Physics-Stable Equilibrium and Oscillations

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Homework Statement



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?




Homework Equations



F(x)=0

The Attempt at a Solution



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.
 

Answers and Replies

  • #2
vela
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You need to redo part a because both of your answers don't make sense. Note that the m in the first term is part of the units and stands for meters; it's not the mass.
 
  • #3
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(3.0N/m^1/2 )x^1/2 - (1.0N/m)x acts on an object of mass m = 2.57kg. <- that is my problem copied and pasted... so Newton/sqrt(meter) i don't get that :/
 
  • #4
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Vela's correct. Since F(x) is force, it has a unit of 1 Newton. For the function to equal that, m must be meters since if m is mass, x would become mass which x isn't according to the text. m must be meters for the units to make sense.
 
  • #5
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So, pretty much you're telling me my life got 10x easier? thank you haha. in that case.... 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.
 
  • #6
vela
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So, pretty much you're telling me my life got 10x easier? thank you haha. in that case.... 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
F'(0) being undefined doesn't imply x=9 is stable. It's the fact that F'(9)<0 that tells you that's a stable equilibrium point.
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 didn't check your actual numbers, but I can verify your method is correct.
 

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