Critically Damped Harmonic System

AI Thread Summary
John, known for his expertise in door-stoppers, claims that a critically damped system is the best for preventing doors from slamming. However, the discussion reveals that if a critically damped door is opened and released without any push, it will not close again due to its equilibrium position being set at the open state. The solution involves understanding the second-order differential equation governing the motion of the door and applying initial conditions correctly. The key to the problem lies in recognizing that the door's initial position and lack of initial velocity lead to it stopping at the open position. This insight clarifies the misunderstanding about the behavior of critically damped systems in this context.
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Homework Statement


John the door-stopper man sells and installs door-stoppers. He prides himself as being the world's best stopper and guarantees your money back if you can install a better stopper than him.

John's secret to door-stopping is that he remembers from his lectures that the best way to stop doors is by fine tuning the stopper such that it is critically damped.


Homework Equations


Trick John (and ask for your money back) by showing him that if you open a critically damed door and release it without also pushing it, it will never ever close again.


The Attempt at a Solution


The equation of motion is a second order differential equation in the form: mx"+cx'+kx=0; then let x=e^(zt);

I realize that for a critically damped system, the determinent for the characteristic equation z^2 + (c/m)z + (k/m) = 0 equals to (c/m)^2 - 4(k/m) = 0;

Then solving for x to find the equation in the form of x(t) = (A + Bt)e^(-t(k/m)^0.5)

However, the problem is that this solution converges to the equilibrium position under critically damped conditions.

How can I show that the door will never close again?? (It seems impossible!)

Your help is much appreciated! :smile:
 
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Careful, you are looking at the discriminant, which is different from the determinate. Also, your solution should be

x(t) = (A+Bt)e^{-\frac{c}{2m}t}

But you might as well save same writing and just call c/2m beta or gamma. Those are just some notational issues, and stupid mistakes. They don't really pertain all that much to the solution, but should be corrected nonetheless. (Edit: Oh wait, I see what you (or your instructor) did. You can leave your solution how you have it.)

You have a couple initial conditions, one is that the door is open some amount x_0 at time = 0. The other is that there is no initial velocity so x'(0) = 0. Then you want to know when the door stops moving, when it is closed.

See if this gets you anywhere.
 
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You have to set the equilibrium position to the open position. In that case the door will stop there.
 
Thanks you guys! I see, so the trick is to have the initial condition as being opened already.
 
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