Question about Damped Oscillations

[LagrangeEuler]

Why are damped oscillation in many books written with equation
$$\ddot{x}+2\delta \dot{x}+\omega^2 x=0$$
$\delta$ and $\omega^2$ are constants. Why $2 \delta$ many authors write in equation?

 

[pasmith]

Why are damped oscillation in many books written with equation
$$\ddot{x}+2\delta \dot{x}+\omega^2 x=0$$
$\delta$ and $\omega^2$ are constants. Why $2 \delta$ many authors write in equation?

Because the solution is a linear combination of exponentials $e^{kt}$ where $$k^2 + 2\delta k + \omega^2 = (k + \delta)^2 + \omega^2 - \delta^2 = 0.$$

 

[malawi_glenn]

Why 2δ many authors write in equation?

it becomes less messy.

The solutions to the characteristic equation $r^2 + 2\delta r + \omega^2 = 0$ becomes $r = -\delta \pm \sqrt{\delta^2 - \omega^2}$

With characteristic equation $r^2 + \delta r + \omega^2 = 0$ solutions becomes $r = -\dfrac{\delta}{2} \pm \sqrt{\dfrac{\delta^2}{4} - \omega^2}$

 

[DaveE]

As others have said, it's just a way of simplifying notation.

However, in my experience, this is a less common way to simplify the problem/solution. There are two very common canonical forms for this problem as below:

1) Using the resonant (natural) frequency $\omega_o$ and the damping coefficient $\xi$. So,
$$ \ddot{x}+2\xi \omega_o \dot{x}+\omega_o^2 x=0 $$

2) Using the resonant (natural) frequency $\omega_o$ and the quality factor $Q$. So,
$$ \ddot{x}+\frac{1}{Q} \omega_o \dot{x}+\omega_o^2 x=0 $$

I suspect that this $\delta$ version is used because when roots of the characteristic equation are complex $\delta$ is the real part as shown previously. Which is also simple. They are all simple from some viewpoint. Usually that viewpoint is about a simple form of the solution, not so much what the original DE looks like.

The bigger point is that these are all just different representations of the same system. It's a personal choice, you could invent your own. Or, you may someday have to decipher someone else's version (like I just had to do with $\delta$).

Wikipedia has a good page about this: https://en.wikipedia.org/wiki/Damping

edit: for what it's worth, I like this form: $\frac{\ddot{x}}{\omega_o^2}+\frac{1}{Q \omega_o} \dot{x}+x=0$

 

[DaveE]

On a lighter note, you can tell the discipline of the writer by their notation (OK, only slightly better than chance).

- Mechanical engineers like $\xi$. I can't fathom why. Since it's not nearly as pretty as...

- Q. Which means you're probably an EE. Although you would grudgingly acknowledge that they are really the same thing.

- If you use $\xi$ or Q but they are an element of a matrix in equations with only one derivative, then you are definitely, absolutely, a controls engineer. You worship the word "state space" as if it was Yahweh.

- If your equations are full of n, (n-1), and (n-2), but no derivatives, then you are a DSP control guy. You don't need $\xi$ or Q, everything is $a_n$, $a_{n-1}$ and such. Half of you thought I should have said k, (k-1), and (k-2). Half of you have never touched the hardware you are controlling with your code. Time is just a constant called $T$, which appears everywhere but is ignored.

- $\delta$: probably a mathematician. Since you never see it in an engineering book.

- If an engineer doesn't even recognize it's a SHO equation, like this, then you are definitely a physicist. Although you use ALL of the other notations too, except the discrete time stuff.

- If you are a biologist, geologist, or chemist, then you don't care and aren't sure where your textbook is with this material. But you know you have it, you knew it, and could relearn it if you can't find a friend to do it for you.

But... It's all the same thing.

 

[vanhees71]

It's just convenience. If you write the equation of motion as
$$\ddot{x}+2 \delta \dot{x}+\omega^2 x=0, \qquad (1)$$
you can first make your life easier by substituting
$$x(t)=\exp(-\delta t) y(t),$$
because then
$$\dot{x}(t) = \exp(-\delta t) [-\delta y(t) + \dot{y}(t)],$$
$$\ddot{x}(t)=\exp(-\delta t) [\delta^2 y(t) -2 \delta \dot{y}(t)+\ddot{y}(t)].$$
Plugging this into (1) you get
$$\exp(-\delta t)[\delta^2 y(t)-2 \delta \dot{y}(t) + \ddot{y}(t)+2 \delta \dot{y}(t)-2\delta^2 y(t) + \omega^2 y(t)] = \exp(-\delta t)[\ddot{y}(t)+(\omega^2-\delta^2) y(t)]=0.$$
This means that
$$\ddot{y}(t)+(\omega^2-\delta^2) y(t)=0,$$
which is easily solved for the three cases $\omega>\delta>0$ ("underdamping"), $\omega=\delta>0$ ("critical damping"), and $\delta>\omega>0$ ("overdamping").