Mass and spring in circular motion.

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Homework Help Overview

The discussion revolves around a mass moving in a circular orbit under the influence of a three-dimensional harmonic potential, specifically focusing on determining the frequency of the orbit and the frequency of small oscillations when a radial kick is applied.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between the spring-mass system and centripetal force, questioning the role of the equilibrium length in the context of the problem.

Discussion Status

Some participants express confusion regarding the assumptions made about the equilibrium length, while others provide insights into the implications of these assumptions. There appears to be a productive exchange of ideas regarding the definitions and relationships involved in the problem.

Contextual Notes

There is a noted absence of specific information regarding the equilibrium length, which leads to discussions about how it should be treated in the context of the problem. Participants suggest that the value of the equilibrium length may need to be defined based on known terms.

AbigailM
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Just preparing for a physics prelim and working through previous exam questions.

Homework Statement


A mass m moving in a circular orbit about the origin is attracted by a three dimensional harmonic potential,

[itex]U(r)=\frac{1}{2}kr^{2}[/itex]

What is the frequency of the orbit? If a small kick is supplied in the radial direction, what will be the frequency of the ensuing small oscillations in r?

The Attempt at a Solution



[itex]k(r-r_{0})=m\omega^{2}_{0}r[/itex]

[itex]\frac{k}{m}\frac{(r-r_{0})}{r}=\omega^{2}_{0}[/itex]

[itex]\mathbf{Frequency\hspace{1 mm} of\hspace{1 mm} orbit\hspace{1 mm}}\omega_{0}=\sqrt{\frac{k}{m}\frac{(r-r_{0})}{r}}[/itex]

[itex]\ddot{r}=-\omega^{2}r \hspace{5 mm} \omega^{2}=\frac{k}{m}[/itex]

[itex]\omega^{2}=\frac{k}{m}=\frac{r\omega^{2}_{0}}{r-r_0}[/itex]

[itex]\mathbf{Frequency\hspace{1 mm} of\hspace{1 mm} oscillation\hspace{1 mm}}\omega=\omega_0\sqrt{\frac{r}{r-r_{0}}}[/itex]

Just wondering if my solution looks correct. Thanks for the help.
 
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Hi AbigailM :smile:

Your method looks correct to me, but you aren't given r0 in the question... :wink:
 
Ok, so where I'm confused is that as the spring-mass is rotated, it extends from its equilibrium length due to centripetal force. So why is [itex](r-r_{0})[/itex] wrong?

Thanks for the help.
 
AbigailM said:
Ok, so where I'm confused is that as the spring-mass is rotated, it extends from its equilibrium length due to centripetal force. So why is [itex](r-r_{0})[/itex] wrong?

Thanks for the help.

It isn't wrong. The 'equilibrium length' r0 isn't a given information in your question, you have assumed it(unless you wrote an incomplete question here :rolleyes:). And normally, we give answers in terms of stuff that's known to us. What do you think would be the value of r0 in known terms? :smile:
 
Ahhh ok I see now. The restoring force is balanced by the centrifugal force from the rotation. We can set [itex]r_{0}=0[/itex]. The reason I included [itex]r_{0}[/itex] is because for a harmonic oscillator, the mass oscillilates around the origin [itex]r_{0}[/itex]. Any many cases it's defined to be zero. And I think your right, if we were to consider it in this problem it would have been given.

Thanks again for the help Infinitum.
 
AbigailM said:
Ahhh ok I see now. The restoring force is balanced by the centrifugal force from the rotation. We can set [itex]r_{0}=0[/itex]. The reason I included [itex]r_{0}[/itex] is because for a harmonic oscillator, the mass oscillilates around the origin [itex]r_{0}[/itex]. Any many cases it's defined to be zero. And I think your right, if we were to consider it in this problem it would have been given.

Thanks again for the help Infinitum.

Looks correct now :smile:
 

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