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Frequency of an oscillator

  • #1

Homework Statement

consider a one dimensional parabolic potential of the form V(z) = 1/2π(√k/m)
What is the oscillation frequency of this mass?

Homework Equations


1/2π(√k/m)

The Attempt at a Solution


So here this is my attempt

1/2π(√10/.5)
1/2π(3.16/.5)
6.32(1/2π)
=9.9 hz?
 

Answers and Replies

  • #2
DrClaude
Mentor
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V(z) = 1/2π(√k/m)
That is not a function of z. And please use LaTeX as this equation is very hard to read. What is in the denominator?

1/2π(√10/.5)
Where do these numbers come from? If you want help, please post the full problem.
 
  • #3
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.
Oh crap sorry man the problem didn't get fully pasted I didn't catch that. Here editing it now

1. Homework Statement

Consider a one-dimensional parabolic potential of the form V(z)=10z[itex]^{2}[/itex], acting on a mass of 0.5kg. What is the oscillation frequency of this mass? Choose the most correct answer.

2. Homework Equations
1/2π(√k/m)

3. The Attempt at a Solution
So here this is my attempt

1/2π(√10/.5)
1/2π(3.16/.5)
6.32(1/2π)
=9.9 hz?
 
Last edited:
  • #4
DrClaude
Mentor
7,273
3,430
V(z)=10z[itex]^{2}[/itex]
I'm guessing that this is to mean ##V(z) = 10 z^2\ \mathrm{J}\,\mathrm{m}^{-2}##, with proper units.

2. Homework Equations
1/2π(√k/m)
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.

1/2π(√10/.5)
Why do you take ##k=10##?
 
  • #5
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##?
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.

I thought to use the 10 from v(z)=10z^2
 
  • #6
DrClaude
Mentor
7,273
3,430
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.
I see. So I guess I won't try to get you calculate the derivative to get the correct result!

I thought to use the 10 from v(z)=10z^2
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.
 
  • #7
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.
Thank you very much ! Just for future reference though , what does the k actually mean in the equation?
 
  • #8
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291
Thank you very much ! Just for future reference though , what does the k actually mean in the equation?
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.
 
  • #9
DrClaude
Mentor
7,273
3,430
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.
That should be ##F(x) = -kx##, as the force is always directed opposite to the displacement.
 

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