3
$\begingroup$

I am not sure if I understand the term damping coefficient correctly (I am a high-school student). Here's the link for the info that I learned:

http://hyperphysics.phy-astr.gsu.edu/hbase/oscda.html

So, as far as I understand, damping coefficient is $\gamma$. It's equation:

$$x/x_0 = e^{\gamma t} \, .$$

And another equation:

$$\gamma = c/2m \, .$$

$c$ as I understand it is decay constant. Although the website doesn't provide with such information.

However, one of the questions on the units of damping coefficient in this forum had an answer that said that $\gamma$ doesn't have units (Dimensonal analysis of damping constant?). How can that be? I mean, I don't understand something. It should be s^-1 in this case. Are the equations flawed?

What I have in mind is underdamped case.

The graph I have looks like this: enter image description here

So can I use the first formula $(x/x_0 = e^{\gamma t}$) to get $\gamma$, which is the damping coefficient, or the website has it wrong?

$\endgroup$
  • $\begingroup$ $\gamma$ definitely does have units of $s^{-1}$ the way you have defined it (and the way it is usually defined). One way you can see this for sure is that the argument of an exponential must be dimensionless, and you have $e^{-\gamma t}$ appearing. Do you have a link to where it was said $\gamma$ doesn't have units? $\endgroup$ – Andrew Jan 5 '17 at 19:04
  • $\begingroup$ Yes, I do understand that from the way I defined it, it is correct. Here's the link: physics.stackexchange.com/questions/9754/… $\endgroup$ – Karolizzz Jan 5 '17 at 19:06
  • $\begingroup$ The problem is that I am not sure whether the info on website is correct $\endgroup$ – Karolizzz Jan 5 '17 at 19:06
  • 1
    $\begingroup$ Welcome to Physics Stack Exchange! First, we need to get some terminology straight. $\gamma$ doesn't have any particular units. It has dimensions of 1/time. You can pick units of 1/hour or 1/day or whatever you want. Second, please go back through your post and make sure you 1) Define all symbols! You haven't defined $c$ so we have no idea what $\gamma = c/2m$ means, 2) When you refer to another post, provide a link! $\endgroup$ – DanielSank Jan 5 '17 at 19:07
  • $\begingroup$ Note that in the answer you link to, $e^{-\zeta \omega_0 t}$ shows up. You should be able to compare that with what you write, $e^{-\gamma t}$, to see why $\zeta$ and $\gamma$ have different dimensions. $\endgroup$ – Andrew Jan 5 '17 at 19:19
0
$\begingroup$

In the case of solving RLC circuits, damping ratio determines the nature of the solution. If damping ratio is smaller than 1, you would have the above graph. It is actually described by this equation (underdamped). $$ i(t)=e^{-\alpha t}(A_1 \cos\omega_d t+A_2 \sin\omega_d t) $$ Damping ratio is often written as $$\zeta = {\alpha\over \omega_0}$$

As you can see from the first equation, it has a exponential component (decaying) and sinusoidal component (oscillates). For more information on how to derive the equation and the graph https://youtu.be/dGc-ozvwnjE

$\endgroup$
  • $\begingroup$ Is that a link to your own video? If so, you should disclose that. $\endgroup$ – Chris Nov 27 '17 at 21:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.