Solving Simple Harmonic Motion: Algebraic Trick Needed?

noodlesofrome
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Hello

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

\ddot{r}-A*\sqrt{r^{2}-B^{2} } = 0

where A and B are constants.

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

\ddot{r} + 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?
 
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noodlesofrome said:
Hello

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

\ddot{r}-A*\sqrt{r^{2}-B^{2} } = 0

where A and B are constants.

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

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

\ddot{r} + 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?

The above equation does not reduce into this desired form unless B=0, so either B=0 or your first equation is incorrect.
 
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)
F = -4/3 G \rho \pi m r
F_{x} = F Cos \theta = F\sqrt{r^{2}-h^{2}}/r
F_{x} = -4/3 G \rho \pi m \sqrt{r^{2}-h^{2}}

\rho = M/V = Constant

F_{x} = m\ddot{r} = -4/3 G \rho \pi m \sqrt{r^{2}-h^{2}}
\ddot{r} = -4/3 G \rho \pi \sqrt{r^{2}-h^{2}} = C*\sqrt{r^{2}-h^{2}}
\ddot{r}-C*\sqrt{r^{2}-h^{2}}=0
I need that last equation to take the form of
\ddot{r}+C_{2}*r=0
so that i could find the period, but I always seem to get stuck on some algebra.
 
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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:

<br /> \stackrel{\rightarrow}{F}_{x} = m\ddot{r} = -4/3 G \rho \pi m \sqrt{r^{2}-h^{2}} <br />
 
I changed it so that we are working with just magnitudes but I still can't see anything wrong. what would you do?
 
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?
 
thank you guys. you are right. this equation is wrong;

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

It should be

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

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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