Solving Simple Harmonic Motion: Algebraic Trick Needed?

In summary, the conversation discusses finding a formula for a Simple Harmonic Oscillator and reducing it to the form of \ddot{r} + C*r = 0. After some algebra, the equation is found to be \ddot{r}-A*\sqrt{r^{2}-B^{2} } = 0, but it is later discovered that this equation is incorrect. Through further discussion and corrections, the correct equation is determined to be \ddot{r}-C*\sqrt{r^{2}-h^{2}}=0, and the conversation ends with the individual being able to solve the problem.
  • #1
noodlesofrome
7
0
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?
 
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  • #2
noodlesofrome said:
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.
 
  • #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:
  • #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]
 
  • #5
I changed it so that we are working with just magnitudes but I still can't see anything wrong. what would you do?
 
  • #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?
 
  • #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.
 

1. What is simple harmonic motion?

Simple harmonic motion is a type of oscillatory motion in which an object moves back and forth in a regular pattern, with the acceleration being proportional to the displacement from a fixed point and directed towards that point.

2. Why is an algebraic trick needed to solve simple harmonic motion?

An algebraic trick is needed because the equation for simple harmonic motion involves a trigonometric function (sin or cos) and it can be difficult to manipulate algebraically. The trick involves using the relationship between sine and cosine to simplify the equation and make it easier to solve.

3. How can I determine the period of a simple harmonic motion equation?

The period of a simple harmonic motion equation can be determined by finding the value of the coefficient in front of the trigonometric function (sin or cos). The period is equal to 2π divided by this coefficient.

4. Can I use the algebraic trick for any simple harmonic motion equation?

Yes, the algebraic trick can be used for any simple harmonic motion equation as long as it follows the form of y = A sin(ωt + φ) or y = A cos(ωt + φ), where A is the amplitude, ω is the angular frequency, t is time, and φ is the phase angle.

5. How does the amplitude affect simple harmonic motion?

The amplitude affects the maximum displacement of the object from its equilibrium position. A larger amplitude means a greater maximum displacement, while a smaller amplitude means a smaller maximum displacement. The amplitude also affects the energy of the system, with a larger amplitude resulting in a higher energy and a smaller amplitude resulting in a lower energy.

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