Checking whether my Diff EQ limit problem is correct

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Homework Help Overview

The discussion revolves around a differential equation related to a spring system that includes a damping force. The original poster is attempting to determine whether the weight associated with the spring passes through its equilibrium position based on their derived equation of motion.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster presents their equation and attempts to analyze the limit of displacement as time approaches infinity to ascertain if the weight reaches equilibrium. Some participants question the appropriateness of focusing on the limit at infinity instead of finite time values, suggesting a need to clarify the problem's requirements.

Discussion Status

The discussion is ongoing, with participants providing guidance on interpreting the problem correctly. There is a divergence in understanding regarding the relevance of limits and the conditions under which the equilibrium position is reached.

Contextual Notes

Participants note that the question is not about behavior as time approaches infinity but rather whether the displacement can equal zero for any finite time value. There is an emphasis on the role of damping in the system's behavior.

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figured it out, thanks

Figured it out, thanks.
 
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lilmul123 said:
I believe I have solved the problem, I just want to make sure of its correctness.

I was to solve a spring differential equation that involved a damping force and to create a equation of motion. I did. Now, I need to figure out whether the weight ever passes through its equilibrium. The equation I ended up with is

x = e^(-8t) * (.5 - 2t) where x is displacement and t is time.

Now, to check whether weight passes through the equilibrium, what I did was set up the limit as x approaches infinity of (.5 - 2t)/(e^8t). Then, I used l'hopital's rule to find the limit as x approaches infinity of (-2)/(8e^8t). This comes out to be that the limit is 0, and therefore, the spring never passes through the equilibrium and there is no t-value.


Is my logic sound and my math correct?


I think you wrote the wrong thing. Don't you want to take the limit as t goes to infinity?
 
lilmul123 said:
I believe I have solved the problem, I just want to make sure of its correctness.

I was to solve a spring differential equation that involved a damping force and to create a equation of motion. I did. Now, I need to figure out whether the weight ever passes through its equilibrium. The equation I ended up with is

x = e^(-8t) * (.5 - 2t) where x is displacement and t is time.

Now, to check whether weight passes through the equilibrium, what I did was set up the limit as t approaches infinity of (.5 - 2t)/(e^8t). Then, I used l'hopital's rule to find the limit as t approaches infinity of (-2)/(8e^8t). This comes out to be that the limit is 0, and therefore, the spring never passes through the equilibrium and there is no t-value.


Is my logic sound and my math correct?

No, your logic isn't sound, assuming your solution for x is correct. Presuming the equilibrium position is x = 0 the question is asking whether that happens for t > 0.
 
lilmul123 said:
Right, and as t goes to infinity, it seems x gets closer and closer to x=0, but will never hit zero, is that correct or am I still lost?

It isn't asking you about the limit as t goes to infinity. It is asking whether there are any finite values of t where x hits the equilibrium position.
 
Stop with the infinitely large bit. The question has nothing to do with infinity. The question is whether x = 0 for any value of t. It isn't a rocket science question. Look at your equation.
 
You're welcome. The point of questions like that is that it depends on the damping. If the system is damped strongly enough it might just ooze down towards equilibrium. Otherwise it might do as in this problem, cross equilibrium once and settle towards equilibrium. Or if it is underdamped, it might oscillate around equilibrium as it settles.
 

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