
x(t)=Asin(wt-α)
v(t)=wAcos(wt-α)
Another student told me to "elimate the t's" to get
x2/A2 +X2/(Aw)2 =1
and that is total energy is E=1/2KA2 and w2=k/m then
x2/(2E/k) +X2/(2E/m) =1First of all, I don't under stand how eliminated his t's. I do get that he found the eq of an ellipse, but how do I go from an eq with X and w to one with x and dx/dt?

b)I have no sweet clue
 
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I suspect that his answer should read

[tex] \frac{x^2}{A^2} + \frac{v^2}{A^2w^2} = 1[/tex]

To arrive at this rearrange the x and v equations so that only the sin and cosine functions are left on the right hand side. Then square both equations and add. \

For part b you will want to add a damping constant to the equation of motion:

[tex] m \frac{d^2x}{dt^2} = -kx - c \frac{dx}{dt}[/tex]

You will need to find solutions to this equation. From there you can find v, and plot x vs v.
 
mjordan2nd said:
I suspect that his answer should read

[tex] \frac{x^2}{A^2} + \frac{v^2}{A^2w^2} = 1[/tex]

To arrive at this rearrange the x and v equations so that only the sin and cosine functions are left on the right hand side. Then square both equations and add. \

For part b you will want to add a damping constant to the equation of motion:

[tex] m \frac{d^2x}{dt^2} = -kx - c \frac{dx}{dt}[/tex]

You will need to find solutions to this equation. From there you can find v, and plot x vs v.


Ok a) makes total sense now
b) In is c the damping constant? (we're using b) we have found in class that the solution to his comes in the fourm Aej(pt+α) is this what you mean?
 
mjordan2nd said:
Yep, that's what I'm talking about.

Ok then but I am still not sure where v is going to come from...
 
Once you solve for x(t) then the velocity is just the derivative.