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Simple Harmonic Motion

  1. Feb 23, 2009 #1
    Hello

    I'm trying to find a formula for a Simple Harmonic Oscillator.
    After some algebra I've got it down to;

    [tex]\ddot{r}[/tex]-A*[tex]\sqrt{r^{2}-B^{2} }[/tex] = 0

    where A and B are constants.

    I'm trying to get it down to be in the form of

    [tex]\ddot{r}[/tex] + C*r = 0

    I am feeling like a dummy cause it seems like it's just some algebra trick I'm not aware of. Does anyone know if it can be done?
     
  2. jcsd
  3. Feb 23, 2009 #2

    gabbagabbahey

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    Gold Member

    That doesn't look right.:wink;.....why don't you show me your derivation of this equation...

    The above equation does not reduce into this desired form unless B=0, so either B=0 or your first equation is incorrect.
     
  4. Feb 23, 2009 #3
    Its a central forces problem where the mass can only move in the x direction along y = h resulting in Simple Harmonic Motion. (h is a constant)
    [tex]F = -4/3 G \rho \pi m r [/tex]
    [tex]F_{x} = F Cos \theta[/tex] = [tex]F\sqrt{r^{2}-h^{2}}/r[/tex]
    [tex]F_{x} = -4/3 G \rho \pi m \sqrt{r^{2}-h^{2}} [/tex]

    [tex]\rho = M/V = Constant [/tex]

    [tex]F_{x} = m\ddot{r} = -4/3 G \rho \pi m \sqrt{r^{2}-h^{2}} [/tex]
    [tex]\ddot{r} = -4/3 G \rho \pi \sqrt{r^{2}-h^{2}} = C*\sqrt{r^{2}-h^{2}} [/tex]
    [tex]\ddot{r}-C*\sqrt{r^{2}-h^{2}}=0[/tex]
    I need that last equation to take the form of
    [tex]\ddot{r}+C_{2}*r=0[/tex]
    so that i could find the period, but I always seem to get stuck on some algebra.
     
    Last edited: Feb 23, 2009
  5. Feb 23, 2009 #4
    You need to be a more careful with your notation. I see vector quantities being equated to scalar quantities, which is a no-no.

    Also, the following equation is wrong:

    [tex]
    \stackrel{\rightarrow}{F}_{x} = m\ddot{r} = -4/3 G \rho \pi m \sqrt{r^{2}-h^{2}}
    [/tex]
     
  6. Feb 23, 2009 #5
    I changed it so that we are working with just magnitudes but I still can't see anything wrong. what would you do?
     
  7. Feb 23, 2009 #6

    gabbagabbahey

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    Your original problem still is not very clear.

    You say that the particle can only move along y=h and then you assume that the only force is directed towards the origin. How could this be possible? If the force is directed towards the origin, then the particle will move towards the origin unless there is some other force which constrains it to the line y=h.

    Could you please state the problem word for word from your assignment sheet?
     
  8. Feb 23, 2009 #7
    thank you guys. you are right. this equation is wrong;

    [tex]

    \stackrel{\rightarrow}{F}_{x} = m\ddot{r} = -4/3 G \rho \pi m \sqrt{r^{2}-h^{2}}

    [/tex]

    It should be

    [tex]

    \stackrel{\rightarrow}{F}_{x} = m\ddot{r_{x}} = -4/3 G \rho \pi m \sqrt{r^{2}-h^{2}}

    [/tex]
    and
    [tex] r_{x}=\sqrt{r^{2}-h^{2}}[/tex]
    so
    [tex]
    \stackrel{\rightarrow}{F}_{x} = m\ddot{r_{x}} = -4/3 G \rho \pi m r_{x}
    [/tex]

    thank you very much. I could solve from here. I always get stuck on a trivial mistake and it costs me lots of time.
     
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