How Is Stable Equilibrium Achieved in a One-Dimensional Force System?

AI Thread Summary
Stable equilibrium in a one-dimensional force system is determined by setting the force equation F(x) = (3.0N/sqrt(m))sqrt(x) - (1.0N/m)x to zero, yielding equilibrium positions at x = 0 and x = 9. The derivative F'(x) indicates that x = 9 is a stable equilibrium point. For small oscillations around this position, the frequency of oscillation must be derived from the potential energy associated with the force, rather than assuming a simple harmonic oscillator model. The calculated frequency for the system, considering the square root term, differs from the frequency obtained by ignoring it, highlighting the impact of the force's non-linear characteristics. Understanding the potential function V(x) is crucial for accurately determining the frequency of oscillations in this context.
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A one-dimensional force F(x)=(3.0N/sqrt(m))*sqrt(x)-(1.0N/m)x acts on an object of mass m = 2.57kg.


a Find the position x0 where the mass is at a stable equilibrium.
b Find the frequency of small oscillations around that equilibrium position. How does this compare to the
frequency if we were to simply ignore the rst term (the square root dependence) in the force?


So for the first part i set F(x)=0 and i got x=0 and x=23.13 but then i did it again and got x=0 and x=64.274 I have noooo idea how they ended up that different. and I'm not even honestly sure how to start b. anything would be helpful.
 
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hello,

your approach to part a is correct, and certainly zero is one solution to F(x)=0, but I don't see how you arrived at either 23.13 or 64.274 as the other solution. I get a much smaller number. could you show the work you did to arrive at those results?

for part b, we need to consider what would happen if the object was displaced slightly form its equilibrium position. well, if it is a stable equilibrium, then the object will oscillate about that point. if the force obey's Hooke's law, then the object undergoes SHM and the frequency is f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}. with the addition of the square root term, this will change. if know how to derive the formula for f for SHM, you should be able to tweak the process and get f for the new case. let me know if you need help with that.

cheers
 
would i be correct in saying that it's at x=0 and x=9? and then take the derivative which would be...
F'(x)=(3/2)x^-1/2-1 so if i use 0 its undefined, so therefore 9 is stable. that's part A.
so F'(9)=-.5
and for part b would i just use... 1/(2pi)*sqrt(k/m) (where k=.5) which would equal... .0702hz? i hope my units are right... and then if the sqrt wasn't there, k=1 so it would be .0992hz. Please anyone, correct me if I'm wrong.
 
i was using m as the mass, not N/M. so my work was doing the right thing, but just wrong numbers.
 
hi, you got part a now, but part b is not quite right. you have given the frequency as if this were a simple harmonic oscillator (that is, as if there where no square root term in the force). this is the frequency they want you to compare with, but it is not the frequency of your system.

the frequency you want will come from the series expansion of V(x), the potential associated with F(x). at least, this is the only way i can think to find it at the moment, there may be a quicker way. do you know how to find V(x)?
 
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