Physics (Classical Dynammics) Exam

[mooeypoo]

Hey guys.

 

Okay, hehe, funniest story ever - I am having a few time issues with my 2 advanced phys classes, I had to end up choosing which one I put my time in and which I am just cruising through, getting my expected F (probably) and doing again (we have an "F Drop" policy, where if u do it again your F is replaced). In any case, we had an exam on the class I "gave up" on. Among other things, I'm having difficulties with the math (I'm just now studying the beginning of it in the math course), so I REALLY didn't expect to do well.

 

As it turns out, no one did very well. I have no idea what my grade is (probly bad) but from my talk with other people in class, I did fairly well compared to some of them. Weird, huh? anyhoo, I have the exam + my answers, and I got seriously stuck in 2 parts of a question, which I am hoping you can help me out with.

 

The point of this entire long intro, other than amusing you, is to put a notice that I might not be too well versed in the math, so, uhm, if I start asking questions that sound amateurish, it's because I *am* an amateur. I have no clue what the professor's doing (math wise) half the time. I'm reading the book by myself and trying to figure things out.

 

Please bear with me here.

 

Okay, first question I got stuck with:

 

A particle of mass m moves in one dimension

[math]-\infty <= x <= \infty[/math]

in a force where the potential energy is given by:

[math]

V(x) = \frac{\lambda}{x^2-6ax+25a^2}

[/math]

where [math]\lambda , a[/math] are positive constants.

 

(a) What are the equilibrium points where the force vanishes?

(b) Expanding the potential around the stable equilibirium point for small amplitude motion, show that the particle undergoes oscillations. What is the frequency of the oscillations?

 

Okay, so what I thought was that since V is with only 1 variable, the equation is

[math]F= - \frac{dv}{dx}[/math]

And then I just take F=0

 

So, hopefully my calculations were right, i got x=3a when F=0, which is the equilibrium point. (and x=0)

 

(b) I didn't even know how to start. I know about the equation

(how do you do x-dot and x-doubledot in LaTeX? I'll just write this as 'primes' of X, but I mean xdot and xdoubledot, that is the dx/dt and its second derivative)

[math]x'' + w_{0}^{2} = 0[/math]

 

but.. x'=0 and so is x''=0

 

so.. uhh.. I have no clue what to do. In my frustration, I tried putting in

[math]x=3a+\epsilon[/math] (epsilon being a tiny number) but then that doesn't work either.

 

What.. how.. who.. where.. how do I approach this?

 

Okay, expect another question in a nother thread (probably tomorrow when I'm after some sleep

 

Thanks!

 

~moo

Edited October 29, 2008 by mooeypoo

[NeonBlack]

Do you know how to do a Taylor expansion?

Expand the force around your equilibrium position, throwing out high-order terms. (Keeping up to the x term)

Now, when you write now Newton's 2nd law, F=mx'', you should get a 2nd order differential equation with constant coefficients.

[mooeypoo]

I so don't. I have no clue.. I'm staring at the book at both Taylor Expansion and something called "Fourier Trick"... I heard of the Fourier trick and I've learned, about 2 semester ago, about Taylor series (I need to review), but it was in the context of a series, and just something about how to construct it (i sucked btw) and so far, I haven't figured out how to *use* them in here.

 

Can you give me an example so I can see steps/method? If you feel uncomfortable "solving my hw question" then just use whatever other parameters, or a generalized version, but I would REALLY appreciate an example or some explanation, I am *SOOO* lost in this, I have no idea how to even start!

[NeonBlack]

Sure.

Let's say we have a force [math]F=\frac{A}{x^2}-\frac{B}{x}[/math] with an equilibrium point at [math]x=\frac{A}{B}[/math]

In general, the Taylor series is:

[math]\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^{n}[/math] where [math]f^{(n)}(a)[/math] means the nth derivative evaluated at a. And a is the point we want to expand around (the equilibrium point).

[math]F(\frac{A}{B})=0[/math] (It's an equilibrium point).

[math]F'(x)=-2\frac{A}{x^3}+\frac{B}{x^2}[/math] so

[math]F'(\frac{A}{B})=-\frac{B^3}{A^2}[/math]

Now we can write the force as a Taylor series up to the x term (n=1):

[math]F \approx 0 -\frac{B^3}{A^2}(x-\frac{A}{B})[/math]

 

Now use this force in Newton's 2nd law and you should get a differential equation that you recognize.

[mooeypoo]

Okay we got the exams back, so I'm going over them AND what you wrote here (I tried reading about this more and doing some preparation excercises to get the hang of it).

 

Just a bit of news (I started laughing in class, seriously) - I got 22 out of 30. One of the higher grades in class. The average was close to 13... I should stop getting ready for exams

 

Thanks for the help btw, it really did help, I just had no time responding yet. THANKS!