Simple Harmonic Motion derivation

[Sclerostin]

Homework Statement

Hookes Law gives: F = -kx. This is SHM. But I cannot see how to get to the sinusoidal expression from this. (In all the explanations, they cheat, and just introduce de novo Omega or Omega^2.)
But how do you get to m. d2x/dt^2 = -x.(omega) ^2

Homework Equations

F = -kx.
m. d2x/dt^2 = -x.(omega) ^2

The Attempt at a Solution

 

[LCKurtz]

Homework Statement

Hookes Law gives: F = -kx. This is SHM. But I cannot see how to get to the sinusoidal expression from this. (In all the explanations, they cheat, and just introduce de novo Omega or Omega^2.)
But how do you get to m. d2x/dt^2 = -x.(omega) ^2

Homework Equations

F = -kx.
m. d2x/dt^2 = -x.(omega) ^2

The Attempt at a Solution

You do know that force = mass x acceleration, right? So you have $F = ma = -kx$ where $a = \frac {d^2x}{dt^2}$. Put these together to get $\frac {d^2x}{dt^2} +\frac k m x = 0$, which is the sine-cosine equation. Is that what is bothering you?

 

[Sclerostin]

Thanks ++. That is exactly what is bothering me. But I am still not getting it!
Show me how to get from your last equation to the sine-cosine format.
I must have forgotten my derivations because its 3rd sentence where I get stuck. If you integrate I can't see where a Cos term appears.

 

[LCKurtz]

Thanks. That is exactly what is bothering me. But I am still not getting it!
Show me how to get from your last equation to the sine-cosine format.
I must have forgotten my derivations because its 2nd sentence where I get stuck. If you integrate I can't see where a Cos term appears.

The equation $x'' + ax = 0$ where $a>0$ is a constant coefficient equation. The standard way to solve such an equation is to try a solution $x = e^{rt}$. This leads to the characteristic equation $r^2 + a = 0$ with roots $r = \pm \sqrt a i$ and a fundamental pair of solution $\{e^{i\sqrt a t}, e^{-i\sqrt a t}\}$, or the easier to work with pair $\{ \sin \sqrt a t,\cos \sqrt a t\}$. Even better, since $a > 0$ let's call $a =\omega^2$ giving $\{ \sin \omega t,\cos \omega t\}$. In your case $\omega = \sqrt \frac k m$. You can find this in any reference to constant coefficient DE's.

 

[Orodruin]

Did you try differentiating $\cos(\omega t)$ twice to see what comes out?

 

[Sclerostin]

Thank you, LCKurtz. I can work with your explanation. I have forgotten my maths, many years. So it helps to "know" some other relationships, allowing things to fall into place.

Just saying things like "call the constant ω^2" is (from my point of view) assuming what you are trying to prove. And that's what I kept seeing.

So fine, thanks!

 

[PeroK]

Thank you, LCKurtz. I can work with your explanation. I have forgotten my maths, many years. So it helps to "know" some other relationships, allowing things to fall into place.

Just saying things like "call the constant ω^2" is (from my point of view) assuming what you are trying to prove. And that's what I kept seeing.

So fine, thanks!

There's a difference between assuming what you are trying to prove and guessing a solution and then proving that it is indeed a solution to your equation. The latter approach is common in integration and differential equations.

 

[LCKurtz]

Just saying things like "call the constant ω^2" is (from my point of view) assuming what you are trying to prove. And that's what I kept seeing.

You are welcome. Just to elaborate on the above, for $x''+ \frac k m x=0$ you get terms like $\sin\sqrt {\frac k m }t$. Remember that in $\sin(bt)$ the angular frequency $\omega$ is $b$. In this case $\omega = b =\sqrt{\frac k m}$ and so $\omega^2=\frac k m$. So we aren't just calling it that for no reason.

 

[Sclerostin]

There's a difference between assuming what you are trying to prove and guessing a solution and then proving that it is indeed a solution to your equation. The latter approach is common in integration and differential equations.

I almost agree: but you aren't guessing. With background knowledge, you are making an informed try. I will accept that (for those with their maths relationships alive) what I was seeing wasn't guessing. But for me, it wasn't clear.

Its not a trivial derivation, as it turns out. No wonder I didn't get it!

 

[Sclerostin]

There's a difference between assuming what you are trying to prove and guessing a solution and then proving that it is indeed a solution to your equation. The latter approach is common in integration and differential equations.

I like the photo. Not Ama Dablam, is it?

 

[PeroK]

I like the photo. Not Ama Dablam, is it?

Yes, it's Ama Dablam.

 

[Sclerostin]

You are welcome. Just to elaborate on the above, for $x''+ \frac k m x=0$ you get terms like $\sin\sqrt {\frac k m }t$. Remember that in $\sin(bt)$ the angular frequency $\omega$ is $b$. In this case $\omega = b =\sqrt{\frac k m}$ and so $\omega^2=\frac k m$. So we aren't just calling it that for no reason.

I have deliberately waited before adding this:
In the initial equation, the trick, of course, is that the second derivative of x equals another f(x). So this suggests that x equals either a cos function, or an exponential function. So we get to your answer which you knew instantly!
Thanks again!