Measuring the natural frequency of a spring-mass system driving force

[haruspex]

$-2e^{(-\gamma(7/2))} / -8e^{(-\gamma/2)}$

But what does that mean?

How you get $Ae^{-\gamma(t+T/2)/2}$

I still not sure why only this part $Ae^{-\gamma/2}$.

$Ae^{-\gamma/2}$ = ?

Having obtained the graph of the transient, we can see it oscillates. That means it is underdamped, and alpha in the transient term $x(t)=Ce^{-\alpha t}$ will be complex. This means the transient can be written in the form of eqn 2.11 at the link:
$x(t)=e^{-\Gamma t/2}(c\cos(\omega t)+d\sin(\omega t))$, where $\omega^2 = \omega_0^2+\Gamma^2/4$.
For our purposes, it is more convenient to write the amplitude as $C^2=c^2+d^2$, i.e. $x(t)=Ce^{-\Gamma t/2}(\sin(\omega t+\phi))$.
The graph starts heading down from the origin, so $\phi=\pi$.
If we focus on the maxima and minima, the sin() will be taking values +1 and -1, and the magnitude is just $Ce^{-\Gamma t/2}$.
If the period is $T=\frac{2\pi}{\omega}$ and there is a trough at time t₀ then the next peak is at time $t_0+T/2$.
$x(t_0)=-Ce^{-\Gamma t_0/2}$, $x(t_0+T/2)=Ce^{-\Gamma (t_0+T/2)/2}=Ce^{-\Gamma t_0/2}e^{-\Gamma T/4}$.
So what is $|\frac{x(t_0+T/2)}{x(t_0)}|$?

 

[haruspex]

It occurred to me that having obtained approximate values for the various frequencies and amplitudes from my crude recreation of the transient, we could model it in a spreadsheet and tune them to get a really good reconstruction of the graph you started with.
So for the transient I plugged in:
- a period of 4 seconds
- a per cycle attenuation of 1/4 (i.e., each complete cycle reduces the amplitude by three quarters),
- an amplitude at one quarter cycle (t=1 second) of -8. (Making C negative meant I could drop the phase shift φ=π.)
and for the steady state portion:
- a period of 1 second
- an amplitude of 2.8 (roughly)
- a phase of 0 (i.e. just sin(w_dt)).

With no tweaking at all, it produced:

Look familiar?

 

[happyparticle]

So what is $|\frac{x(t_0+T/2)}{x(t_0)}|$?

Even with all your explanation I don't know what is that.$x(t_0) = -Ce^{\Gamma t_0/2}$

If C is the amplitude (-8). Where's x at t=0. In my head at t=0 x = 0.

With the values above did you find $\omega_0$ ?

 

[haruspex]

Even with all your explanation I don't know what is that.

Just plug in the expressions I gave for x(t₀) and x(t₀+T/2) in the preceding line.

$x(t_0) = -Ce^{\Gamma t_0/2}$

If C is the amplitude (-8). Where's x at t=0. In my head at t=0 x = 0.

With the values above did you find $\omega_0$ ?

I defined t₀ as the time of the first minimum. It is not t=0.
Yes, I found $\omega_0$.

 

[happyparticle]

Just plug in the expressions I gave for x(t₀) and x(t₀+T/2) in the preceding line.

$\frac{C^e{-\Gamma t_0/2}e^{-\Gamma T/4}}{-Ce^{-\Gamma t_0/2}} = 4/-8$

$\Gamma = 0.69$

Like that?

 

[haruspex]

$\frac{Ce^{-\Gamma t_0/2}e^{-\Gamma T/4}}{-Ce^{-\Gamma t_0/2}} = 4/-8$

$\Gamma = 0.69$

Like that?

Yes.

 

[happyparticle]

So basically, $\Gamma$ is how the system decrease "speed"?

 

[haruspex]

So basically, $\Gamma$ is how the system decrease "speed"?

It's how the transient dies away, allowing the period to be controlled by the driving force. It continues to affect the amplitude.

 

[happyparticle]

I love you! Seriously big big thanks

one more thing if you want. $A(\omega_d) = 2.8$ is that correct?

 

[haruspex]

I love you! Seriously big big thanks

one more thing if you want. $A(\omega_d) = 2.8$ is that correct?

That's how it looks on the graph you posted.
I encourage you to try to repeat the path I took: observe that the general equation is the sum of a decaying oscillation and a steady oscillation; deduce the steady oscillation parameters from the right hand part of the graph; subtract the equation for the steady part from the graph as a whole to reveal the transient oscillation; from the period and attenuation of the transient deduce its parameters; recreate the curve in a spreadsheet using the parameters calculated.