The discussion revolves around calculating the oscillation frequency of a mass in a one-dimensional parabolic potential, specifically given by the equation V(z) = 10z². Participants are exploring the relationship between the potential energy and the oscillation frequency, which involves the parameters k (spring constant) and m (mass).
Some participants have provided guidance on the correct equation for oscillation frequency and have pointed out potential errors in the original setup. There is an ongoing exploration of the definitions and implications of the parameters involved, particularly k and its role in the equation.
Participants note that the original problem statement was incomplete, leading to misunderstandings about the parameters and the equation's structure. There is a recognition of the challenges faced by the original poster, who is navigating complex concepts in their studies.
That is not a function of z. And please use LaTeX as this equation is very hard to read. What is in the denominator?AdrianHudson said:V(z) = 1/2π(√k/m)
Where do these numbers come from? If you want help, please post the full problem.AdrianHudson said:1/2π(√10/.5)
DrClaude said:That is not a function of z. And please use LaTeX as this equation is very hard to read. What is in the denominator?
Where do these numbers come from? If you want help, please post the full problem.
I'm guessing that this is to mean ##V(z) = 10 z^2\ \mathrm{J}\,\mathrm{m}^{-2}##, with proper units.AdrianHudson said:V(z)=10z[itex]^{2}[/itex]
That is not an equation. An equation needs an equal sign somewhere. I take it you meanAdrianHudson said:2. Homework Equations
1/2π(√k/m)
Why do you take ##k=10##?AdrianHudson said:1/2π(√10/.5)
DrClaude said:I'm guessing that this is to mean ##V(z) = 10 z^2\ \mathrm{J}\,\mathrm{m}^{-2}##, with proper units.That is not an equation. An equation needs an equal sign somewhere. I take it you mean
$$
\nu = \frac{1}{2\pi} \frac{\sqrt{k}}{m}
$$
As I said previously, it is hard to understand what is in the denominator and what is under the square root. If this is the equation you used, which seems to be the case considering your numerical result: there is an error in the equation.Why do you take ##k=10##?
I see. So I guess I won't try to get you calculate the derivative to get the correct result!AdrianHudson said:Sorry about if I'm really not giving you the stuff you need to help. I am taking a shot in the dark here because I am in grade 11 and I'm trying to do stuff that is way above my head, I have been following along in my course but sometimes this stuff throws me in a loop I haven't properly learned all this stuff.
The thing is, the number in there is not ##k##. The potential energy of a harmonic oscillator is expressed asAdrianHudson said:I thought to use the 10 from v(z)=10z^2
DrClaude said:I see. So I guess I won't try to get you calculate the derivative to get the correct result!
The thing is, the number in there is not ##k##. The potential energy of a harmonic oscillator is expressed as
$$
V(x) = \frac{1}{2} k x^2
$$
As for the other equation, there was a problem with the square root. The correct equation for the oscillation frequency ##\nu## is
$$
\nu = \frac{1}{2 \pi} \sqrt{\frac{k}{m}}
$$
With this, you should be able to calculate the answer to your problem.
If you differentiate ##V(x) = \frac{1}{2} k x^2## wrt x you get ##F(x)=kx##. So k is the spring constant - ie how much force it takes to change the distance by x.AdrianHudson said:Thank you very much ! Just for future reference though , what does the k actually mean in the equation?
That should be ##F(x) = -kx##, as the force is always directed opposite to the displacement.Mentz114 said:If you differentiate ##V(x) = \frac{1}{2} k x^2## wrt x you get ##F(x)=kx##. So k is the spring constant - ie how much force it takes to change the distance by x.