Calculate quality factor of a damped oscillation from a graph

[Redwaves]

Homework Statement

Calculate quality factor of a damped oscillation from a graph

Relevant Equations

$x(t) = A_0 e^{\frac{-xt}{2}}cos(\omega_d t - \alpha)$

I'm trying to find the quality factor of a damped system.
I know 3 points from the graph, $(t,x): (\frac{\pi}{120},0.5), (\frac{\pi}{80},0), (\frac{\pi}{16},0)$

From this I found that $T = \frac{\pi}{20}$
$\omega_d = \frac{2\pi}{T} = 40 rad$

Then, from the solution $x(t) = A_0 e^{\frac{-xt}{2}}cos(\omega_d t - \alpha)$
$cos(\omega_d t - \alpha)$ must be 0 when x = 0
$\omega_d t - \alpha = \frac{\pi}{2}$
$\alpha = 0$

I also know that $Q = \frac{\omega_0}{\gamma}$
and
$\omega_0^2 = \omega_d^2 + \frac{\gamma^2}{4}$

I need some help to find $\gamma$

 

[haruspex]

Homework Statement:: Calculate quality factor of a damped oscillation from a graph
Relevant Equations:: $x(t) = A_0 e^{\frac{-xt}{2}}cos(\omega_d t - \alpha)$

I'm trying to find the quality factor of a damped system.
I know 3 points from the graph, $(t,x): (\frac{\pi}{120},0.5), (\frac{\pi}{80},0), (\frac{\pi}{16},0)$

From this I found that $T = \frac{\pi}{20}$
$\omega_d = \frac{2\pi}{T} = 40 rad$

Then, from the solution $x(t) = A_0 e^{\frac{-xt}{2}}cos(\omega_d t - \alpha)$
$cos(\omega_d t - \alpha)$ must be 0 when x = 0
$\omega_d t - \alpha = \frac{\pi}{2}$
$\alpha = 0$

I also know that $Q = \frac{\omega_0}{\gamma}$
and
$\omega_0^2 = \omega_d^2 + \frac{\gamma^2}{4}$

I need some help to find $\gamma$

This is a bit confusing. You cannot mean $x(t) = A_0 e^{\frac{-xt}{2}}cos(\omega_d t - \alpha)$ since that has an x both sides of the equation.
If I change it to $x(t) = A_0 e^{\frac{-\beta t}{2}}cos(\omega_d t - \alpha)$ then the point at $t=\frac{\pi}{120}$ with $\omega_d = 40 rad/s$ would be $(\frac{\pi}{120},0.5A_0 e^{\frac{-\beta 0.5}{2}})$

 

[Redwaves]

This is a bit confusing. You cannot mean $x(t) = A_0 e^{\frac{-xt}{2}}cos(\omega_d t - \alpha)$ since that has an x both sides of the equation.
If I change it to $x(t) = A_0 e^{\frac{-\beta t}{2}}cos(\omega_d t - \alpha)$ then the point at $t=\frac{\pi}{120}$ with $\omega_d = 40 rad/s$ would be $(\frac{\pi}{120},0.5A_0 e^{\frac{-\beta 0.5}{2}})$

I make a mistake. It is
$x(t) = A_0 e^{\frac{-\gamma t}{2}}cos(\omega_d t - \alpha)$

Why t = 0.5 and not $\frac{\pi}{120}$

In this case $A_0e^{\frac{-\gamma t}{2}} = 1$

If $\alpha = 0, A_0 = 1$?

$x(t) \neq 0.5A_0 e^{\frac{-\gamma t}{2}}$

Because, $\gamma$ or t must be 0.

However, the answer is $\gamma = 18$ and $Q = \frac{41}{18}$

In this case $\omega_0 = 41$

Using $\omega_0^2 = \omega_d^2 + \frac{\gamma^2}{4}$ where, $\gamma = 18, \omega_d = 40$
$\omega_0 = 41$ which is correct.

 

[haruspex]

I make a mistake. It is
$x(t) = A_0 e^{\frac{-\gamma t}{2}}cos(\omega_d t - \alpha)$

Ok.

Why t = 0.5 and not $\frac{\pi}{120}$

Yes, I should have written:
If I change it to $x(t) = A_0 e^{\frac{-\gamma t}{2}}cos(\omega_d t - \alpha)$ then the point at $t=\frac{\pi}{120}$ with $\omega_d = 40 rad/s$ would be $(\frac{\pi}{120},0.5A_0 e^{\frac{-\gamma\pi }{240}})$

In this case $A_0e^{\frac{-\gamma t}{2}} = 1$

Having t on the right must be wrong. Do you mean
$A_0e^{\frac{-\gamma \pi }{240}} = 1$

If $\alpha = 0, A_0 = 1?$

Did you mean, If $\gamma = 0, A_0 = 1$?
If so, yes.
$A_0=e^{\frac{\gamma\pi }{240}}$

Then, from the solution $x(t) = A_0 e^{\frac{-xt}{2}}cos(\omega_d t - \alpha)$
$cos(\omega_d t - \alpha)$ must be 0 when x = 0
$\omega_d t - \alpha = \frac{\pi}{2}$
$\alpha = 0$

I didn't get to this before. I think you mean
$cos(\omega_d t - \alpha)$ must be 1 when t = 0, so $\alpha = 0$

The information in post #1 is insufficient to determine $\gamma$.
You need data points with two nonzero values of x. Do you know x at t=0?

 

[Redwaves]

I didn't get to this before. I think you mean
$cos(\omega_d t - \alpha)$ must be 1 when t = 0, so $\alpha = 0$

The information in post #1 is insufficient to determine $\gamma$.
You need data points with two nonzero values of x. Do you know x at t=0?

If x(t) = 0 , $A_0 = 0$ or $cos(\omega_d t - \alpha) = 0$
Thus, $\omega_d t - \alpha = \frac{\pi}{2}$
$40 \cdot \frac{\pi}{80} - \alpha = \frac{\pi}{2}$
$\alpha = 0$

I had a another t on the graph $\frac{\pi}{10}$ I spent the entire day to figure the x position using a software and I found $(\frac{\pi}{10}, e^{\frac{-33\pi}{40}})$

However, I still don't see how to get $\gamma$

I tried $\frac{-33\pi}{40} = \frac{-\gamma\pi}{20}$
$\gamma = 16.5$

 

[haruspex]

If x(t) = 0 , $A_0 = 0$ or $cos(\omega_d t - \alpha) = 0$
Thus, $\omega_d t - \alpha = \frac{\pi}{2}$
$40 \cdot \frac{\pi}{80} - \alpha = \frac{\pi}{2}$
$\alpha = 0$

I had a another t on the graph $\frac{\pi}{10}$ I spent the entire day to figure the x position using a software and I found $(\frac{\pi}{10}, e^{\frac{-33\pi}{40}})$

However, I still don't see how to get $\gamma$

I tried $\frac{-33\pi}{40} = \frac{-\gamma\pi}{20}$
$\gamma = 16.5$

You have an equation for $x(\frac{\pi}{10})$ and an equation for $x(\frac{\pi}{120})$. You can eliminate $A_0$ between them and find $\gamma$.
Please post your working as far as you get.

I get the expected answer.

 

[Redwaves]

You have an equation for $x(\frac{\pi}{10})$ and an equation for $x(\frac{\pi}{120})$. You can eliminate $A_0$ between them and find $\gamma$.
Please post your working as far as you get.

I get the expected answer.

It works! it was obvious. I feel like I wasted my time and your time.

There are the 2 equations

$A_0e^{-\gamma \frac{\pi}{10}} = e^{\frac{-33\pi}{40}}$
$A_0e^{\frac{-\gamma \pi}{240}} = 1$

Once again, thanks!

One more thing.
If I don't have any point only the height of some maximums, I see that I can get Max1/Max2 and then
$\frac{e^{\frac{-\gamma t_1}{2}}}{e^{\frac{- \gamma t_2}{2}}} = \frac{Max_1}{Max_2}$
$\omega_d(t_2 - t_1) = 2\pi$
After some steps I have something like $\omega_d = \pi \gamma$

I can do that because the maximums are controlled only by the damping $e^{\frac{-\gamma t_1}{2}}$

It is correct?

 

[haruspex]

If I don't have any point only the height of some maximums, I see that I can get Max1/Max2 and then
$\frac{e^{\frac{-\gamma t_1}{2}}}{e^{\frac{- \gamma t_2}{2}}} = \frac{Max_1}{Max_2}$
$\omega_d(t_2 - t_1) = 2\pi$
After some steps I have something like $\omega_d = \pi \gamma$

I didn't understand the last step. What happened to the maxima?
If the attenuation over consecutive maxima is $\beta=Max_2/Max_1$ then $\beta=e^{-\frac \gamma 2 T}$, $\gamma=-\frac 2T\ln(\beta)=-\frac {\omega_d}{\pi}\ln(\beta)$.