Why Do Harmonic Motion Equations Differ Across Sources?

[Cog77]

I was hoping that someone could explain why these different equations can be found from different sources please.

The time dependent position, x(t), of an underdamped harmonic oscillator is given by:

$x(t)=e^{-\gamma t}acos(\omega_{1}t-\alpha)$

where $\gamma$ is the damping coefficient, and $\omega_{1}$ is the frequency of the damped oscillator. This is the equation I am familiar with and can be found in many explanations of harmonic oscillators, e.g. here.

In contrast, Wolfram gives the following equation (note that this is copied from their website, and it uses different notation to the equation above. In particular $\gamma$ is NOT the damping coefficient):

$x(t)=e^{-\frac{\beta t}{2}}[Acos(\gamma t)+Bsin(\gamma t)]$

I believe that changing this equation to use the same notation as the first thing I posted gives this:

$x(t)=e^{-\gamma t}[Acos(\omega_{1}t)+Bsin(\omega_{1}t)]$


So here is what I'd like to understand:

1) Why are these different? I tried to see if there was a case when $A>>B$, which I found happens if the initial velocity is zero, $\dot{x}(0)=0$, and $\zeta<<\frac{1}{\sqrt{2}}$. Where $\zeta$ is the damping ratio. In other words, when there is very little damping the Wolfram equation reduces to the shorter equation. Is that correct?

2) Whether or not that's correct, I don't understand why there would sometimes be two terms in the equation. What does that physically represent?

3) I actually started looking at this after getting confused about the overdamped case in a similar way. It also has two terms, and I believe that it reduces to a single term when there is a lot of damping ($\zeta>>1$). Again, what does it physically mean when there are two terms? Although hopefully understanding one case will lead to an understanding of both, so maybe this question is redundant.

Thanks for any help!

 

[tiny-tim]

Hi Cog77! Welcome to PF!

$x(t)=e^{-\gamma t}acos(\omega_{1}t-\alpha)$

$x(t)=e^{-\gamma t}[Acos(\omega_{1}t)+Bsin(\omega_{1}t)]$

A = acosα, B = asinα ​

 

[dauto]

All the equations are equivalent to each other

 

[jtbell]

Hint: to your first equation, apply the trig identity

cos (u - v) = cos u cos v + sin u sin v

 

[Cog77]

Aah thanks guys! Should have spotted that...

So physically the two terms account for the amplitude and the phase?

What about the overdamped case?

$x(t)=Ae^{r_{-}t}+Be^{r_{+}t}$

with

$r_{\pm}=\frac{1}{2}-\beta\pm\sqrt{\beta^{2}-4\omega_{0}^{2}}$

Does the fact that the expression contains faster and slower decaying terms mean something physically? This was where my confusion originally came from as I always assumed there was just a single exponential term in the overdamped case.

 

[Philip Wood]

An intriguing question. I can't think of any significance in there being a fast term and a slow term, but would love to be shown to be missing something.

Instead, I can, perhaps remove some of the motivation for the search. It is easy to show that in the case when the body is displaced and released from rest, A and B are equal and opposite. In that when you pull out $e^{- \frac{\beta t}{2}}$ as a factor you're left with a multiplied cosh function. So it doesn't look like two added terms any longer!

This is somewhat similar to writing $A\ \mbox {cos}\ \omega t + B\ \mbox {sin}\ \omega t$ as $a\ \mbox {cos}\ (\omega t + \epsilon)$