Simple Harmonic Motion

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?
 

gabbagabbahey

Homework Helper
Gold Member
5,001
6
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.
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

[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?
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)
[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:
242
0
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]
 
I changed it so that we are working with just magnitudes but I still can't see anything wrong. what would you do?
 

gabbagabbahey

Homework Helper
Gold Member
5,001
6
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;

[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.
 

Want to reply to this thread?

"Simple Harmonic Motion" You must log in or register to reply here.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Top Threads

Top