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Mass and spring in circular motion.

  1. Jun 29, 2012 #1
    Just preparing for a physics prelim and working through previous exam questions.

    1. The problem statement, all variables and given/known data
    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?

    3. 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.
     
  2. jcsd
  3. Jul 1, 2012 #2
    Hi AbigailM :smile:

    Your method looks correct to me, but you aren't given r0 in the question..... :wink:
     
  4. Jul 1, 2012 #3
    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.
     
  5. Jul 2, 2012 #4
    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 :uhh:). 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:
     
  6. Jul 2, 2012 #5
    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.
     
  7. Jul 3, 2012 #6
    Looks correct now :smile:
     
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