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

  • #26
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I'm trying to find a way to use the equation 2.10 but I need ##\theta## and ##A_0##

##\theta## and A are determined by the inital position and velocity. Thus, if at x = 0 the velocity = 0, then A and ##\theta## = 0 ?
 
  • #27
haruspex
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I'm trying to find a way to use the equation 2.10 but I need ##\theta## and ##A_0##
No, you don't. You are only concerned with how the amplitude changes across one cycle, so you can choose the peaks and troughs, i.e. the cos term is +/-1.
If the amplitude at the first trough is ##Ae^{-\Gamma t/2}## then at the next peak it will be ##Ae^{-\Gamma (t+T/2)/2}##, where T is one period of the transient.
What is the ratio of those two expressions? What is that ratio according to my graph?
 
  • #28
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Do you mean that for low damping ##A = A_0 e^{(-\gamma/2)t}##

The ratio according to the graph seems to be ##-8/-3##, but what this ratio means?

Is it the attenuation per cycle.

Is the attenuation per cycle gamma?
 
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  • #29
haruspex
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The ratio according to the graph seems to be ##-8/-3##, but what this ratio means?
Looks to me like the successive extremes are -8, +4, -2, +1... so for one complete cycle the amplitude is multiplied by 1/4.
As to the meaning algebraically, you did not answer this question:
What is the ratio of those two expressions?
... where the two expressions are for the amplitudes half a cycle apart.
 
  • #30
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##-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}## = ?
 
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  • #31
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##-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)}|##?
 
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  • #32
haruspex
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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(wdt)).

With no tweaking at all, it produced:
1613092053044.png

Look familiar?
 
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  • #33
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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## ?
 
  • #34
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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##.
 
  • #35
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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?
 
  • #36
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##\frac{Ce^{-\Gamma t_0/2}e^{-\Gamma T/4}}{-Ce^{-\Gamma t_0/2}} = 4/-8##

##\Gamma = 0.69##

Like that?
Yes.
 
  • #37
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So basically, ##\Gamma## is how the system decrease "speed"?
 
  • #38
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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.
 
  • #39
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I love you! Seriously big big thanks

one more thing if you want. ##A(\omega_d) = 2.8## is that correct?
 
  • #40
haruspex
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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.
 

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