Solving Simple Harmonic Motion: Algebraic Trick Needed?

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Homework Help Overview

The discussion revolves around finding a formula for a Simple Harmonic Oscillator, specifically transforming an equation involving constants A and B into a standard form. Participants are exploring algebraic manipulations and the implications of their equations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the transformation of the equation \ddot{r}-A*\sqrt{r^{2}-B^{2}} = 0 into the form \ddot{r} + C*r = 0, questioning the validity of their algebraic steps and the conditions under which this transformation is possible.

Discussion Status

There is an ongoing examination of the original equations and their derivations. Some participants have pointed out potential errors in notation and assumptions, while others are seeking clarification on the physical setup of the problem. The conversation reflects a mix of attempts to clarify misunderstandings and explore different interpretations of the equations.

Contextual Notes

Participants are grappling with the implications of constants A and B, and the conditions under which the equations hold true. There is also a focus on ensuring that the equations accurately represent the physical scenario described, particularly regarding the motion constraints along a specific line.

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.
 
Last edited:
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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