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Find the resistive constant in a critically damped system

  • #1
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Homework Statement


This problem is taken from Problem 2.3, Introduction to Vibration and Waves, by H.J. Pain and P. Rankin:
A critically mechanical system consisting of a pan hanging from a spring with a damping. What is the value of damping force r if a mass extends the spring by 10cm without overshoot. The mass is 5kg. (g=9.81ms^(-2)).​

Homework Equations


$$m\ddot{x} = F_{spring}+F_{damping}+F_{gravity} = -sx -r\dot{x} + mg$$


The Attempt at a Solution


ob0zv.png
[/B]
Using second Newton law, I can write the equation of the system:
$$m\ddot{x} = F_{spring}+F_{damping}+F_{gravity} = -sx -r\dot{x} + mg$$. Rewriting the equation, we obtain:
$$ 5\ddot{x} + r \dot{x} + s x = 5g$$
Because the system is critically damped, we have r2 - 4ms = 0 so we can remove the s from the equation (s=r2/(4m) = r2/20):
$$ \ddot{x} + \frac{r}{5} \dot{x} + \frac{r^2}{4} x = g$$
Supposing what I've written above is correct, I can obtain the equation of motion. The homogeneous solution is
$$ x_h(t) = e^{-\frac{r}{2m}t} (A + Bt) = e^{-\frac{r}{10}t} (A+Bt)$$
where A, B constant determined by the initial conditions. The particular solution is:
$$ x_p(t) = \frac{4g}{r^2} $$
So the complete equation of motion is:
$$ x(t) = \frac{4g}{r^2} + e^{-\frac{r}{2m}t} (A + Bt) $$
Actually I have two questions:
- What are the initial conditions? I have supposed $$x(0)=0, \dot{x}(0)=0$$ but this leads to A = B = 0;
- Where do I have to plug the Δx = 10cm data in order to obtain r? Is it possible I need to use the energy formula?
 

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Answers and Replies

  • #2
tnich
Homework Helper
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It is natural to want to write out and solve the differential equation of motion, but many problems can be solved much more simply by considering potential and kinetic energy. In this case the damping force is not conservative, so you need to use a point where the damping force is zero to do your energy calculations.
 
  • #3
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I agree with tnich. That cannot be the easiest place to start this problem. Your textbook must have defined critical damping and I suspect in doing so they also derived some relations that would be useful here.
 
  • #4
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Thank you both for your replies. I think I can use the energy formula
$$ E = \frac 1 2 m \dot{x} ^2 + \frac 1 2 s x^2$$
At time 0 there is no kinetic energy (is the pan in rest position?) so the energy is, according to my coordinate system
$$ E(t=0)=0$$
At time infinity there is no kinetic energy (the pan has stopped) so the energy is
$$E(t=\infty) = \frac 1 2 s (10cm)^2 = \frac 1 8 r^2 10^{-2}$$
The energy lost during the damping is $$E(t=\infty) - E(t=0)$$
I can plug here the equation of motion in order to obtain
$$E(t) = e^{-rt/2m} (A+Bt) \left[\frac 1 2 m (B - \frac{r}{2m}) + \frac 1 2 s\right] $$
And now I still need to know A, B. Any suggestions?
 
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  • #5
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Another idea. If I change the coordinate system in order to have x(t=0) = 10cm and x(t=infinity) = 0 I can have initial conditions
$$A=10cm=10^{-1}m; \quad B=\frac{r}{2m}$$
Edit: with these initial conditions, I obtain that the velocity is zero constant. How is it possible I don't see useful initial condition in this problem?
 
  • #6
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Nothing you have tried makes use of the stated premise that the system is critically damped. They would not have given you a problem where the premise is that the system is critically damped without at least defining critical damping. Critical damping is “critical” to solving this problem and you must use the fact that it is critically damped to solve it. Look up critical damping in your text book and see if they don’t tell you some things about what it means to be critically damped.
 
  • #7
tnich
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$$ E = \frac 1 2 m \dot{x} ^2 + \frac 1 2 s x^2$$
This equation should include potential energy due to gravity. Once you put that in (and assume zero initial velocity), you can solve for s in terms of m, g and x.
Using the critical damping relationship you can solve for r in terms of s.
[/QUOTE]
 
  • #8
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@Cutter Ketch The only thing I found there - the 300th time I read the chapter - is the formula (which does not consider gravity) of maximum displacement in horizontal damped motion:$$x_{max} =x(t=1/\omega_0)=2mV/(re)$$ where V is the initial velocity. What formula are you referring to?

@tnich thank you, I forgot about potential gravitational energy. Still, I obtain the formula energy
$$E(t=\infty) = - \frac 1 2 s x^2 + mgx$$
And I can have r from
$$r = \pm \sqrt{4ms} $$
However, I still don't know how to deal with energy formulas. The other thing I can think is that at time $t=\infty$ the force of gravity is equal to the force of the spring (is it true?) so $$F_g = F_s \rightarrow mg = s \Delta x \rightarrow s = \frac{mg}{\Delta x} = \frac{5 \cdot 9.81}{10^{-1}} = 5\cdot 9.81 \cdot 10$$
 
  • #9
tnich
Homework Helper
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This equation should include potential energy due to gravity. Once you put that in (and assume zero initial velocity), you can solve for s in terms of m, g and x.
Using the critical damping relationship you can solve for r in terms of s.
[/QUOTE]
@Cutter Ketch The only thing I found there - the 300th time I read the chapter - is the formula (which does not consider gravity) of maximum displacement in horizontal damped motion:$$x_{max} =x(t=1/\omega_0)=2mV/(re)$$ where V is the initial velocity. What formula are you referring to?

@tnich thank you, I forgot about potential gravitational energy. Still, I obtain the formula energy
$$E(t=\infty) = - \frac 1 2 s x^2 + mgx$$
And I can have r from
$$r = \pm \sqrt{4ms} $$
However, I still don't know how to deal with energy formulas. The other thing I can think is that at time $t=\infty$ the force of gravity is equal to the force of the spring (is it true?) so $$F_g = F_s \rightarrow mg = s \Delta x \rightarrow s = \frac{mg}{\Delta x} = \frac{5 \cdot 9.81}{10^{-1}} = 5\cdot 9.81 \cdot 10$$
Right. So now you have a value for s, and you have written r as a function of s. Looks like you have everything you need to solve for r.
 
  • #10
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@tnich Thank you very much. The last thing I miss is about the initial conditions. The coefficients are
$$m=5,\quad s=\frac{mg}{\Delta x} = 490.5, \quad r=\sqrt{4ms}=99$$
So the differential equation describing the system is
$$5\ddot x+99\dot x+490.5x=490.5$$
[not required by the exercise] I still don't understand the initial condition of the homogeneous solution. The homogeneous solution should be
$$x(t) = e^{-rt/(2m)t} (A+Bt)=e^{-9.9t}(A+Bt) \qquad \dot{x}(0)=x(0)=0$$
Initial velocity should be zero because the mass is in rest position, and starts to move due to the effect of gravity force. Initial position is zero according to my coordinate system. I get:
$$x(0)=e^{... \cdot 0}(A + B\cdot 0) =0 \to A=0$$
and
$$\dot{x}(t) =e^{-9.9t}(A+Bt)(B-9.9)$$
So
$$\dot{x}(0)=1\cdot 0\cdot(B-99)=0 \quad\text{nothing can be said} $$
This is obviously wrong because leads to a constant displacement. However, the formulas seems to be correct: the gravity force should not influence homogeneous solution since its an external force. Am I missing something?
 
  • #11
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It doesn’t work because the generic solution to the ODE effectively changed coordinates on you. You forgot about the 490 on the right. Substitute your solution back into the ODE. You’ll see there are no time independent terms on the left. That solution equals zero. However that solution plus a constant will result in a constant on the left which you can set equal to your constant on the right.

Alternatively you can change coordinates such that the equilibrium point is zero. At that point the spring force cancels the force of gravity and, guess what, you get the same ODE but with zero on the right side. This is the system your solution matches, and that is why you found no movement. Your initial condition was the equilibrium point.
 
  • #12
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@Cutter Ketch I think I've understand the concept.

I have written the exercise in Latex, presented in a clearer way. You can see it in the attachments. If you find any error please pm me.
 

Attachments

  • #13
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Still not a completely correct solution, here's an update:
 

Attachments

  • #14
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m=5,s=mgΔx=490.5,r=√4ms=99​
Just to be sure we are on the same page, the title of your post is “find the resistive constant” and although the text of the problem says “find the force” it also says “r”. Taking those together I was sure and remain confident that your one line above is the entire solution to the problem. All of your work to claim some initial condition and then solve the differential equation is entertainment, right?

I looked through your solution in detail, and it looks right to me, so I’m not sure why you say it isn’t quite right.
 
  • #15
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Ehm, I just wanted to find (1) r, the solution of the problem, but also (2) the equation of motion. I wasn't able to find the solution. Now both the points seems clear to me. I produced a well written document for anyone who want to read the solution without having to read the whole thread which is a little mess. Thank you very much for your support.
 

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