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

[RockenNS42]

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

 

[mjordan2nd]

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.

 

[RockenNS42]

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 Ae^j(pt+α) is this what you mean?

 

[mjordan2nd]

Yep, that's what I'm talking about.

 

[RockenNS42]

Yep, that's what I'm talking about.

Ok then but I am still not sure where v is going to come from...

 

[mjordan2nd]

Once you solve for x(t) then the velocity is just the derivative.

 

[RockenNS42]

Oh right. I get it. Thanks a lot :)